Rèn luyện Toán nâng cao Đại số 9 - Nguyễn Cam

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TS.

N G U Y E N

CAM (Chu bien)

^ ThS.

N G U Y E N V A N

PHUCfC

REN LUYEN TOAN NANG CAO

DAI

SO

9

* Phan loai cac dang toan nang cao dinh cho hoc sinh khi gi6i.

* Tuyen chon cac de thi tuyen sinh Idp 10 vao cac triTdng chuyen.

THLT VIENTIfvHBINH THUAN

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LdlNOIDAU

Cung vdi viec bien soan bo sach Ren luy^n todn can ban LOP 9,

chiing toi gidi tliieu cuon Ren luyen todn ndng cao DAI SO 9 nham giup

cac em hoc sinh nang cao trinh do de tham gia cac ki thi tuyen vao Idp 10

thuQC cac trirdng chuyen, Idp chuyen.

Cuon sach bao gom 9 chifcfng :

ChUdng 1 : Dang thiJc

ChiTdng 2 : Bat dang thiJc

Chufdng 3 : So hoc

ChiTdng 4 : Gia tri Idn nhat va gia tri nho nhat cua ham so

ChiTdng 5 : Phifdng trinh

ChUdng 6 : He phifdng trinh

ChiTdng 7 : Do thi ham so

ChiTcfng 8 : Mot so bai toan ve td hdp ; dpng tiJ

ChiTdng 9 : Mot so de thi vao Idp 10

Vdi muc dich h5 trd dac life viec tuT hoc ciia minh, cuon sach diTc^c

trinh bay theo bo cue nhiT sau : md dau mSi chufdng la phan torn t^t cac

kien thilc can ye'u ma hoc sinh phai nam vffng; tiep theo la mot he tho'ng

b^i tap CO hiTdng din each giai diTdc sap xep theo thiJ tif kho dan.

De viec hoc cd ket qua to't, sau khi doc va tim hieu that ky cac bai

tap CO hiTdng din hoc sinh nen co gang tif minh giai cac bai tap ay mot

cdch doc lap cho den khi thuan thuc. Vdi each lam ay, hy vong cac em se

tif minh kham pha ra nhieu each giai khac mot each thu vi.

Raft mong nhan diJdc cac y kien phe binh cua quy doc gia de cuon

sich ngay cang diTdc hoan thien va phuc vu ban doc tot hdn.

TM.nhdm tdc gid

TS. Nguyen Cam

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f

CHl/OfNGI:

DANG THlfC

I . HANG DANG THlfC

1)

(a + bf =a'+2ab + b\

2)

{a-bf=a'

-2ab + b\

3)

{a + bf =a' -h^a'b + ^ab^ +b\

4)

{a-bf=a'

-3a'b +3ah^-b\

5)

a' +b' ={a + b){a'-ab + b').

6)

a'~b'={a--b)(a^+ab + b^).

7)

a"-b"={a--b\a"-' +a"-'b + a"-\b' +... + ab"-' +b"-').

8)

{a + b + cf = +b^ + 2(ab + bc + ca).

9) (a + Z) + c)' =

a'+b'+c'+ 3(a + b)(b + c){c + a).

n.

CAN THlfC

1)

^A^ =\A\=<

AneuA>0

1)

^A^ =\A\=<

-AneuA<

0 2)

A+B±lylAB=(^±4B^{A>0,B>Q)

3) ^ + 5

' ± 2 5 V I = ( V I

± 5 ) ' ( ^ > 0 )

1

-IA-JB

-j-j-(A>0,B>0,A^B).

1

•SIA+JB

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1.1: Tinh cac bieu thitc :

2 - V 3 2 + V3

3) C = ^ + . '

4) D = - l ^ . . ' 
3 - V 2 3 + V 2

1) A = - i ^ + '

2 + V3

2 - ^

2) B =

3) C =

2-S 2 + V3 f2

-V3

)(2

+ ^/3]

(2

+ ^3)

(2

- ^/3)

. 2 + ^ + ^ = 2 + 7^ + 2- 7^ ^ 4
4 - 3 4 - 3

1 ^ / 5+ V 2 V 5 - V 2 2V5

• +

+ •

^5-42+ V 5 + V 2 5 - 2 . 5 - 2 
1 V 3+ I V 3- I

V 3- I V 3+ I 3 - 1 3 - 1
= 1.

1 1 _3 + V2 3->/2 ^ 3 + V 2+ 3 - N ^ _ 6

3- V^ ^ 3 + V 5^ 3 ^ - 2^ 3^ - 2 = 7 7'
De 1.2

1) Riit gon bieu thiJc M = 
a ;t 1 va a > 0.

I - A/ ^ l + ^/ajv ^/« 
1 - vdi

2) Tinh gia tri cua M khi a = - .

1) M = 1

-,i->/^

i+V^Jt

(i

-V

^)(i

+V

^)'~vr~

2Va V a- 1 - 2

( l - V a ) ( l + V a ) ' V a 1 + V«

- 2 -3
2) Khi a = - . T a c 6 : M = — ^

1.3: Rut gon (Loai bo dau can va dau gia tri tuyet do'i)
1) A = ^Ja^ - 6 a + 9.

2) 5 = Vx^ + Vx' - 4 x + 4.

Giii

1) A = ^Ja^ -6a + 9 =^(a-3y =\a-3\. 
a-3 neu a-3>0 f a- 3 new a> 3
3 - a n ^ M a- 3 < 0 3- a neu a<3 
2) 5 = V7 + V x' - 4 x + 4 = V7 + ^ ( x - 2 ) '

= | x | + | x - 2 |

-x-x + 2neu x<0 {-2x + 2 ne'ux<0 
x-x + 2 neu0<x<2 = l2 neu 0<x<0 
x + x-2neu x>2 2x-2 neu x>0

De^ 1.4: T i i
1) ^ =

2) B = 
ih:

V 4 8 - 2 V 7 5

L ^/a+l)

+ V 1 O 8- - V
7 '147.

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Gi§i

1) yl =

V48

-2V75 +

Vr08-iVT47.

= - 2

V i ^ +

V 6 ^ - - V T ^

2) B =

1 +

1 +

-= ( l + V ^ ) ( l - V ^ ) -= l - a .

1.5: Tinh :

rV216 2^-yf6^

[ 3

V 8

- 2

1

7216 2 V 3 - V 6 '
~ 3 V 8 - 2

V ^ ( ^ / ^ - l )

6^6 2 V 3 - V 6

•3

- 2 ^ - 2 ) yfe

2 > / 6

-2

( v ^ -

1 _ 2 > / 6 - 1 2>y6 V6
V6 76 2^/6

2 2

1.6: Tinh:

5 = 713-7160 - 7 5 3 + 4790.

= 78-2715 - 7 8 + 2715 = 7 5 - 2 7 1 5 + 3 - 7 5 + 2715+3

5) - 27^7^ + (73) - ^(7?) + 27^^^ + ( 7 ^ ) '

= ^ ( 7 ? - ^ f - ^ ( 7 ? + 7 ^ f

= 75->^|-|75 + 73| = 75 - 73 - 7 5 - ^ ( v / 7 5 > 7 3 )
= 2 - 7 1

5 = 7 l 3 - 7 l 6 0 - 753 + 4790 = 78 - 7 l 6 o" f 5 -7 4 5 + 4790+ 8

8 ) ' - 2 ^ 7 5 + (75 ) ' - -^(745 ) ' + 2745 78 + (78 ) '
= ^ ( > ^ - 7 5 ) - ^ ( 7 4 5 - 7 8 ) =|78 - 75|-|745+78
= 78 - 75 - 7 4 5 - ^ (v/ 8 > 5 ^ 7 8 > 7 5 )

= - 7 5 - 3 7 5 = -475

1.7: Cho bieu thitc:

M =

7^ •

1 + 7 ^ 27a

7 a - l (a + l ) ( 7 a - l )
1) T i m dieu kien cua a de M CO nghia.
2) Rut gon bieu thu-c M .

3) V d i gia tri nguyen nao cua a thi M c6 gia tri nguyen ?

M = 1:

1:
1

-1 + 7 ^

1 27a

\ yfa

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1: l + \/aj (a + l)(x/a-l)

a>0

1) Bieu thu'c M c6 nghia <=>

\

4a^0 \a>0

<=> {

2) Vdi dieu kien a > 0 va \a c6:

' 1 ^ ( v / ^ - l j (l + V^)(-v/^-lj

a > 0

1: 1 + VaJ (a + l)(Va-l) (a + l)(Va-l)

a + \

a +

1

3) Ta c6: M = -^-^ = 1 - — B i e u thu'c M nhan gia tri nguyen

a +

1 a +

1 khi

a > 0

a

;t 1 <z>

2;(a

+

l)

a > 0

a

1 o

a + l = l;-l;2;-2

a > 0

a = 0;-2;l;-3.

Vay khi a = 0 thi bieu thu'c M nhan gia tri nguyen..

1.8: Cho bieu thu'c:

M =

4^a'~a'b'

M =

b'

vdi

|a| > |^>| >

0. Rut gon bieu thu'c M.

Vdi dieu kien a\>\b\> 0. Tacd:

10 a' + l a ^ l a ^ + a^-b^-a^+ la^a"-b"-a" + b" 4|a|Va^-6'

a^-[a'-b') b'

Aa^a'-b' Aab'^a^-b' _ fl mu a>Q

'A\a\4^^ 4\a\b^yja^-b^ [-1 neu a<Q

DC 1.9: Riit gon bieu thu'c:

3 (yfai) - b) (yfa - f + 2av'a + b ^

M=-^ ^- + ^ ^= P

a-b ay/a+by/b

(vdia>0, b>Ova a;tb)

V d i a > 0 , b>Ova a^^b, ta c6:

3(V^-Z>) (4a-4b^ +la4a+bylb

M =

a-b aja+byjb

• + •

3V^(V^-V^) ayf^-3ayfb+3by[^-by[b+2ayl^ + b ^

(V^-V*)(N/^ + V ^ ) ^

[ ^ + yfb)(a-yf^ + b)

3yfb ^ 3ay[a-3ay[b+3by/a

(yf^ + yfb) (^/^ + ^ / ^ ) ( a - ^ / ^ + Z>)

3jb

3yfa(a-yfab +b)

(V^ + VA)

(^ + y[b^(a-4ab+b^

34b , 3ra _3V^ + 3 V ^ _ 3 ( ^ + ^ ) _ 3

y/a+yfb yfa+yjb yja + yfb yfa + yfb

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1.10: Rut gon bieu thrfc sau:
1

- yfa a\[a + a + -Ja

1) ^ = 1

- >/a ayfa + a + ^/a

'a>0

>/a + 1 0 yfa ^\

V d i dieu k i e n a > 0 va a;t 1 ta cd:

1 >/a+l

B i e u thitc A cd nghia o a> 0

A =

^/a(a + ^ / a + l j j

2) B = ^Jx^ +2x + \-ylx^ -2x + \.

=^{x+\f-^{x-\y = \x+\\-\x-\\

- x - l - ( - x + l ) neu x < - l

x + l - ( - x + l ) neu - l < x < l =
x + l - ( x - l ) neu x>\

- 2 neu' x < - l
2x neu' - l < x < l
2 wew' X > 1

D e 1.11: Cho M = 1 Vx + 1

X^ -4x Xyjx + X + y[x

1) T i m dieu k i e n cua x de M cd nghia.
2) Rut gon M .

1) M cd nghia o \

Gidi 
X > 0

x ^ - ^ ^ 0 
yfx+l^O

x-Jx + x + ^fx ^ 0

x > 0

^ / x ( x + ^ / x + l ) ^ 0

x> 0

{•fx)'-wo

x> 0 x> 0

x;tl

2) V d i d i ^ u k i e n : x > 0 va x ^ 1 . Ta cd :

1 Vx + 1 1 > / x + l

X^-yfx xJx+X + yfx >/x(x + Vx + l)

1 ^[x + yf^ + l) ^ ^ 1 ^ 1

V^(V^-l)(x + V^ + l)' + \^ + l x-\

D e 1.12: Cho M = 4 x - \ - i 4 ^ 
V x^ - 1

1) T i m dieu k i e n c\ia x de M cd nghia.

2) T i n h M l
3) Rut gon M .

1) M e d nghia

GiSi

x - 2 > 0

x - l - 2 ^ / x ^ > 0 o

V T ^ - l ^ O

x > 2

( x - 2 ) - 2 > / x ^ + l ^

yfx^^l

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x>2

( V x - 2 - l ) ' > 0 <=> x>2 
x^3 
x - l ^ \

2) V d i d i ^ u kiSn x > 2 va x 3, ta c6 :
V X- 1 - 2A / X ^

Vx - 2 - 1

\

X -1 - 2Vy - 2

( x- 2 ) - 2 V ^ + l
• - l- 2 V > : - 2

= 1.
x - l- 2 7 x ^ 
3) V d i = 1 ^ M = ±1

M a t khac v i : y]x-\-2y[x^ > 0 nen :

• Ne'u : V x - 2 - 1 > 0 <::> V x - 2 > l o x - 2 > l < : > x > 3
thi M > 0 ; d o d 6 : M = 1.

• Ne'u : V x - 2 - l < 0 < r > v ' x - 2 < l < = > 0 < x - 2 < l < : : > 2 < x < 3
t h i M < 0 ; do d d : M = - l .

1 (neu : X > 3)

- 1 ( n^ M: 2 < x < 3 )

vonghia {neu: x = 3;x < 2).
V a y M =

D e 1.13: Rut gon cac bieu thiJc:

x V x ^ + 2
Gi§i

3- J ( 2 V 5- 3 )

= ^ / ^ / ? ^ V 5- 2 ^ / 5 + l = J V S - J V s ) -2V5.1 + 1^

= ^ V 5 - ^ p ^

=

^ V 5 - | V 5 - l | = ^ V 5 - ( V 5 - l )

(V/

V5 >1)

y =VV5

-V5

+i=vr=i.

2) 5 =

+ 3 x ^ + 4 ( x ^ + 4 x ' ' + 4 ) - x ^
x ' + x' + 2 x ' + x' + 2

( x V 2) ' - ( x ^ ) ' ( / + 2 + x^ ) ( x V 2- x ^ )
x V x' + 2 " x V x + 2

= x ^ - x ^ + 2 .

1.14:

u-^ , ^ 1 3 V 2- 2 ^

1) R , U g p n b . e u t h . c M = - ^ ^ ^ ^ ^ .
2) T i m gia t r i nho nha't cua:

y = -1 - 2 V x ^ + Vx + T - W x ^ . 
Gi^i

1) M = -

I3V2-2V3

V 2- V 3 V3V2 + 2V3 V 2 - V 3^ V 6( > / 3 + >/2)

V 2- V 3
^ - 1 .

3- 2

. ( > ^ - V 2 ) ( v i V 3 > V 2 )

2) Ta cu :

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= ylx-2-2y/x-2+l+yjx-2-6^fx^ + 9

Dieu kien: x>2.

Ap dung \A\>A. Da'u "=" xiy ra o ^ > 0. 
Ta c6 y> •Jx-2 -1 + 3 - V x - 2 = 2 •

V x ^ - 1 > 0

+

Da'u xay ra o <^

3 - V x - 2 >0

o l < x - 2 < 9 o 3 < x < l l .

Vay Miny = 2,datdu'c{ckhi:3<x<ll.

<=>\<sfx^ <3

Bi 1.15: Phan tich thanh nhan IvC M = x'' +x' +1.

M = x"'+x'+l.

= x'° +x' +x'-x'-x'-x'. + x' +x' +x'-x'

-X^ -X* +x^ +x^ +x^ -x^ -x^ -x + x^ +x + \ 
= x'(x'+x + l)-x'(x'+x + \)+x'(x'+x + l)

-x" (x^ + X +1)+ x'(x^ +x + l)-x(x^ +x + l)+l(x^ +x + l)

= (x'+x + l X x ' - x U x ' - x V x ' - x + l )

1.16: Chitng minhr^ng ne'u: -ab-bc-ac = 0, thi: 
a = b = c.

GiSi

Tacd: -ab-bc-ac = 0

O l(a^ - lab + b') + ^{b'-2bc + c') + l(c^ - lac + a')

1 1 1

<>-(a-bf+-(b-cf+^c-ay=0

2. ^ 2 
16

a-b = 0

b-c-O <;i>a = b = c. 
c - a = 0

Vay, nSu a^ +b^ -ab-bc-ac = 0 thi a = b = c. 
)e 1.17: Cho ba so' a, b, c sao cho c ;t b, c ?t a + b va

=2iac + bc-ab) (1) 
^-{a-cy _ a - c 
ChiJng minh rkng:i

b^+{b-cf b-c

GiSi

TO (1) ta c6: + 2aA - 2ac - 2^)C = 0 
nen: = a^ + {c^ + lab ~ lac - 2bc) 
s = (a^ - 2ac + ) + 2Z)(a - c)

= {a-cf+lb{a-c) = {a-c){a-c + lb)

Wngtuf: b'^ =b^+(c^+lab-lac-lbc^

= [b^ -lbc + c^) + la(b-c)

= (b-cy +laib-c) = {b-c)(b-c + la) 
a' + {a-cf _ (a - c){a-c + 2A)^+ (a - cf 
Do do:

b'+ib-cf ~ ib-c)(b - c + la) + {b- cf 
(a-c)(la-lc + lb) _a-c

~ (b-c)(2b-lc + la) b-c (vi a+b^c). 
D6 1.18: Cho a, b, c doi mot khac nhau va thoa

a b c ^ 
+ = 0 
b-c a - c a-b

(1)

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TO (l)ta

c6:

Gi§i

a b c ab-b^ -ac + c^

b-c a-c a-b (a-c){a-b) 
a ab-b^ -ac + c^

(b-cf {a-c){a-b){b-c)

Lam Wdng W nhtf tren, ta se c6:

b _ bc-c^ -ba + a^

(c-af " (a-b){a~c)(b-c) 
c ca-a^ -bc + b^

{a-by ~ (a-b)(a-c)(b-c)

Cong ve theo ve (2), (3) va (4) ta diTcJc:

a b c . 
• + r + r- = 0.

(2)

(3)

(4)

(b-cf {c-af (c-bf

De 1.19: Cho ba so' a, b, c thoa a + b + c = abc.

Chu-ng minh: a(A' - l)(c' -1) +

b(a^ -

l)(c' -1) + c(a' -

\)(b^

-1) =

4abc.

GiSi

Ta

CO: a(b^ - l)(c'

-

1)

+

b(a^ - l)(c'

-

1)

+

c{a^ - \)(b^

-

1)

= ia + b + c) + (ab^c^ + baV + ca^b^) - (ab^ + ba^ +ca^+c^a+be" + b'^c)

= abc + {ab^c^ + ba^c^ + ca^b'^) - [ab{b + a)+ ac{a + c) + bc{b + c) 
= abc + (ab^c^ + ba^c^ + ca^b^) - [ab(abc -c)+ac{abc -b) + bc{abc - a)

(vi a + b + c = abc nen a+b-abc-c,b + c-abc-a,a + c = abc-b) 
= abc + (ab^c^ + ba^c" + cc^b^)

-

(a^Z»'c+a'c^i+

ab'^c^)+l>abc

= 4abc.

Vay ddng thiJc da diTdc chiJag minh.

1.20:

Cho - = ^ = - = w

(m>0).

Chiang minh ring :

a b c

^a'+b'+c'

18 Tacd:

— = — = — = m

cho

ta: m

=—j--a b c =—j--a b a''+b^^c'

nen:

m = ^a'+b'+c'

Bi 1.21: Choa,b,c, a',b',c' la cac so diTdng thoa — = — =

a' b' c' 
CMng

minh:

^|aa' + ylbb* + yfcc' ^ yj(a + b + c)(a'+b'+c'j.

GiSi

Dat: w = — = — = — thitac6:

a' b' c'

a + b + c (a + b + c)(a'+b'+c')

~ a'+b'+c'~ (a'+b'+cf

^ , . . aa'. bb' cc' r- 4aa' JbT' yfcc'

Talaico:

m =

—- = —-

= —-=> y^m^ ^- =

a''

b'^ c ' ' a' b' c'

[— yfaa'+ ^/bT'+ yfcc'

=>ylm

a'+b'+c'

TOCl)

va (2)tasuy ra:

{a + b + c)(a'+b'+c') yfaa' + \fbb' + \[cc

(a'+b'+cf

nen {a + b+ c){a'+b'+c') = (4^'+4bb'+ 4^'^ 
a'+b'+c'

2

hay

4aa'+ y]bb'+y[cc'= ^{a + b + c)ia'+b'+c')

(dpcm)

(1)

(2)

De

1.22:

Cho — = —= ^

vd

— + ^ + — = 1.

X y z a^ b^ c^'

Churng mmh: — + — + = —

5 — ^ .

a" b^ c" x^+y^+z"

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Dat / = — = —= — thico x = —;)> = —;z = —

X y z t t t

a^t^ b^t^ c^f

2 rn'

TO / = — = - = — thi cd: =

X y z

TO (1) va (2) ta suy ra r^ng:

a'

x'

/

z'

(1)

x^ ^y^ -v z^ (2)

p^ + p^

(dpcm)

1.23: Cho ba so' a, b, c thoa - a'+b'+c'^l (1)

[a^+b'+c^=\) 
my iiDhtSng so S ^a + b^+c\

Gidi

TO(2)tac6 a' <l,b^ < l , c ' <1 nen: |a| < 1, \b\ 1, |c| <1

=>a' <a\b' <b\c' <c'

m a ( l ) v a (2) cho ta: a' +b' +c' +b^ = 1 
nen phai xay ra b'=b'

\6i =a^ => a^(a-l)-0=> a = 0 hay a-l=>a^=a

vadodd S = a + b^+c^ =a^ +b^ = \. VayS = l .

CHLfOfNGll:

B A T DANG THtfC

§1. PHEP BIEN DOI TUdNG DLfdNG.

TINH CHAT CUA BAT DANG THLfC

I. PHEP B I E N DOI Tl/dNG DlJdNG

1)
2)

a>b <:i> a-b >Q 
a>b <^ a-b>0

Chu v; Vdi moi so' thrfc X, ta cd: > 0.

II. C A C TINH C H X T C d BAN CUA BAT DANG THLfC 
1) a>b va b>c^a>c.

2) a>boa + c>b + c. 
3) a>b + c<^a-c>b.

a>b

4)

c>d

5) a>bO'

'a>b>0

6) c>d>0

•a + o b + d. 
ac > be ni'u c > 0

ac<bc ne'u c<Q. 
ac > bd.

7) Neu:ab>0 thi:a>bo-<-.

a b

Bi 2.1: Cho a > 0, b>0. CMng minh: a^ +b^ >a^b + ab^

Giii

Tacd: a' +b^ >a^b + ab^oa'-a^b + b'-abJ>Q

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<:>a\a-b)-b\a-b)>0<>(a^-b^)(a-b)>0

o(a + b)(a-b)(a-b)>0<:>ia + b)(a-by >0 (*)

V i a > 0 , 6 > 0 nen a + b>0 va (a-bf >0 n^n (*) diing. 
Vay +b^ >a^b + ab^.

Bi 2.2: Cho a > 0, b > 0. Chitng minh: + > ^/a + V ^ .

yjb

Va

T a c d : -^^-^>4a + y[b<:>a4a + b4b >(y[a+y[b)^fab.

yfb ^ ^ '

i^^J + ( V * ) ' - ( V ^ +V A) V o ^ > 0.

[4^+y}b)[4^-y[^+Jh'Y[4^+4b)4ab>o •

o ( + ^/6 ) ( - 2 + V * ^ ) > 0
(N/^ + V ^ ) ( V « - V * ) ' > 0 (diing)
Vay: - ^ + - ^ > y/a + yfb.

yjb yja

Bi 2 . 3 : Chitng minh ba't dang thitc sau: +l>ab + a + b v d i

moi so'thifc a, b.

GiSi

T a c d a ' +Z»' + 1 > aA + a + A o 2a'+2Z>^ + 2 > 2aA+ 2a + 2Z>
<=>2a' + 2 6 ' + 2 - 2 a 6 - 2 a - 2 Z ) > 0

o ( a ' -2a^)) + ( a ' + l - 2 a ) + (6' + 1 - 2 Z > ) > 0
o ( a - 6 ) ' + ( a - 1 ) ' +{b-iy>0 (diing) 
Vay a ' + 6 ' + l > a Z » + a + 6.

f)i 2.4: Chitng minh cac ba't dang thtfc:

1) a " * > a^6 + aZ>^ vdi moi a, b. 2) a ' + 6 ' > - v d i a + 6 > 1
2

GiSi

1) T a c d a*+b^ >a'b + ab' <^ + b^ - a'b - ab' > 0

<^a'(a-b)-b'(a-b)>0^{a-b)(a'-b^)>0

o (a - ^»)(a - Z))(a' + aA + A') ^ 0 <^ (a - 6)' a ' + a 6 + — + 36 2 A > 0 
< : > ( a - 6 ) ' a + - + -36' > 0 (dung).

Y&y: a'+b'>a'b + ab\

2) T a c d a ' + 6 ' + 2a6 = (a + 6 ) ' > l ( l )
va -lab = [a-bf >0 (2) 
Cong (1) va (2) ta cd 2 a ' + 2 6 ' > 1 a ' + 6 ' > i

V a y a ' + 6' > i .

Bi 2 . 5 : Chu-ng minh bat dang thiJc:

a ' + 6 ' + c ' > a 6 + 6c + c a .

GJdi

Ta cd: a ' + 6 ' + > a6 + 6 c + ca
O 2 a ' + 2 6 ' + 2 c ' ^ 2a6 + 26c + 2ca

<:>(a' - 2 a 6 + 6 ' ) + (6' - 2 6 c + c ' ) + (c' - 2 c a + a ' ) > 0
O (a - 6 ) ' + (6 - cf + (c - a ) ' > 0 (diing)

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D6 2.6: Chu'ng m i n h rang vdi m o i a, b , c, d, e ta c6:

(a^ >a{b + c + d + e).

Gidi

T&cd: + b ^ > a { b + c + d + e) 
<^a^+b^+c'+d^ + e^-a(b + c + d + e)>0

<^a^ + b^ + c^+d'^ + e'^-ab-ac-ad-ae>0

t:. + b^-ab

+

a • + c'^-ac

+

- + d^-ad

+

Ve —ae

4

> 0

^—b

+

a

c
2

2 / \1

+

a e

2-2 '

> 0 ( d u n g )
V a y : + + + c/^ + > a(A + c + J + e ) .

D d 2.7: Chtfng m i n h rang vdi m o i a, b, khac 0 ta c6:

b' 
a'+b'<^ +

a ,2 •

a* b^

T a c d : a ' < — + — •

b' a'

O a^t' + a'b' <a'+b'<^a'- a'b' + b'- a'b' > 0

O a'(a' - fc') - b\a^ -b')>Oo(a'- b')ia' -b')>0 
o(a' -b')ia' -b'){a' +a'b'+b')>0

<:>ia'-b')\a'+a^b^ + b')>0 (dilng)

V a y : a +b' <^ + ^ . 
b . a

6 ^ 6

D d 2.8: C h o a.b>0. Chu'ng minh rang: (a + bfxy<{ax + by)(bx + ay).

24

T a c(5: (a + b)"" xy<(ax + by)(bx + aj)

<=> a^xy + 2aZ7jc>' + b^xy < abx^ + aby^ + c^xy + b^xy 
<:> abx^ + aby'' - 2aZ?x>' > 0 <::> ab{x^ + -2^:^) > 0

<^ab{x-yf>Q.

Ba't d^ng thi?c n a y diing vi ab > 0,(jc - y)^ > 0 .
V a y : {a + bf xy <{ax-\- by)(bx + ay)

Bi 2.9: C h o a + > 2 . Chu'ng m i n h rang: a^ +b' <a' +b\

Gidi

TsiCd: a'+b'>a'+b^ 
oa'-a'+b'-b'>0

c>a\a-l) + b\b-l)-{a-\)-ib-l) + a + b-2>0 
<^(a-l)(a'-l) + (b-\)(b'-l) + a + b-2>0

<:>(a-l)\a^ + a + l) + (b-l)\b' + b + l) + (a + b-2)>0

Bat d d n g thiJc tren d u n g vi: •

* (a-lf >0,(b-\)^ >0.

a^+a + \

2 4
* a + ^ - 2 > 0 (do:a + b>2)

V a y : + Z?^ < a' + Z?^ da du'dc chu'ng m i n h .

D d 2.10: C h o ba so a, b, c thoa a' + Z?' + = 1.

Chu'ng m i n h rang: abc + 20- + a + b + c + ab + ac + bc)>0.

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Gidi

Tac6: a' = 1 nen <\,b^ < l , c ' < 1 . 
^ - 1 < a < l , - l < ^ < l , - l < c < l

^ l + a > 0 , l + Z 7> 0 , l + c > 0 => il + a)il + b)(l + c)>0

=^l + a + b + c + ab + bc + ca + abc>0 ' (1)
Mat khac, ta lai cd: (1 + a + + c)^ > 0.

^\ a^ +b^ + +2(ab + bc + ca) + 2(a + b + c)>0

=>2 + 2(ab + bc + ca) + 2(a + b + c)>0 (vi a^ + b^ =1) 
=^l + ab + bc + ca + a + b + c>0 (2)

Cong ve theo ve' (1) va (2) ta difdc:

abc + 2(l + a + b + c + ab + bc + ca)>0(dpcm)

De 2.11: Cho 0 < a < l , 0 < Z ? < l va 0 < c < LChu-ng minh:

+ b ^ < l + a^b + b^c + c^a (1)

GiSi

Tacd: (l)<::>a^(l-Z?) + Z>^(l-c) + c ' ( l - a ) < l
ma 0 < a < 1,0 < < 1,0 < c < 1 nen :

a\l-b)<ail-b) 
b\l-c)<b(l-c) 
c\l-a)<c(l-a)

Do do de chiing minh (1) ta chi can chiJng minh bat ddng thrfc sau:

a(l-b) + b(\-c) + c(\-a)<\)

That vay, ta c6: 1 - a > 0,1 - ^ > 0,1 - c >,a^c > 0
nen (1 - a)(l - ^)(1 - c) + a^c > 0

=^l-{a + b + c) + ab + bc + ca>0 
^l>a + b + c-(ab + bc + ca) 
=^ a(\ + b(l-c) + c(l-a)<l

Do d6 (2) dufcJc chiJng minh. Vay bat d^ng th-ic (1) diTcJc chu-ng minh.

1 1 2

pi 2.12: Cho ac> 0 va thoa : - + - = -a c (1)

. , a + b 'c + b ^ .

Chiang mmh : + > 4

2a —b 2c —b

Giii

2ac

. 2 c + a

Tuf(l)tac6: - = ^b =

b ac a + c

do dd: - + •

2ac

a + c c + 
lac 
a + c 
2a-b 2c-b 2a- 2ac

a + c

2c-2ac 
a + c 
a^+3ac c'^+3ac 1

2a' 2c'

( 3c' 1

r

3a' , 3 c a 
l + —

+ -

\ —

+

-[ a J 2 [ c J 2 [a cj 
V i a c > O n e n - > 0 va - > 0 .

Tacd: J - - J - c a r (1

> 0 = > - + - - 2 > 0 ^ - + - > 2 a c a c

>1.2

= 3. Vay : ^ + ^ > 4 .

~2 2a-b 2c-b~ 
Bi 2.13: Cho a > 0 , b >0, c>0thoa a + b = c. Chitngminhr^ng:

I

Giii

Ta cd: a > 0, b > 0 va a + b = c nen 0 < a < c v a O < b < c .

a a , b

> - 7 = vd —!=>

a b a b a" b' a + b c , . , . . + i—>—;=- = - = (vi a + b = c) 
'4b 'Tc fc \a "ib 4~c fc

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(1)

2.14: Cho ba s6' a, b, c thoa:

ab + bc + ca>0 . (2)

aboO

(3)

Chiang minh rang a > 0, b > 0 va c> 0.

Giii

Ta chiJng minh bang phan chitng:

Gia su* c6 : a < 0.

Tir (1) => + c > - a > 0 nen b + c> 0

Tv[(2) ^a(b + c) + boo

=>bc>-a(b + c)>0 (vi a<0,b + c>0)

Do a < 0 va b o 0 nen suy ra abc < 0. Dieu nay trai vdi gia- thi6'l

la abc > 0.

Trirdng hop b < 0 hoac c < 0, ta ly luan trfdng tif.

Vay ta da chiJng minh r^ng: a > 0, b > 0, c> 0.

Bi 2.15: Cho 0 < a < l , 0

dang thtJc sau cd it nha't mot bat dang thiJc la sai:

ail-b)>]-, b(l-c)>]-, c(\-a)>]-.

4 4 4

Gi^i

Ta chiJng minh bang phan chiJng:

Gia su" ca ba bat ding thiJc deu dung, liic dd nhan ve'theo ve ta cd:

a(l-a)bil-b)c(\-c)>

. ^ 3

(1)

Mat khac ta cd:

V ^ - V T ^ ) ^ >0=>a + ( l - a ) - 2 ^ a ( l - a ) > 0

28 V a ( l - a ) < i = ^ a ( l - a ) < i

TuTdng tir ta cd :

b(l-b)<

c ( l - c ) <

]_

4 Nhan ve'theo ve ba ba't ding thtfc vuTa n6u thi:

a(l-a)b(l-b)cO.-c)<

Dieu nay mau thuan vdi (1)

Vay ta phai cd: Trong ba bat dang thiJc trong di bai cd it nhat

mot bat dang thiJc sai.

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§ 2. BAT DANG THLfC COSI (Cauchy)

I . BS't d a n g thufc C6si cho h a i so' k h o n g fim
C h o a > 0 , & > 0 :

Ta c6: a + b > 2 ^ 
Da'ii dang thiJc xay ra ^ a = b .

I I . Ba' t d a n g thuTc C6si cho n s6' k h o n g &m

Cho n so'khong am: a^,a^,ããã,ô ã
T a c 6 :

Dau dang thiJc xay ra <^ = = ... = a „ .

D I 2 . I 6 : Cho a, b, c > 0. Chitng minh:

2) (a + fe + c)
Da'u dang thiJc xay ra k h i nao ?

1)

Ut>2.

b a a b c > 9 .

GiSi

1) A p dung bat dang thiJc Cosi cho hai so' khong a m - va - , ta c6:
b a

b a

a b a b ^ ^

- . - ^ - + ->2 (dpcm)
\b a b a

Da'u d i n g thiJc xdy ra <^ — = -<F^a = b. 
b a

2) A p dung B D T C6si cho ba so' khong am, ta c6:

• a + b + c>3y[abc (1)

i , U i > 3 f i . i

a b c \ b c

(2)

30

Nhan (1) va (2) ve theo ve', ta dtfcJc:
(a + b + c)

a b c

> 9 3 L c

~ V

a b c

.i.i.i

Da'u dang thiJc xay ra <^

a b c > 9 (dpcm).
a = b = c

1 1 1 a = = c.
a b c

D e 2.17: Cho a, b, c > 0. Chitng minh:

[• + -1

> 8 . 
c, I a) 
Da'u dang thiJc xay ra k h i nao ?

A p dung B D T Cosi cho hai so'khong am, ta c6:
(1)

l + - > 2 J l . ^

b \

c V c

. l + - ^ > 2 i . £

a \

(2)

(3)

Nha.i (1), (2), (3) v e t h e o ve'ta dUdc:

1 + ^ 1 + ^' [1 + - ] 1 + ^

c K "1 \b c a

i

b] c,

> 8 (dpcm)

D f u d i n g thiJc xay ra <^

1 = ^

\ a^b = c.

a

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D

(J

2.18: Cho a > 0, b > 0. Chitng minh: (a + bf +

Dau dang thiJc xky ra khi nao ?

-A

\

Gidi

Ap dung BDT Cosi tac6:

• a + b> lyfab =^ (a + bf > 4ab

a b ab

in

>

Cong (1) va (2) ve theo ve, ta dufcJc:

(1)

(2).

(3)

Lai ap dung BDT Cosi cho hai so": 4ab va

ab

Tac6: Aab + -->2.

ab \

Aab.

—

ab

= 8

(4)

'-A

Tir (3) va (4) suy ra: {a + b) +

Dau ding thtfc xay ra <^ a = ^ = 1.

>8.

2.19: Cho a,b, c > 0 va a + b + c = L Chtfng minh:

>64.

a

A

c

Dau ding thiJc xay ra khi nao ?

GiSi

. Taco: 1 + 1 = 1 + ^ + ^ = 1 + 1 + - + - (*)

a a a a

Mat khac ap dung BDT Cosi cho bo'n s6': 1,1,

32 Ta drfdc: ! + ! + - + £ >

4 4 / ^

a a V a

Do dd tir (*)

• TuTdngtu":

a \a

l +

i

> 4 4

b

ca

1 1 +

- > 44

c

ab

V

Nhan (1), (2), (3) ve theo ve ta dufdc:

(1)

(2)

(3)

1 +

1

a.

A A

c

s

c.

>64^

>64

'fee ca aZ?

(dpcm)

Dau dang thiJc xay ra

a a

b b ^a^b = c =

-1 = -1 = ^ = ^

c c

a+b+c=l

Bi 2.20: Cho a, b, c > 0 va abc = 1. Chifng minh:

- + b + c

a

- + c + a

b

- + a + b

c

>27.

Dau ding thiJc xay ra khi nao ?

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a

be

• - + c + a>3l

b

ca

• - + a + b>3l

c \ 
b

Nhan (1). (2), (3) ve'theo v e ' f dtfd^:

b

(1)

(2)

(3)

c

be ca ab 
a • b c

/ - \

- + ^ + c f l

1

> iV^abc

a \b )

Mat khac theo gia thie't: abc = 1

Vay:

Dau dang thifc xay ra <f=^ a = ^ = c = 1.

f l

1

- + fc + c - + c + a > 27. (dpcm)

\a } \b ) c ,

De 2.21: Cho a > b > 0. Chiang minh: a +

' DS'u dang thiJc xay ra khi nao ?

GiSi

(a-b).b. 1
ia-b).b

(BDT Cosi)

i(a-b).b

Dau d i n g thtfc xay ra <^

> 3 . (dpcm)

a = 2 
b = \.

B6 2.22: Cho a > 0, b > 0. Chu-ng minh: < 4 / ^ ,

Gidi

0, b > 0. Ap dung baft dang thiJc Cosi cho hai s6' du'dng, ta c6:

2

\fja4b

<4a+^

=^2i/ab

<4a

+ yfb

2^b-j^<(^a + S ) ^ ^

4a^4b ^ '4a+4b

2 ^

4a

+ y[b

< '4ab

(dpcm

).i

D d 2 . 2 3 :

1) Cho a>\,b>\. Chtfng minh: a^Jb-X + b^a- \

2) Cho ba s6' a, b, c doi mot khac nhau. ChiJng minh: 
{a-Vbf , ib^-cf . {c + af

{a-bf"^\b-cf • (c-a) 2 — >2.

Gi§i

1) Ap dung bat dang thtfc C6si cho hai so' khong am, ta c6:

V r ^ = 7( Z j - l ) . l < b-l + \

= ^^a4b^\<^.

(1)

ab 
2 2 ~ 2

Chitng minh tu'dng tu" cflng c6 :

Cong (1) va (2) va' theo ve' ta diTcfc :
a — 1 <ab. (dpcm)

b4^\<±.

(2)

2) Da,t: r = ^'T = =
2-a —o o —c c — a

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2a 2b 2c

Khid6:* (x + l)iy + l)(z + l) =

* (x-l)(y-l)(z-\) =

a — b b — c c — a

%abc

' ia-b){b-c){c-a)

lb Ic la

a — b b — c c — a

Sabc

' ia-b){b-c)(c-a)'

Dodd:

(X

+ \)(y + l)(z +

\)

= (x- l)(y - \)(z

-1)

<!F^xyz

+

xy

+

xz

+

yz

+ x + y + z + 'i=xyz —

xy

—

xz

—

yz

+ x + y +

Z'

^

xy

+

xz

+

yz

=

- \)

Matkhactacd:

(x + y + zf>0^x^+y^ + z^+2{xy + xz + yz)>0 (2)"

(1) va(2) = ^ A : ' + / + Z' - 2 > 0

= ^ x ^ + / + z ' > 2 .

(a — b) {b — c) {c

—

a)

De 2.24: Vdi a >0, b > 0, c > 0. Chiang minh cdcbat dang thiJc sau:

1) ±^^S>2b.

c a

ab be ca ^ , , ,

2) — + — + -—>a + b + c.

c a b

2ab 2bc lea •

1) Ap dung bat ding thifc Cosi cho hai so' dtfdng ta cd

c a \ a

± , ^ ^ ^ + ^ > l b (dpcm)

36 2) Tacd: ^ + ^>2b

c a

ChiJng minh tu'dng tuf; ta cd:

1 >la

c b

be ca ^ ^

— + — > 2 c

(1)

(2)

(3)

(1) +(2) + (3): — + — + — + — + — + — > 2 6 + 2a + 2c

c a c b a b ~

ab be ca

+ — + — > a + Z7 + c.(dpcm)

c a b

3) Tacd: {a + b){a-bf >0

=i

^{a^b)[{a^-ab^b')-ab

=^a^ +b^-ab{a + b)>0

^a' +b^> ab{a + b)

> 0

lab

ChiJng minh tu'dng tu" ta cung cd:

Ibe

c'+a'

lea

(4) + (5) + (6) :

b' +c' ^ b + c

>

1

c + a

a'+b' , b'+c' c'+a'

lab

• +

Ibc

lea

> ^ (4),

(5)

(6)

> a + + c (dpcm)

Be 2.25; Cho a>Q, b>O-Chitng minh: 3a' + lb' > 9ab^

Giii

Ap dung BDT Cosi cho ba so'khong am 3a';3b';4b\

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Tac6: 3a' + 3 ^ ' +Ab' >^l]3a\3b\Ab\

^3a' +lb' >3ab\lj33A (1)

Mat khac: IJ^JA > ^ 3 3 3 = 3 (2)
(1) va (2) ^ 3a' + 7b' > 9ab^ (dpcm).

§ 3. BAT

D A N G

THUfC TRONG TAM G I A C

Cho tam giac ABC. Dat: BC = a; AC = b; AB = c.

Khido: A
a > 0

1) b>0

OO B

\a-b\<c<a + b 
\b-c\<a
\c-a\

Bi 2.26: Cho a, b, c la do dai canh ciia mot tam giic. 
ChiJng minh rkng: +b^ +c'^ < 2{ab + ba + ca).

GiSi

Theo tinh chat trong tam giac ta c6:

0 < a < + c < a(Z? + c)

Q><b<a^-c^b'^ <b{a^-c) 
Q<c<a^-b=>c^ <c{a^-b)

C6ng v6' theo ve ba ding thtfc tren ta dufdc:
a^ <2(a^ + fea + ca).

2.27: Cho a, b, c la do dai ba canh ciia mot tam giac.

Chiang minhr^ng: abc>(a + b-c)(b + c-a)(a + c-b).

Giii

Tac6: a' >a^ -{b-cf ={a + b-c)(a + c-b)>0

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b"" >b^ -(a-cf =(a + b-c)(b + c-a)>0 
. '^c^ -(a-bf =ia-\-c-b)ib + c-a)>0 
Nhan ve' theo ve'ba bat dang thiJc tren ta cd:

a

'feV

>ia + b-c)\a + c-b)\b + c-af 
=^ abc>ia + b- c){b + c - a){a -^c-b)

2.28:

1) Cho x > O v ^ y > O . C h u ' n g m i n h r ^ n g : - + - > — ^ . D i n g

X y x + y

thtJc xay ra khi nao ?

2) Trong tam giac ABC cd chu vi 2p = a + b + c (a, b, c la do dai
ba canh tam giac). ChiJng minh rang:

. D i n g thu-c xay ra khi
1 1 1

• + > 2
p—a p—b p—c

tam giac ABC la tam giac gi ?

a b c

1) T a c 6: i + i > - i - ^ « ^ > "

X y x-\-y xy x + y

'^{x-'ryf > 4 x y

^{x-^yf -Axy>0 
^{x-yf>Q (diing)
Vay: — + - > ^ . Dau dang thiJc xay ra o x = y.

X y x + y

a+b+c -a+b+c

2) p-a = a = > 0

^ ^ 2 2
V i a, b, c la ba canh cua mot tam giac nen b + c > a

=^'-a + b + c>0^p-a>0. 
Tu'dng tu* : p - b > 0; p - c > 0.

Ap dung 1) ta cd:

p—a p—b p—a+p—b 2p—a—b c 
1

Tu'dng

tu-Do dd

' >1

p-b p — c a' p-a p-c b

p-a p — b p — c 
1

> 4

\a b c 
> 2

a b c

Dau xay ra <^

tam giac deu.

p-a p-b p-c

p — a = p — b

p-b = p-c^a=^b = c^ AABC la
p—c=p—a

De 2.29: Cho a, b, c la do dai ba canh cua mot tam giac vdi chu vi 2p.

Chiang minh rang: (p - a){p - b){p -c)< abc

Gi^i

T^ a^ x * a + b + c b + c-a 
T a c o : * p-a = a =

2 2
^ , a+b+c , a+c-b 
* p-b = b =

* p — c — a+b+c ^_a+b-c 
2 2 
Nhan (1), (2), (3) ve'theo veta du^dc:

(1)

(2)

(3)

{p-a){p-b){p-c)^ {a + b-c) (b + c-a) (a + c-b) 
2 2 2

Mat khac: (a + b-c)(b + c-a)(a + c-b)< abc.( De 2.27 )

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Vay: {p - a){p - b){p - c) < abc

~8~ (dpcm).

2.30: a, b, c la do dai cua mot tam giac. Chitng minh rang:

< 1 .

a ^ c a c b

b c a c b a

Ta c6: V T =

GiSi

<^ a c b 
b c a c b a

a b' b c' c a]

+

b a,

+

c b,

+

a c)

1 b'-c' 1

ab 1 be 1 ca

abc c.(a^ -b^) + aib^ - ) + Z?(c' - a')
Mat khac: c(a' -b^) + a(b^ - ) + b(c^ - a')

= c c'-b'

Do d6 :

= c(a' - ) + c(c' -b^) + a(b^ - ) + &(c' - a')
= ( a ' - )(c - Z?) + ( c ' - 6')(c - a)

= ( a - c ) ( c- Z 7) . ( a + c - c - ^ )
= (a-c)(c-b)ia-b).

(a-c)ic-b)(a-b) b.a.c

V T =

<

abc abc 
V\:\a-c\<b;\c-b l< a;l a-b\<c.

= 1 (dpcm)

42

§4. PHUdNG PHAP LAM TRQI

Dilng tinh chat cua bat dang thtfc de du'a bat dang thrfc can chiJng
minh ve dang c6 the tinh diTdc theo phrfdng phap tinh tong hffu han.
Gia su" can tinh tong: S„=U,+U^+U^+... + [/, + . . . + 1 / „ .

Ta phan tich: — a,^ — . 
K h i d o :

S„ -a^) + (a^-a,) + (a,-a,) + ... + (a„- ) = - a „ + , •

2.31: Cho n la so' nguyen thoa n > 1. ChiJng minh:

i . + ^ + i + . . . + _ J _ < , .

1.2 2.3 3.4 n(n + \)

T a c d : — = 1- 1 
1.2 2

1 1 1
2.3 2 3

3.4 ~ 3 4

1 1 1

(n — l)n n — 1 n 
1 1 1
n(« + 1) n n + l

Cong ve' theo ve va ddn gian cac ding thiJc ta diTcJc:
1 . 1

1.2 2.3 •... + • = 1 - -n + l

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Do do: i . + J , + . . . + ^ < , , 
1.2 2.3 n(n + l )

2.32: Cho n la so' nguyen thoa n>2. ChiJng minh:
1 , 1 , , 1 n - 1
- ^ + ^ + ... + -T< •

Giii

1
2^

1
< —

1.2

= 1 - 1
2 '
1

3'

1
< —

2.3
_ 1
~ 2

1
3
1 1

S

. 4

_ 1 1
4^

S

. 4

3 4
1 1

< —

1 1

n (n — \)n n — l n

Cong ve theo ve' cac bat dang thiJc tren ta difdc:
1 , 1 , , 1 1 1 n-\

— + r r + ... + - r < l •

n n n

De 2.33: Cho n la so'nguyen du'cfng. Chitng minh r^ng: 
3n + l . , . . 1 , 1 . . 1 ^« 1

n

2 ( n - l ) S l - + 5 r + ^ + - . . + - < 2 - . n

Giii

1
T a c o : 1 +-1 + ^ + . . . + ^ < 1 + ^ + J- +

2 ' 3 ' 1.2 2.3 {n-\)n

< 1 + f1 1 ' 1 — 1

f ^ M

+

+ ...+

2 ; 2 3, rt — 1 n

T a l a i c o : l + 4r + i + - + -V>l • ^

2.3 3.4

< 2 - i .

n(n + l )
44

1 r 1 1 1 1

+

+...+

2 ' 3, 3 4, ,« « + i .
> 1 +

> l + i - '

1 ^ 3n + l

V a y :

2 n + 1 2 n + 1 2 ( « + l )

2 ( n - l ) 1' 3' .2 — • n

D^ 2 . 3 4 :

1) Cho k > 0 . Chitng minh:

<

1 1

(2Jt + l ) ' Ik 2k + 2 
2) Cho so' nguyen dicing n > 1 . Chitng minh :

9 25 49

(2n + l ) ' ' ^ 4 '

(1)

1) T a c d (1) <^

<

2 
{2k + \f 2k(2k + 2)

4^ (2;t + If > 2k{2k + 2) <^ 4 ^ ' + 4;k + 1 > Ak^ + Ak

<^ 1 > 0. Die 11 nay luon diing.

V a y : 1

<

(2^ + 1)' 2k 2k + 2

2) A p dung cau 1 ta c6:

2 1 1
V d i k = 1 thi <

-9 2 4

k =

2 t h i A

< i _ i ,

25 4 6

k = 3 M A < i _ i .

49 6 8

k = n t h i 1

<

1
(2n + l ) ' 2n 2n + 2

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C6ng vd" theo v<S' n ba't ddng thitc tr^n ta duTdc:
2 2 2

5+ 2 5 + 4 ? ^ - + - ^ '

1 1

(2« + l ) ' ^ 2 2n + 2 ' ^ 2 '

Do 66:

9 25 49

1 1

<-.

(2n + \f 4

2.35: 1) Cho so'nguyen k>\. Chitng minh:
1

< 2 1 1

2) Cho s6' nguyen n > 1. Chiang minh:
1

• +

1 + ... + • 1
2 3^2 4^/3

(n + l)V^

• < 2

(1)

(2)

1) Tacd (1) 4=^

< 2

Giii

(k + l)

< 2

( ^ + 1 ) 7 ^

4k + \-yJk

^ / ) t . ^ / ^ f l

< 2 .

yfk + l

^^/k < yfk + l. Bat dang thiJc nay lu6n diing.

Vay : 1

< 2

(A: + 1)V^
2) Ap diing cau 1) ta c6:

k = 1 t h i - < 2

2

1 1

4k V T H ,

k =

2

thi

k = 3 thi
1

<

1 1

• < 2

172

1 1

4V3

V4,

46

k = n thi < 2

Cong ve theo ve cac ba't dang thitc tren ta du-dc:
1

Vay :

2 + 3 . ^
1

<2 1

-...

+ •

2

3 ^ "• (rt + l)V^ < 2 .

< 2 .

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CHlTOfNG III: SO HOC

I.TINHCHIA HEX

1) Dinh nghia: Cho a, b la cac so nguyen. Ta noi a chia bet

cho b neu ton tai mot so nguyen q sao cho:

a = bq

Ki hieu: a:b. Khi do ta cung noi b chia het a va ki hidu: b|a.

2) Tinh chat:

a) Neu: a chia het cho b va b chia he't cho c.

thi: a chia he't cho c.

b) Ne'u: a, b cung chia he't cho c

thi: a + b va a - b chia he't cho c.

c) Ne'u: a chia he't cho c hay b chia het cho c

thi: a b chia he't cho c.

3) Dau hieu chia he't:

a) Dau hieu chia he't cho 2: chi? so tan cung la chan ;

b) Dau hieu chia he't cho 3 (cho 9): tong cac chi? so chia

he't cho 3 (cho 9).

c) Dau hieu chia he't cho 4 (cho 8): hai (ba) so' cuo'i chia

he't cho 4 (cho 8).

d) Dau hieu chia he't cho 5 : chu" so' tSn ciing 1^ 0 hoSc 5.

e) Dau hieu chia he't cho 11 • hieu giffa tdng cac chiJ so d

vi tri chan va tdng cac chff so'd vi tri le chia he't cho

11 .

II. SO NGUYEN TO

1. Mot so nguyen du'dng du'cJc ggi la so nguyen to ne'u no chi

chia he't cho 1 va cho chinh no.

• Nhtf vay theo dinh nghia nay, so' 1 la so nguyen to.

48 Tuy nhi^n trong cic s6' nguyen t<5' tham gia l^m thiYa

s6' ciia mot tich, ngtfdi ta qui rfdc khong ki d6'n so' 1.

2. S6' nguyen dufdng kh6ng la s6' nguySn t6' duTcJc goi Ik hdp

s6'.

3. Gia siJ so' nguyen difdng N diTdc phan tich ra thilfa s6'

nguy6n t6':

N = a'".b\c'.

Khi dd tdng cac u'dc s6'khdc nhau cila N Ik:

(m + l)(« + l)(^ + l).

III. U(5C CHUNG L(5N NHAT - BOI CHUNG NH6 NHXT

1) Binh nghia 1: Udc chung Idn nhS't (l/CLN) ciia hai s6' a b

khong ddng thdi bkng 0 la s6' nguySn Idn nhat chia ha't ca a

vkb.

Kihi6u: (a ,b) d^ chi UCLN cila a vk b.

2) Binh nghia 2: c^c s6' nguyen a, b diTdc goi Ik nguySn t6'

cing nhau n^'u: (a, b) = 1.

3) Tinh chat: ra. b) = ra. \a + h\\= (\a + h\h)

4) Cach tim UCLN cua a vk b:

• Phan tich a va b thknh thiTa s6' nguyen t6'.

• Lay tich cac thu'a s6' chung vdi s6' mu be nhat.

5) Binh nghia 3: Boi chung nhd nhat (BCNN) cua hai s6' a va b

la so' h6 nhat chia he't cho ca a va b.

Kijiieu: [a;^] de chi BCNr cua a va b.

6) Cich tim BCNN cua a va b:

• Phan tich a vk b thanh t'lii'a so' nguyen t6'.

• Lay tich cac thil'a s6' chung va rieng vdi s6' mu Idn

nha't.

IV.DONGDUTHLTC

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ta duTdc cilng dif so'.

Khi 66, ta k i hien: a = b (mod m).

2) a = b(mod m)-^ a — b + km vdi k la so' nguy6n
3) Tinh chat:

* a = a(modm)

* a = ^(mod m)=> b = a(mod m)

a = b{Kodm)

b = c(modm)

a = ^(modm)

a = c(modm)

c = c/(mod m)

* a = Z>(mod m) =^

a±c = b± c(mod m)

ac = b.d{modm) 
na = nb(modm) 
a" =b''(modm).

V. SO C H I N H PHUdNG

1) Binh nghia: Neu a la so' nguyen thi a^dmc goi la so' chinh
phifcfng.

2) Tinh chat: Tan cung cua so'chinh phu'cfng 'a 0, 1, 4, 5, 6, 9.

V I . N G U Y E N L Y D I R I C H L E T

Khong the nho't 7 chit tho vao 3 chiec long, sao cho
moi long c6 khong qua hai chit tho.

* Nhu" vay neu nho't 7 chit tho vao 3 chiec long thi it nha't phai
c6 mot chiec long c6 tu" 3 chu tho trd len.

V I I . NHJ T H l f C NEWTON

( a +b f =

cy +

cy-'b

+

cy-^b^

+...+

cy.

De 3.1: Chifng minh r^ng vdi moi s6'tur nhien n > 1 ta c6:
3'"+' + 4 0 « - 2 7 ; 6 4 .

Gidi

Ta C O: 3 ' " + ' + 40« - 27 = 3^3'" + 40n - 27.
= 27.9" + 40n -21 = 27.(8 +1)" + 40n - 27

= 27 (8" +

Cl^"-'

+ . . . + C"-^8' + C;-' .8 +1) + 40A7, - 27

= 27 (8" + C' .8"^' +

... + Cr

'.S') + 27

.Cr'

.8 + 27 + 40n - 27

= 27.8'.^ + 27.8rt + 40n - 27.64.^ + 256n = 64(27^ + An). 
Vay: 3 ' " + ' + 4 0 n - 2 7 ; 6 4

Ghi chu: C ° = l

;Cr'=n.

Tong quat: =

A:!(n -A:)!

Trong do: n! = 1.2.3...n (vdi n6 iV*) va 0! = 1.

3.2: Chifng minh rang tong 2p + 1 so tu" nhien lien tie'p chia he't
cho 2p + 1.

Gidi

Gia su" 2p + 1 s6' tuT nhian lien tie'p la: k; k + 1; k + 2 , . . . , k + 2p.
Khi d6: S = k + (k + 1) + (k + 2) + ...+(k + 2p)

= A: + ^ + A:+... + /: + ( l + 2 + 3 + ... + 2p)
2p+l so'

= {2p + l)k + 2p. 2p + l

=^i2p + l)k + p{2p +1) = (2/7 + \){k + p). 
Vay: S\2p + \.

Ghi chit: V d i moi so'tif nhien n. ta cd: l + 2 + 3 + ... + n = «(n + l )

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3.3: Cho n e N'. Tim UCLN cua hai s6': 2n + 3 va n + 7.

Gi§i

Ap dung tinh chat: (a,b)= {\a±b\,b)^{a,\a± b l).
Ta c6: (2n + 3;n + 7) = (l n - 4 l,n + 7) = (l n - 4 1,11).

Nhtfng 11 Ik so' nguyen to' nen so' can tim la 1 hoSc 11.
Khi do: * Khi In - 41 la bpi so cvia 11 thi:

UCLN(2n + 3;n + 7) = 11.

* Khi In - 41 khong la boi so' cua 11 thi:
UCLN(2n + 3;n + 7 ) = 1.

3.4: Mot so' nguy6n du'Png N c6 dung 12 tfdc s6' (du'dng) khac
nhau ke ca chinh n6 va 1, nhiTng chi c6 ba tfdc so' nguyen t6' khac
nhau. Gia su" tong cua cac ufdc so' nguyen to' la 20, tinh gia tri nho
nhat C O the cd ciia N .

Gi^i

Goi cdcirdcnguy6n to'ciia so'N la p, q, r va p < q < r

f^ = 5;r = 13
q = l;r = U

[A^=2".7Mr

V d i a,b,c e N va (a + l)(b + l)(c + 1) = 12.
T a c d : 12 = 2.2.3

Do d 6 N c 6 thela : 2.5.13^2.5^13 ; 2^5.13 ; 2.7.11^ ; 2 . 7 l l l ; 2^7.11
N nho nhat nen N - 2^5.13 = 260.

3.5: Chitng minh r^ng c6 s6' nguyen drfdng chi chiJa cac chi? s6' 0
va chi? so 1 va so do chia het cho 1999.

Xet 2000 so'nguyen du-png sau: 1; 11; 111; . . . ; H L ^

2000 chuso'X 
Cac s6'tr^n khi chia cho 1999, ton tai it nhat hai sd'chia cho
1999 cd cilng so'du" (nguyen ly Dirichlet). Goi 2 sd'dd la :

11 • 1. va 11 1 (2000 > m > « > 1).

m chusd'l n chusd'l

» 1 1 1 - ,11 1 chia het cho 1999.
m chusd'l n chusd'l

^ 11 100 0 chia h^'t cho 1999 (dpcm).

m—n chiisd'i n chuso'O

Bi 3.6: Cho n la so' nguyen diTdng Idn hcfn 1. ChiJng minh
n" + + 1 la mot hdp so'.

Gidi

Tacd: n ' + n ' + l - n ' + 2 n ' + l - n '

= (n^ + lf-n^ = (n^+n + l)(rt' - « +1).
V d i n > l , t a c d : n ^ + n + 1 ^ 1 va n ^ - n + 15^1.

Vay: + n ' + l l a m p t h d p s o .

Bi 3.7: Tim s6'nguy6n dtfdng n de: -n^ +n-l la s6'nguyen to.

Gi§i '

Tacd: A = n'-n^ +n-l = n ' ( n - l ) + ( « - l ) - ( « - l ) ( n ' + 1 ) .
• Khi n = 1, A = 0: khdng la so' nguyen t6'.

• Khi n = 2, A = 5: Ik so' nguyen t6'.

• K h i n > 2 : T a c d : n - l > 2 va n ' + l > 1 0 n a n A l a
hdp s6'.

Tdm lai: Khi n = 2 thi -n^ + n-\a so'nguyen to'.

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3.8: Cho p va + 2 la hai s6' nguyen t6'. Chitng minh: + 2
cung la s6' nguy6n to'.

Gi^i

Dat: p = 3 m + r ( v d i meN';r = 0;1;2 )

K h i d d : + 2 = (3m + r ) ' + 2 - 3(3m' + 2mr) + r^+2.

• V d i r = 0 : T a c d p = 3m.

V i p la so' nguyen to' nen m = 1

=^ p = 3 =^ + 2 = 29 + 2 la so nguyen to.

• V d i r = 1: Ta CO

+2 = 3(3m' + 2m) + 1 + 2 = 3(3m' + 2m + 1 )
V i : m G A^* =^ 3m^ + 2m + 1 > 1

=^ + 2 khdng la so' nguyen t6'.

D i e u nay trai v d i gia thie't.

V a y khong xay ra tru'dng hdp r = 1.

• V d i r = 2 : T a c 6 / ? ' + 2 = 3 ( 3 m ' + 4 m ) + 4 + 2

= 3 ( 3 m ' + 4 m + 2)

- , V i : 3m^ + 4m + 2 > 1 nen + 2 la hdp so.

D i e u nay trai v d i gia thie't.

vay cung khong xay ra tru-dng hdp r = 2.

T d m l a i : K h i p va /? V 2 la so' nguyen to t h i : p = 3.

K h i d d : + 2 = 29 cung la so' nguyen to' (dpcm).

3.9: ChiJng minh rang:

1) V d i m o i so'nguyen du'dng n, phan so'sau la to'i gian: ^ ^ " ^ ^ .
14n + 3

2) N^ l a s o v o t i .

Giii

1) Ta se chtfng n i n h : 21n + 4 va 14n + 3 la hai so' nguyen to' ciing
nhau.

^ That vay: ne'u goi d(d > 1) la U C L N cua hai so' tren.
i i 2 1 n + 4 = pd

• va 14n + 3 = qd ^2)

trong dd p, q la nhifng sd'nguyen, du'dng.

( l ) - ( 2 ) : 7 n + l = ( p - q ) d = ^ 2 1 « + 3 = 3 ( / ; - 9 ) J (3)

( l ) - ( 3 ) : l = p.d-3(p-q)d = (3q-2p).d=^d = 3q-2p = l.

V a y : 21n + 4 va 14n + 3 la hai so' nguyen to' cung nhau

td-c la: i l ^ i ± l la phan so t o i gian.
14n + 3

2) Ta chiJng minh bang phan chiJng.

G i a i su- rang V2 e g v d i ^ = ^ (viet du^di dang to'i gian nghia la

1

p va q la hai so' nguyen to' ciing nhau).

T a c d 4- = 2 = ^ / =2q^

^ p^ la so'nguyen chan =^ p Ik so'nguyen chSn

^ p = 2m ( v d i m G Z ) =^ 4m^ = 2^^ =^ = 2m^

q^ la so'nguyan chan =^ ^ la so'nguydn c h f n

=^q = 2k

V i p = 2 m va q = 2k nen chiing khong nguyen td' cung nhau (cd tfdc
so' chung la 2). D i e u nay mau thuan.

V a y b ^ t b u d c yf2 l a s o v o t y .

t>i 3.10: Chu-ng minh rllng: 3" = - l ( m o d l O ) . K h i \'k chi k h i
3"^' = - l ( m o d l O ) .

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Giii

Tac6: r =-l(modlO) 3" -10A:-1

<^3"+' =(10/:-l).81

.<i4

.3''^'' =810A;-81.

^3"+' =(8U-8).10-l.<f::^3"+'' =10^-1.

<!=J^3"+'* =-l(modlO).

3.11: ChiJng minh rling vdi moi s6' nguyen difdng ta c6:

1376" =1376(mod 2000).

Ta dung phufdng phap qui nap de chiJng minh. .

• Vdi n = 1: De thay: 1376^ = 1376(mod 2000).

Ttfc menh dd dung khi n = 1.

• Gia su" menh de ddng tdi n = k. Ttfc la:

1376* =1376(mod2000). .

=^ 1376*+' = 1376'(mod 2000) (1)

Matkhac: 1376^=1893376 -1892000 + 1376

= 946.2000 +1376 =^ 1376' = 1376(mod 2000) (2)

Tit (1) Yk (2) 1376*+' = 1376(mod2000)

Nghia la mdnh de diing tdi n = k + 1.

Vay vdi moi s6'nguyen difdng n, ta cd: 1376" = 1376(mod2000)

Bi 3.12: ChiJng minh r^ng s6' jc = 10^" +10'" +10^" +1 chia het

cho 7, 11 va 13 neu n le. Tim so' dif cua phdp chia x Ian liTdt cho 7,

l l , 1 3 v a l l l .

Giii

.x = 10'"+10'"+10'"+l

TH,: Vdinle.tacd: * 10'" +1 = (10')" -(-1)"

chia het cho

56 10'-(-1) va: 10'-(-1)-1001 =

7.11.13-Do d6: 10'" + 1 chia het cho 7 ; 11 ; 13.

Matkhdc: ;c = 10'"(10'"+l) + 10'"+l = (10'"+l)(10'"+l).

Do dd:

X

chia het cho 7 ; 11 ; 13.

11 ^2:

Vdi n chan

10'"+l = [lO'"-(-l)"] + 2 chiacho7 ; 11 ; 13dir2

10'"+l = [lO'"-(-l)'"] + 2 chiacho7;ll ;13dir2.

* Do dd

X

chia cho 7 ; 11 ; 13 deu cd sd'diT la 4.

* Ngokiratacd: 10'"+1 = [(10')"-l"] + 2.

Vi: (10')" - 1 " chia ha't cho (10' -1) nhiTng: 10' - 1 = 999 = 3.111

chia he't cho 111.

Do dd 10'" +1 chia cho 111 du" 2 vdi moi neN.

Tu-dng ttf 10'" +1 chia cho 111 dtf 2 vdi moi n e iV.

Suy ra x chia cho 111 dtf 4 vdi moi neN.

Tom lai: x chia cho 7 ; 11 ; 13 duT 0 n^'u n le ; dtf 4 na'u n chan.

X

chia cho 111 du" 4 vdi moi neN.

Bi 3.13: Cho so' nguyen n > 1. ChiJng minh ring: n" -«'+«-1

chia he't cho (n-lf.

Giii

Nhan x^t rkng vdi n = 2 thi : n" -n^ +n-l= (n-\f= 1 nan bai

, toan hien nhian dung.

Bay gid ta xet n > 2:

Tacd: n"-n" + n - l = («""' -l)n' + ( n - l )

= (n - l)(n"-' + n"~' +... + « + l)n' +(«-!)

= (n - l)(n"-'+ n"-'+... + ) + (n -1)

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=l + k^(n-l) (vdi k^=n + l).

n"^' = 1 + k„_^ ( « - ! ) ( vdi k„.i = n"-^+.. .+n +1 )

Cong ve theo ve ta c6:

+ ... + + 1 = (n - 1 ) + (k, +... + ).(n - 1 )
= ( n ^ l ) ( l + k i + . . . + k , , i )
nen n"~' + . . . + + 1 chia he't cho n - 1.

Do do tu" (1) ta s«y ra rang n" -n^ +n-\a het cho (n - 1 ) l

De^ 3.14: Xet ba so t\i nhien a, b, c th6a he thiJc: ^b^+c\ 
ChiJng minh r i n g neu a, b, c nguyen to' cung nhau thi a la so' le va

trong hai so' b, c c6 mot so' le va mot so' chfn.

Giii

Gia su" a chfn .

V i a, b, c nguyen to' cung nhau nen b, c le.
• a chan chia he't cho 4.

• h,c le =>b + c chia he't cho 2 va be khong chia he't cho 2
^ib + cf - 2bc khong chia he't cho 4.

=^b'^ khong chia he't cho 4.
Nhtfng theo giai thie't: = ZP^ + c^ V6 l i !

Vay a le ma a' = + =^ c6 mot sd' le va m6t so' chJn
^b,c CO mot so' le va mot s6' chfn. (dpcm)

Y)i 1.15: Chu-ng minh rling tong cac binh phu-Ong cua ba so' nguyen

hen tie'p khong the la lap phu'dng cua mot so' nguyen dtfcfng.

Giii

Tru'dc he't ta chitng minh neu n e Z thi:
• chia cho 3 dtf 0 hoac 1.

* chia cho 9 dtf 0 hoac 1 hoac 8.

That vay: Dat n = 3m + r{n,m eZ;r=- 0;±1)
* = (3m + rf - 9m' + 6mr + r '

r ' = 0 ;1 nen n'chiacho 3 du-Ohoac 1 (1)
• * ==(3m + r)^ = 2 7 m ^ + 2 7 m V + 9 m r ' + r \

= 0 ;1 ; - l nen chia cho 9 du" 0 hoac 1 hoac 8 (2)
* Goi ba so nguyen hen tie'p Ian lu-dt la a - 1 ; a ; a + 1.

Ta CO (a - 1 ) ' + a' + (a + 1 ) ' = a' - 2a + 1 + a' + a' + 2a + 1
= 3 a ' + 2 .

Tu" (1) ta CO 3a' + 2 chia cho 9 drf 2 hoac 5 ma chia cho 9 du" 0
hoac 1 hoac 8 (tu" (2)).

Do do 3a' + 2 khong the la lap phu'dng cua mot so'nguyan drfdng.

3.16: ChiJng minh rang tich bo'n so'nguyen du-dng l i ^ n tie'p khong
the la so' chinh phu'dng.

Giii

Giai siJ ton tai so' nguyen m de:

m ' = nin + l)(n + 2)(n + 3) <^ m ' = ( n ' + 3n)(n' + 3n + 2)
= k{k + 2) (vdi k = n ' + 3n )

Tacd: y t ' < ) t ( ^ + 2 ) < ( ^ + l ) '

Dieu nay mau thuan vdi k{k + 2) = m ' .
TO do suy ra dieu phai chifng minh.

3.17: Tim so' nguyen n sao cho: n' + 2n' + 2n^ +n + l la mot so
chinh phu'dng.

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Giii

Theo de bai ta c6: n* + 2n^ + 2n^ + n + 1 = (vdi m e Z )
^ A{n' + In" + 2n^ + rt + 7) = Am"

4=^ ( 4 « ' ' + 8n'. + 8 n N 4n +1) + 27 = W
<i=^ (2n^ + 2n +1)^ - 4m^ = - 2 7 .

^ {In' + 2n + 1 - 2m)(2n^ + 2n + 1 + 2m) = - 2 7 .
Do do c6 cac trifdng hdp sau day xay ra:

2 n ^ + 2 n + l - 2 / n = l
2 « ^ + 2 n + l + 2m = - 2 7 '

2 n ^ + 2 n + l - 2 m " = - l
!)•

5)

2n^+2n + l + 2m = 27

2 n H 2 n + l- 2/ n = - l
« = —3 /la}' n = 2

-2n^+2n + l - 2 m = 3
2n^ +2n + \+2m = -9 
2n^ + 2rt + l- 2 m = - 3

2/i^ + 2n

+ l +

2m = 9

2 n ^ + 2 « + l- 2 m = 9

2n^+2n + l + 2/n = - 3

4n^ + 4n + 28 = 0 (vdnghiem) 
2 « ' + 2 n + l + 2m = - 2 7

2n^+2rt + l - 2 m = - l

/n = 7 m = 7

n = - 3

n = 2

2n^+2Ai + l- 2 m = 3

4n^ + 4n + 8 = 0 (vo nghiem) 
2n^+2n + l- 2 m = - 3

+ n 4-1 = 0 {vo nghiem) 
2 n ^ + 2 n + l - 2 m = 9
n ' + n - l = 0

V d i + n - 1 = 0 <!:^ n = i ( - 1 ± ^/5): khong la so nguyan (loai)

2n^ + 2 n + l - 2 m = - 9
2n^ + 2 n +

H-2m

= 3
2n^ + 2 n + l - 2 w = 27
2n^ +2n + \ 2m = -l

2 « ' + 2 n + l - 2 m = - 9
+n + 2 — 0 {vo nghiem) 
2 n ^ + 2 « + l - 2 m = 27

n ' + n - 6 = 0

n = 2

m = 7

n- - 3

2 n ^ + 2 n + l - 2 m = : : - 2 7
2 n ^ + 2 n + l-|-2m = l

2 n ^ + 2 n + l - 2 m = - 2 7
n ^ + n - | - 7 = : 0 {vo nghiem) 
, Vay cac s6'c^n tlm la: n = 2, n = -3.

f)i 3.18: Tim ba so' t\I nhien m, n, k d6i mot khac nhau sao cho
J- + i + - la mot so' tu" nhien.

m n

Ta CO the gia su" r^ng m < n < k

Dat / = — + - + - vdi i la s6' tu* nhian
m n k

V i m < n < k -

>i>i=»/ = l + l + i < A

m n k m n k m

= ^ I < 3 (vi m > 1 =j> — < 3 ) =j> / = 1 hoac i = 2.
m

* V d i i = 1 thi c6 1 < — =^m < 3 =j> m = 1 hoac m = 2.
m

• V d i i = m = 1 thi khdng th6a

m n k [ I n k ,

• V d i i = 1, m = 2 thi dieu kien

m n k 2 n k n k 2 
2 1

Suy ra —>-=^n>4 
n 2

vt - > —

n k

Ngoai ra 2 = m < n < 4 =^ n = 3 va do dd:

/ = - + - + - . < ^ ^ = 6
m n k

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* V d i i = 2 tl>i c6 I 2 < — =4> m < - =4> m = 1
m 2

' L t i c d d d i l u kientrd thanh: 2 = 1 + - + ^ = ^ - + ^ = ! .

n k n k

M a m = l < n < k n e n n > 2 , A : > 3 - ^ - + | < ^ + ^ - 7 < l .
n A: 2 3 6
Do d6 khong thoa dieu kien neu tren.

Ket luan: m = 2, n = 3, k = 6 la ba so' can tim.

3.19: T i m cac so' nguyen difclng m, n sao cho: 2m + 1 chia he't
cho n va 2n + 1 chia he't cho m.

GiSi

Trirdc he't ta xet triTdng hdp \

Ta c6: m < 2n + 1 < 2m + 1 (vi 2n + 11a boi so' cua m)
2n + 1 = m hay 2n + \ 2m hay 2n + \ 3m.

(vi 4m > 2m + 1 nen kh6ng the nhan 2n + 4 = 4 m , . . .)
• 2n + 1 = m:

Ta c6: 2m + 1 = 2(2n + 1) + 1 = 4n + 3 chia he't cho n nen suy
ra 3 chia h6't cho n, v i the phai c6: n = 1 hoac n = 3.

V d i n = 1 t h i m = 3 .
V d i n = 3 t h i m = 7.
• 2n + 1 = 2m:

Ta CO 2m + 1 = 2n + 2 chia ha't cho n nen suy ra 2 chia he't cho
n va do do: n = 1 hay n = 2.

V d i n = 1 thi m = - (loai)
V d i n = 2 thi fn = ^ (loai)

• 2n + 1 = 3m. .

62

Ta cd 2m + 1 > 2n + 1

2m + 1 > 3m =^ m < 1 =^ m = 1.
V d i m = 1 t h i n = 1.

Do do ta thu du-dc: (n = 1, m = 3),(n = 3, m = 7), (m = n = 1)
jsfhan xet rang vai tro ciia m va n nhu* nhau trong de bki, do vay khi

1 < m < « ta tim them du-dc: (m = 1, n = 3), (m = 3, n = 7).

D6 3.20: M o t so' chinh phu-dng cd dang abed. Bie't ab-cd = l. Hay
tim so' a'y.

Dat abed = vdi a, b, c, d la cac chff s6'. Ta phai cd a^O,

ngoaira, do ab-cd = l n€n : e^O.

T a c d : = 1 0 0 a i + ^ = 1 0 0 Q + l ) + ^ .

Suyra: - 1 0 0 = 1 0 1 ^ - ^ ( n + 1 0 ) ( « - 1 0 ) = 1 0 1 ^

. V i n^cd bo'nchu'so'nen n < 100, do dd n + 10 = 101, hay n = 91.
Vay 8281.

Bi 3.21: T i m so' tu* nhien n = ab cd hai chu" so' sao cho:

n + ab = (a + hf

T a c d : n = ab = \Oa + b.

Theo gia thiet: n + ab = ( a +b )"

^I0a + b + ab = a^ +b^ +2ab 
^b^ +ia-\)b + a(a - 1 0 ) = 0

^b^ +(a-\)b = a(lO-a) (1)

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Ta lai bi^'t r i n g (x + yf > 4xy

n€n (a + ( 1 0 - a ) ' ) > 4 a ( 1 0 - a) = i- 1 0 0 > 4 a ( 1 0 - a )

= ^ a ( 1 0 - a ) < 2 5 (2)
TO (1) va (2)suy ra:

+(a-\)b<25^b^ <25 (•v$a-$b>0) 
= ^ ^ < 5 , d o d 6 b = 0, 1,2, 3,4,5
* Neu b = 0: (1) a = 0 : kh6ng phu hdp.

* N e u b = l : ( l ) = ^ l + a - l = 1 0 a - a '

=^ - 9 a = 0 a = 9 (vi phai c6 a>l) 
* N e u b = 2: (1) 4 + 2 a - 2 = l O a - a '

= ^ a ' - 8 a + 2 = 0 = ^ a = 4±N/ l 4 (loai)

* N a u b = 3 : ( l ) = ^ 9 + 3 ( a - l ) = 1 0 a - a '

= » a ^ - 7 a + 6 = 0 = ^ a = l hay a = 6.

* N e u b = 4: ( l ) = ^ 1 6 + 4 ( a - l ) = 1 0 a - a ' .
=^ - 6a + 1 2 = 0 (v6 nghiem)
* N e u b = 5 : ( l ) - » 2 5 + 5 ( a - l ) = 1 0 a - a '

I =^ a ^ - 5 a + 20 = 0(v6 nghidm)

Ket ludn: n = 91, n = 13, n = 63 la cac so' can tim.

64

CHL/CfNGIV: GIA T R I L

6 N NHAT

VA GIA T R I NHO NHAT CUA HAM SO

Neu CO hang so'M sao cho:

N6u CO h i n g s6'm sao cho:

I. DINH NGHIA. Cho ham so' y = f(x) x^c dinh vdi x e D.

/ W < M , V ; c e D

3x^eD:f{x,) = M 
thi M la gia tri Idn nhat (GTLN) cua f(x)

K i h i e u : M = max f{x).

'f(x)>m,WxeD 
3x,eD:f(x,) = m 
thi m la gia tri nho nha't (GTNN) cua f(x)

K i hieu: m = min f{x).

Ghi chu: Tap xac dinh D la tap cac gia tri x sao cho f(x)

CO nghia.

II. C A C H T I M G T L N VA GTNN CUA H A M S d

1) Loai 1: Dilng tinh chat: |A| > A. Dau "=" xay ra

2) Loai 2: Gia su' A, B la cac hang so, B > 0 va g(x) > 0.

Cho: f{x) = A^ B

Khi do: * f(x) Idn nha't ^ g(x) nho nhat.
* f(x) nho nhat ^ g(x) Idn nhS't.

B 
Cho: f(x) =

A-g(x)

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3) Loai 3: Dung dieu kien c6 nghiem cua phtfOng trinh bac

hai.

Ghi chu: Cho ham so y = f(x) xac dinh tren tap D.

Neu ph^dng trinh y = f{x)c6 nghiem

xeD ^ a<y<b.

thi min f(x) = a va max f{x)^b.

4) Loai 4: Diang cac linh cha't cua bat ding thiJc.

4.1: TimGTNN cua : j = yjx-l-lyf^^ + ^Jx + l-6y[7^.

Tac6: y = y]x-\-2y/x-2 +^lx + l-6yf7^

^ y = ^(x-2)-2yfI^ + \+yl{x-2)-6sf^ + 9

• Di^ukien:

A

:>2. Khidd:

y =

yJx-2-l + ylx-2-3

^y = ylx-2-\ 3-^x-2 >i^yJx-2-\) + {3-^x-2

Dau "=" xay ra ^ yJx~2-\>0

3-ylx-2>0

^ l

< V x

- 2 < 3 < ^ l < x - 2 < 9

^ 3 < ;c < 11 (thoa dieu kien)

Vay: min y = 2.

Dat dtfcfc khi: 3 <

A

: < H .

66 p i 4.2: Cho M = yjx + 4sjx -4 + ^x-A^x-A . Tim x de M nho

nhS't va xac dinh gia tri nho nha't cua M.

Gi^i

Dau "=" xay ra <^

Tac6: M = ^jx + Asfx^ + yjx-Ayjx - A

^ M =

4{X-A)

+ A^X-A^A +

SJ{X-A)-A4I^^A

M =

^ ( V

^ ^ + 2 ) '

+^^(V7r^-2)'

Dieu kien: x > 4 . Khidd:

y

; c

- 4 + 2

+

V ; c

- 4 - 2

V

^ - 4 + 2 + 2

- V x

- 4

> ( V x

- 4 + 2 ] + f2

- V ; c

-4W4.

V A

: - 4 + 2 > 0

2 _ V ; c

-4 > 0 .

4=^0<

V;c

-4 <4<i=>0

< A

:-4<4

<^ 4 < ;c < 8 (thoa dieu kien)

Vay khi 4

< A: <

8 thi M dat gia tri nho nhat va min M = 4.

4.3: Tim GTNN cua ham so: y = + -4;c + 4.

Giii

Td c6: y = yfx^ + ^x^ -4

;c

+

4. 4^ y^ \[x^ + ^[x-2'f (hilm so'xac dinh vdi moi x).

^y =

+

x-2

Da'u "=" xay ra

< ^ y = : A : + 2- A : > A : + (2- A : ) =

2 ;c

>0

^Q<x<2

2-x>0

Vay min y = 2 dat dtfdc khi 0

< A: <

2,

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4.4: Tim GTNN cua bieu tMc: A =

2 + yj2x -x^ +l'

Ta c6: A =

Gi^i

2 + V 2 x- ; c ' + 7 2 + V- ( ^ - l ) ' + 8

D i l u kien: ^(x-lf +S>0 ^ (x-if < 8

4=^-2 V2 < X - 1 < 2 ^ 4^ 1 - 2V2 . < A: < 1 + 2 v ^ .

Khi d6: A nho nhat ^ 2 + ^-(x-l)^+S Idn nhat

^ -(x-2)' + 8 Idn nhat ^ (x-lf = 0 ^ x = 1, 
Vay khi X = 1 thi A dat GTNN va

3 3

minA = , = j=.

+

V

- 0 '

+ 8 2 + 2V2
Bi 4.5: Cho M =

GTNN c u a M .

x^-2x l_

x'+\ \ ^x + 2 l- V ^ + 2,

. Tim

Dieu kien xac dinh cua M la:

Tacd: M = x{x-2)

Gi^i

A: + 2 > 0

VJC + 2 ^ 1
2

A: > - 2

(;C + 1 ) (A:'-X + 1) • 2 l - ( x + 2)

4^ - 2 ) 1 _ Jc(jc- 2)-(.y'-;c + l )
{x + Viix"-x + \) x^\~ + -;c + l )

68

-(;c4-l) - 1

(jc + l ) ( ; c ^ - x + l ) x'^-x^\ 
X —

Vay M nho nha't khi

-^2 Idn nhat.

X 
2

n

X 
2

H— dat gia tri nho nhat

1 1

4=> = 0 < ^ =

-2 -2

1 4

Vay khi X = - thl min M = .

• 2 3

D l 4.6: Tim GTNN va GTLN cua ham so: y = x" +3x-\ 
x^ +2X-V5'

Gi^i

Ta cd y xac dinh vdi moi x vi + 2;c + 5 = 0 la v6 nghiem
jc' + 3 ; c - l

Tacd: y =

x" -^2x^-5 ^ x'y + 2A;>' + 5y = A;' + 3A: - 1

^ ^ ^ ( j - l ) ; * : ' + ( 2 y - 3 ) x + 5>' + l - 0 (*)
• V d i y = 1: (*) <^ - X + 6 = 0 ^ X = 6

(phu'dng trinh cd nghiem) (1)
• V d i } ' 1 : ( * ) CO nghiem khi:

A > 0 4^ ( 2 j - 3 ) ' - 4(3; - I X S j +1) > 0

^ 4 / -12>' + 9- 4 ( 5 / - 4> ' - l) > 0

4=^16/ - 4> ' - 1 3 < 0

2-2V53 2 + 2N

/53

4=> < J <

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(36)<div class='page_container' data-page=36>

1 - V53 ^ ^ 1 + 753 ,

^ < > ' < r ( y ^ l ) (2)
o o

TO (1) va (2) suy ra: (*) c6 nghiem khi: ^ ^ <y< . 
1-V53 1 + V53

Vay: min y=^ ,max y = .

8 8

De 4.7: TiiriGTNN va GTLN cu& y =

x'-5x + l

Wi x^ - 5x + l ^0 v6 nghiem nen y xac dinh vdi moi x.
x^

Tac6:y = — — ^ y(x^-5x + l) = x^ 
x^-5x + l

^(y- \)x^ -5yx + ly = 0 (*) 
7

V d i y = l : (*) <s=^-5x + 7 = 0 ^ ;c = - ( p t c6 nghiem) (1;
V d i y ^ 1: (*) CO nghiem khi:

A > 0 <^ ( - 5 ^ ) ' - 4iy - l)(7y) > 0
< ^ 2 5 / - 2 8 7 ' + 2 8 y > 0
4^ - 3 / - + 2 8 > ' > 0

^0<y<=j. {y^\) 
28
TO (1) va (2) suy ra: (*) c6 nghiem khi: 0 < y < y .

28
Vay: min y = 0 va max y = — .

Dd4.8: Tim GTNN cua ham so: y = -^'^^

x'+2x^+\

Giii

x" +2x^+\ 'x^ + I f > 0 nen y xac dinh vdi moi x.
x' +2x^+\)-(x^+\) + \

cd: y

x'+2x'+\

(x'+\f (x'+\f = 1

1 1

x^+l '

at t =

x ' + l

{hi y = t^ -t + \^y = 1

/

\2

Xet ? = - 4 ^ + 1 = 2 4=^ jc^ = 1 <^ jc = ± 1 .
; 2

Vay ta cd khi ;c = ± 1 thi f = - / - - i - = 0 <s^ y = - .
• 2 2 ^ 4

3
Vay y dat gia tri nho nha't la: min y = —.

4.9: Tim gia tri nho nha't cua ham so': y = x^ + \ - 2

X

vdi x^O.

1

X 
X

+ 5,

Dat t = x--thi t' = x'+\-2^x'+-l- = t^+2.

X X X

Tacd: y = r' + 2 - 2 / + 5 = / ' - 2 ? + 7 = ( / - l f + 6 > 6
Xet t = \ x - - ^ \ x^-x-\ 0^x = -(\±^f5]

X 2^ I

Vay khi = - ( l ± thi y = 6 nen y dat gia tri nho nha't la: min y = 6.

4.10: Cho y = (x --l)(x -- 2)(x + 4)(x + 5). Tim x de y dat gid tri
nhd nha't. Xdc dinh min y.

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GiSi

Tac6: y ^[(x-l){x + 4)].[{x-2)(x + 5)] ^(x^ +3x-4Xx^ +3x-\0) 
Dat t^x^ +3x-4^ x + -3

2

\2

_25 ^ _ 2 5

4 ~ 4

thitadifdc: = r ( r - 6 ) = - 6 r ^y = {t-3f-9

Khi t = 3^x^+3x-4 = 3 ^x^ + 3 x - 7 = 0 jc = i( - 3 ± 73?)

Vay khi x = - ( - 3 ± thi y dat gia tri nho nhat min y = - 9.

4.11: Cho hai so' thiTc x, y thoa dieu kien x^ =1. Tim gia tri
Idn nhat va gid tri nho nha't ciia x + y.

Giii

Ta c6 (x + yf + (x - yf > (x + yf

^ 2(x^ +y^)>(x + yf (Dau " = " xay ra <^ x = y )

x

'+Z

=1

Do do (x + yf <2^-y/2<x + y < ^ 
• Y6ix + y<^j2

x = y 
Dau " = " xay ra ^

Vay X + y dat gia tri Idn nhat la

x = y = .

x-^y = ^

2 Vdi

x + y

>-42

Dau " = " xay ra x = y 
Vay X + y dat gia tri nho nhat la

<^ x =

y^-72

p d 4.12: Tim gia tri Idn nha't cua y = \x\.yj4-x^ .

GiSi

TAC6: y^ \x\.M-x^ = 4x^.^4 - x^ = y]x\4-x^)
Xac dinh vdi -2 < x < 2.

Ta CO x^ +(4-x^)> 2^x\4-x)- (BDT Cosi)

. ^ 4

> 27

]^^(4^^

<^ 2 > 7 7

( 4 ^ ^ ^2>y

D a u " = " x a y r a ^x~^4-x^ ^ x" = 2^ x = ± ^ . 
Vay max y = 2 dat du-dc khi ;c = ±sl2.

DC- 4.13: Cho y = ^Jx^ +x + l + yjx^ -x + \ Tim x de y dat gia tri
nho nha't. Xac dinh min y.

V i x^+x + l =

x' -x + l^

\2

n

X

2 Giii

4

+ — > 0 n6n y xac dinh vdi moi x.

4

Dung ba't dang thiJc Cosi cho hai so'khong am a, b:
a + b>2s/ab

Taco:

>2^{x'+lf-x'

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D d 4.14: Cho y = — . T i m x de y dat gia t r i Idn nhat. Xac dinh

X + 1
max y.

Gi§i

Ta biet rang + > lab nen suy ra: / + 1 = {x'^f + 1 ' > lx\

Ix" 
^ 1 >

X e t 4 ^ = 1 ^X' - 2A: ' + 1 = 0

1
Do do k h i x = ±.\i max y = —.

2

4.15: Cho X > 0 va y = x^^ - + 6 . T i m gia t r i nho nhat ciia y .

Gi^i

T a c d : y = ( x ' ' - ) - 4 ( x ^ - l ) + 2

= x ' ( x " " - 1 ) - 4 ( x ' - 1 ) + 2 = x ' [ ( x ' ) ' - 1 ] - 4 ( x ' - 1 ) + :

= x ' ( x ' - l ) [ ( x ^ ) ' + ( x ' ) ' + x ' + l ] - 4 ( x ^ - l ) + 2 ^

= ( x ^ - l ) ( x ^ + x ' ^ + x ' ° + x = - 4 ) + 2.

V d i x > l t h i c d x' > l , x ' ° > l , x " > l , x ' ° > 1

nen ( x ' - l ) ( x ' ° + x=' + x ' ° + x ' - 4) > 0 do d6 ta c6 y > 2

V d i 0 < X < 1 t h i x ' < l , x ' ° < l , x " < l . x " " < 1,
nen ( x ' - l ) ( x ' ° + x " + x'° + x ' - 4) > 0

Do dd ta cd y > 2. V a y m i n y = 2 iTng v d i x = 1.

D e 4.16: T i m gia t r i nho nhat cua >' = x^* - 5 x ' + 9 .

T a c d : ); = ( x ' ° - x ' ) - 4 ( x ' - - l ) + 5 = x ^ ( x ' ^ - l ) - 4 ( x ^ - l ) + 5

= x' \{x')' - 1 ] - 4 ( x ' - 1 ) + 5 = ( x ' - l ) ( x " ^ + x ' ' + x ' + x ' - 4 ) + 5

V d i |x| > 1 thi cd x " > x " > x ' > x ' > 1.

=^ x ' - 1 > 0 va x'^ + x " + x ' + x ' - 4 > 0 nen y > 5

V d i |x| < 1 thi cd x'^ < x'^ < x ' < x ' < 1.

x ' - 1 < 0 va x " + x " + x ' + x ' - 4 < 0 nen y > 5.

Do dd m i n y = 5 k h i x = 1.

Gi§i

V i X + y = 1 nen:

T a c d : M = 1 - 1 1

-x ^ /

( X + l ) ( x - X)iy ^\\y-1) (X + \){-y){y + l ) ( - x )

x ^ / x ^ /

(x + l)()^ + l ) _ x y + x + )^4-l _ ^ ^ 2

x>' xy xry

• T a c d : ( x - y ) ^ > 0 ' > 4x3;

\

^ (-^ + yf > ^xy ma X + y = 1.
1

Do d d : — > 4 . V a y A > 1 + 2.4 = 9 .
xy

Dau " = " xay ra X = y = ^ . V a y gia t r i nho nhat cua M la 9.

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Dat diTdc khi: x = y = —.
2

Dd 4.18: Cho x, y > O v a x + y < l . Tim gia tri nho nhat cua bleu thitc:

A =

+ xy -Axy.

Gi^i

Ap dung bat dang thiJc Co si cho hai so' du'cfng a va b:
a-\-b

(a + fef , 1 1 4
a b a + b

va

cib [a + b)

Ap d u n g ( l ) , (2), (3) ta c6:

(1)

(2)

(3)

A = 1 1

x' + r Ixy)

+

'^xy-\-4xy

5 1

+ -.

^—2 T — ^ + 2,

x^+y^+2xy V 
4

4xy. h—. 
4 xy

Vay : A > 1 1

4xy 4 (x + y)

> 4 + 2 + 5 = l l .

Dau " = " xay ra ^ X = y - - . Vay A dat gia tri nho nhat la 11.

De 4.19: Tim gia tri Idn nha't ciia cac bieu thitc sau:
1) M = 2 x y - x ' - 4 y ' + 2 x + 1 0 y - 3 .

2) A^ = x y - x ^ - y ^ + 2 x + 2y.

76

3) P = — ( y z. V x -1 + xzsjy -2+ xy.y/z - 3"

xyz ^ '

G i i i

1) Tacd M = - 3 - ( x ^ 2xy + 4 y ^ - 2 x - 1 0 y )

= - 3 - [ ( x - y - l ) ' + 3 y ' - 1 2 y - l

= - 3 - [ ( x - y - l ) ^ + 3 ( y ' - 4 y + 4 ) - 1 3

( x - y - l f + 3 ( y - 2 r

= 1 0

-Xet ; c - y - l = 0
y - 2 = 0

X = 3 >

y = 2

Vay khi X = 3, y = 2 thi M dat gia tri Idn nhat la max M = 10.
2) Tac6A^ = - ( x ^ - x y - 2 x + y ^ - 2 y )

\

+ | y ^ - 3 y - l

x-^—X

\2

= 4 - x - l - l

+ 7 ( y ' - 4 y + 4 ) - 4
4

Xet 2 ^ x = 2

y = 2

y - 2 = 0

Vay khi X = y = 2 thi N dat gia tri Idn nha't la max N = 4.
3) Dieu kien xac dinh la: x > l , y > 2,z > 3 .

x y z

Dung ba't d i n g thiJc: a + b> 2yfab thi ta c6:

X = l+(X1)> 2y/x \ ^ <

-X 2

(1)

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y^2 + (y-l)>2^2(y-2)^^l^-^<~

I

__2

1 • =:3 + ( z - 3 ) > 2 V 3 ( z - 3 ) ^ ^ ^ ^ < ^

(2)

(3)

Khi X = 2, y = 4, z = 6 thi (1), (2), (3) trd thanh dang thiJc. Do d6

ta diTdc:

max P = — +

2 2^/2 2 ^ 3 '

4.20: Cho ba so' khong am x, y, z thoa man:

2x + y + 2z = 5 (1)

3x + 2 > ' - 2 z - 4 (2)

Hay tim gia tri Idn nhat va nho nha't cua M = 4x - 5y + 8z. •

Tir(l)tac6 :2x + y = 5 - 2 z

Tvlf(2)tac6 : 3x + 2y =:4 + 2z

Suy rax = 6 - 6 z v a y = lOz-7.

Vi x>Q^6-6z>0^z<\.

10 Talaico M = 4(6-6z)-5(10z-7) + 8z =69-66z.

1 44 1

nen maxM = 69-66.—= — itng vdi mmz = —.

min M = 69 - 66 = 3 tfiig vdi max z = 1.

78 CHlJOfNGV: P H I / O N G

TRi]NfH

§1.

PHaONGTRlNH BAG HAI - PHl/dNGTRlNH BAG BA

DINH LY VIET

I. PHl/dNG TRINH BAC HAI ax^+bx + c = 0 {a^O)

Lap biet thitc: A = 6^ - 4ac.

1) Neu A < 0 thi phifcfng trinh v6 nghiem.

2) Neu A = 0 thi phiTdng trinh cd nghiem kdp:

X,

=x^= —

la

3) Nd'u A > 0 thi phifcfng trinh cd hai nghiem phan biet:

_ -^-VA

_ - Z P

+

7 A

la

Trirdng hdp dac biet:

• Neu: a + + c = 0 thi phu^dng trinh cd nghiem :

— 1 - £ 
CI

• N6u: a - ^ + c = 0 thi phu^dng trinh cd nghiem :

x^-=^-{,x^^~-.

a

Ghi chu: 1) Co the dung cong Ihitc nghiem thu gon de giai

phUcfn I tilnh bac hai.

2) N6u phu-dng trinh: ax^ + fejc + c = 0 cd hai nghiem

x^;x^ {x^ < j:2)thi:

a > 0 | : * ax^+bx + c<0<^ x^ <x<x^.

X < X.

* ax'' +bx + c>0^

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a<0 * ax +bx + c<0<!^

* ax^ + bx + c>0'!F> X, <x<x^.

JC < JCj

x> x^

I I . D J N H L Y V I E T

1) Binh ly thuan: Neu phifdng trinh bac hai:

ax'^ +bx + c = 0 ( a ^ 0) CO hai nghiem x^,x^. T h i :

* S =

x,+x^=--a * P = x^.x^ = —. a

2) Binh ly dao: Neu x + y = S va x.y = P thi x, y la nghiem
cua phu-dng trinh: -SX + P = 0

I I I . P H l T d N G T R I N H B A C B A ax' + bx^ +cx + d = 0 (a ^ 0)
Cach giai: 1) Nham nghiem x^ = a .

Dac biet: * Neu: a + b + c + d = 0 thi phu'cfng trinh c6 nghiem

•.x^=l.

* Neu: a - b + c - d = 0 thi phu'cfng trinh cd
nghiem : X o = — 1.

2) Chia ax' + bx' +cx + d cho x — adi du'a 
phrfdng trinh ye dang:

(x-a)(Ax'' +Bx + C) = 0 • (*) 
x-a^O

Ax^ + Bx + C = 0'

T a c d : ( * ) ^

iS.l: Cho phu'cfng trinh: x^ -2(m - 2)x - 2 m - 5 = 0.

1) ChiJng minh r^ng phu'cfng trinh cd hai nghiem phan biet v d i
moi m.

2) Goi x^,X2 la nghiem cua phu'cfng trinh. T i m gia tri m sac 
cho: xf +xl = 1 8 .

1) T a c d : A ' = Z?''-ac = ( m - 2 ) ^ - ( - 2 m - 5 )
= m ^ - 2 m + 9 - ( m - l ) ' + 8
=!^A'>OVm.

Vay phufdng trinh da cho ludn cd hai nghiem phan bi6t v d i moi m.

2) Theo dinh ly Viet ta cd:

x^+X2= = 2(m - 2)

a

X , .x^ = — — —2m — 5.

a

Mat khac theo gia thiet:

xf+xl=lS<:^(x^+x^f-2x^x^=lS

^ [2(m - 2)f - 2 ( - 2 m - 5) = 18 <^ 4 m ' - 16m + 1 6 + 4m + 1 0 = 18 
<^ 4 m ' - 1 2 m + 8 = 0 <!=J> m ^ - 3 m + 2 = 0

m = 1V m = 2 (pt cd dang a + b + c = 0).
Vay khi m = 1 hoac m = 2 thi xf + A:^ = 18.

Bi 5.2: Cho ph^dng trinh bac hai: x^ - 2mjc + 2m - 1 = 0 (1)

1) Dinh m de phu'cfng trinh (1) cd nghiem kdp.

2) V d i gia tri nao cua m thi phu'cfng trinh (1) cd hai nghiem cung
dau. K h i dd hai nghiem mang dSu gi ?

Gi§i

1) T a c d : A ' = ^ ' ' - a c = m ' - ( 2 m - l ) = m ' - 2 m + l = ( m - l ) '

Phu'cfng trinh (1) cd nghiem kep <i=^ A ' = 0

< ^ ( m - l f = 0 < ^ m = l .
2) PhiTdng trinh (1) cd dang dac biet:

a + b + c = l - 2 m + 2 m - l = 0
nen cd nghiem: = 1

</div>
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— — = 2m

—

1. a

• Phifcfng tnnh (1) CO hai nghiem cung dau

X j j T j > 0 <^ 1 .(2m -1) > 0.

2m > 1 <^ m > -2

y

• Khi do: 5 = Xj + jc^ = = 2m > 0 nen J:, ^"^^8 difdng.

Tdm lai: Khi > ~ '^hi phu'dng trinh (1) cd hai nghiem cung

dau va khi dd hai nghiem ciing du'dng.

Ghi chu; Cd the nhan xet: x^ va x^ cung dau va = 1 > 0

n6n x^>0.

5.3: Cho phufdng trinh cd an x: - 2(m - - 3 - m = 0.

1) Chitng to phufdng trinh cd nghiem so* vdi moi m.

2) Tim m sao cho nghiem so' x^,x^ ciia phtfdng trinh thda dieu

kien: xl + x] >10.

1) Tacd: A ' = (m-1)^+3 + m m ^ - 2 m + 1+ 3 + m

= m-m-\-A = m-m^

1 .

4 4

1 m

2 ' 15

+ — > 0 Vm.

4 Vay phrfdng trinh cd hai nghiem phan biet x^,x^ vdi moi m.

x^ -\-x^ =2(m-l)

x^x^ = —3 — m

2) Theo he thiJc Viet ta cd:

Matkhacdo: xf +xl > 10

<^ (JC, + ^2)^

-2x^x^ >10

82 <^[2(m-l)f-2(-3-m)>10<^4(m'-2m + l) + 6 + 2m>10

<^4m^ -8m + 4 + 6 + 2m >10<^4m^-6m>0

m < 0

4^

m

" 2

ay khi: m < 0 hoac m > - thi xf + x\ 10.

De 5.4: Giai phufdng trinh : - (2m + 1)J: + m^ + m - 6 = 0.

1) Dinh m de phu'dng trinh cd hai nghiem deu Sm.

2) Dinh m de phufdng tiinh cd hai nghiem x^,x^ thda x^,-xl = 5C

Giii

Tacd: A = (2m + l ) ' - 4 ( m ' + m - 6 )

= 4m^+4m + l^-4m^-4m + 24 = 25

Phu'dng trinh cd hai nghiem phan biet:

2m+ 1-5 ' 2m+ 1 + 5 , .

X, =

=:m-2

;A:,

= = m + 3

' 2 ' 2

1) Phu'dng trinh cd hai nghiem deu am

4 ^

•ill

x,<0

m - 2 < 0 m <2

•^2 < o

m + 3 < 0 m < - 3

2) Tacd:

= 50<^ ( m - 2 ) ' - ( m +

3y

= 50

m' - 6m' +12m - 8 - m' - 9m' - 27m - 27 = 50

-15m?-15m-35 = 50

<!=^ 3m'+3m + 7 =10.

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3m^+3m + 7 = 10
3 m ' + 3 m + 7 = - 1 0

3/n^ + 3 m - 3 = 0 (1)
3 m ' + 3 m + 17 = 0 (2)
• Giai (1) : 3m' + 3 m - 3 = 0 < = > m ' + m - l = 0

A = l + 4 = 5

. - I + VS - l - ^ / 5
2 ' ' 2

^2

• Giai.(2) : 3 m / + 3 m + 17 = 0
A = 9 - 2 0 4 = - 1 9 5 < 0
Phifdng trinh v6 nghiem.

- l + ^/5 , „ -\-yl5 , . , ^_ 
Vay m = ^ hoac m = ^ thi phifdng trmh da

cho C O hai nghiem x^,x^ th6a man x^^ -x] = 50.

5.5: Cho phifdng trinh: x^ - 2(m + 1)A: + m ' - 4m + 5 = 0 (c6 an

so l a x ) . '

1) Dinh m de phifdng trinh c6 nghiem.

2) Goi x^,x^ la hai nghiem neu c6 cua phifdng trinh. Tinh

xf + ^ 2 theo m.

1) T a c o : A ' = 6 ' ' - a c = (m + l ) ' - ( m ' - 4 m + 5)
= + 2m + 1 - m ' + 4m - 5 = 6m - 4.

2
Phifdng trinh da cho cd nghiem 4 ^ A ' > 0 < ^ 6 m - 4 > 0 < ! ^ m > - .

Vky khi m > J thi phifdng trinh da cho cd nghiem.

2) Theo he thiJc Viet ta cd: S = x^+x^= 2(m + 1) = 2m + 2

P = x^x^ = m ' —4m + 5.

Do do: xf +xl =(x^ + x^f -Ix^x^^ (2m + 2f-2{m^-4m+ 5) 
I = 4 m ' + 8 m + 4 - 2 m ' + 8m - 1 0 = 2 m ' + 1 6 m - 6

Vay: A : , ' + 4 2 m ' + 1 6 m - 6.

Bi 5.6: Cho phifdng trinh: x ' - 2(m + l)x + m ' + 2m - 3 = 0 (1)
1) Chitng minh phifdng trinh (1) lu6n cd hai nghiem phan biet

vdi moi m. T i m hai nghiem dd.
2) T i m m de phifdng trinh (1)

a) Cd hai nghiem trai dau.
b) Cd hai nghiem deii difdng.
c) Cd hai nghiem deii am.

G i ^ i

1) T a c d : A ' = Z?''-af = (m + l ) ' - ( m ' + 2 m - 3 )

= m ' + 2 m + l - m ' - 2 m + 3 = 4.

V i A ' = 4 > OVm n^n phifdng trinh da cho lu6n cd hai nghiem phan
biet.

K h i d d : X =

^ 2 = •

= m + l - 2 = m - l

= m + l + 2 = m + 3.

) a) R6 rang: x^ < x^ nen phifdng trinh (1) cd hai nghiem trai 
dau.

m . - l < 0

^ x^ <0 < x^ 4^

m < 1

x,<0

X , > 0 m + 3 > 0
- 3 < m < 1 .

m > - 3

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(44)<div class='page_container' data-page=44>

X , > 0
X , > 0

m - 1 > 0 m > 1

m > - 3 m > 1

[m + 3 > 0

V a y k h i : m > 1 thi phifdng trinh (1) c6 hai nghiem deu difcfng.
c) PhiTcfng trinh (1) c6 hai nghiem deu am

x,<0 
x^<0

m - K O
m + 3 < 0

m < 1

m < - 3 <^ m < - 3
V a y k h i : m < - 3 thi phu-dng trinh (1) c6 hai nghiem deu am.

5.7: Cho phtfdng trinh : - (2m - 3)x + - 3m = 0 .

1) ChiJng minh rang phu-dng trinh luon luon c6 hai nghiem k h i m
thay d d i .

2) D i n h m de phiTdng tiinh c6 hai nghiem x^,X2 thoa:

\<x^<x^<6.

GiSi

1) Ta c6 : A = (2m - 3 ) ' - 4 ( m . ' - 3m)

= 4 m ' - 1 2 m + 9 - 4 m ^ + 1 2 m = 9 .

A = 9 > 0 Vm . V a y phtfdng trinh l u o n l u o n c6 hai nghiem phan
biet k h i m thay d d i .

2) V A = 3. H a i nghiem cua phu'dng trinh la:
2 m - 3 - 3

= m — 3;x. =

2 m - 3 + 3
= m .

2 ' 2

R6 rang < x^: N e n : \ x, < x^ <6 ^ Km-3 <m <6 
K m - 3

m<6 
<^ 4 < m < 6.

4 < m
m < 6

5.8: Cho phu'dng trinh: - (2m - 5)x + m^ - 5m = 0.

1) Chi?ng minh rang phu'dng trinh luon c6 hai nghiem phan biet
k h i m thay d d i .

2) D i n h m de phufdng trinh c6 hai nghiem thoa:
a) D e u am

b) D e u diMng
c) T r a i dau
d) < 2 < 6 < ^ 2

e) x^<2<x^<6

Gi§i

1) T a c o : A - Z ? ' - 4 a c = ( 2 m - 5 f - 4 ( m ' - 5 m )
= 4m^ - 20m + 25 - 4m^ + 20m = 25.

V i A > OVm nen phu'dng trinh lu6n c6 hai nghidm phan biet:
-b-sfA 2 m - 5 - 5

K h i 66 : X, =•

la = m - 5

x^ = -b + y[A 2 m - 5 + 5
2a 2 
2) a) Phufdng trinh cd hai nghiem deu a m

b) PhiTdng trinh c6 hai nghiem deu dtfdng

• = m .

X < 0 m - 5 < 0
< 0 m. < 0

m < 5

m < 0 <^ m < 0.

x,>Q m-5>0 
x^>0 m > 0

m > 5

m > 0 <^ m > 5.
| c ) R6 rang: x^ < x^ nen: Phu'dng trinh c6 hai nghiem trdi d ^ .

m < 5

X. < 0 m - 5 < 0

< ^

x^>0 m > 0 m > 0

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X, <2 m - 5 < 2

6 < ^2 6<m

m <1

6 < m < 7.
6 < m.

e) Phifdng trinh c6 hai nghiem X j, thoa: x^ <2<x^ < 6.
m - 5 < 2 < m < 6 m - 5 < 2

2 < m < 6

m < 7 t

2 < m < 6 4 ^ 2 < / n < 6 .
De 5.9: Cho phiTdng trinh: x^ - 3(m + l ) x + 2 m ' - 1 8 = 0 (c6 an x ) .

1) T i m m de phifdng trinh c6 hai nghiem phan biet deu am.
2) G o i x^,X2 la hai nghiem cua phtfdng trinh. T i m m de c6:

< 5 .
x^ X2

Gi^i

1) T a c 6 : A = & ' - 4 a c = 9(m + l ) ' - 4 ( 2 m ' - 1 8 )
= 9 m ' + 1 8 m + 9 - 8 m ' + 7 2

= m ' + 1 8 m + 81 = (m + 9 ) ' .

V i A = (m + 9 ) ' > 0 Vm nen phu'dng trinh luon c6 hai nghiem:
3(m + l ) - ( m + 9)

•^1 = • = m — 3

^2 = •

3(m + l ) + (m + 9)

= 2m + 6.

Phu'dng trinh c6 hai nghiem phan biet deu am
A > 0

X i < 0 <^

X^KO

(m + 9 ) ' > 0

m - 3 < 0 ^
2m + 6 < 0

m < 3 < ^ - 9 ^ m < - 3 .
m < - 3

D6 5.10: Cho phu'dng trinh: x^ - 2{m + l ) x + m ' - 4m + 5 = 0.
1) D i n h m de phu'dng trinh c6 nghiem.

2) Dinh m de phu'dng trinh c6 hai nghiem phan biet deu dtfdng.

Giii

1) T a c o : A ' - ( m + l ) ' - ( m ' - 4 m + 5) = 6 m - 4 .

Phu'dng trinh da cho c6 nghiem • ^ A ' > 0 < ^ 6 m - 4 > 0 < ^ m > —
~ 3 "
2

2) V d i dieu k i e n > - phu'dng trinh da cho cd hai nghiem:
X, = m + 1 - V 6 m - 4 ; = + 1 + V 6 m - 4 .

Phu'dng trinh da cho c6 hai nghiem phan biet deu dufdng

A ' > 0
x,>0 
^2 > 0

2
m >

-3

1 ^ 0

6m - 4 > 0

m + 1 - V6m - 4 > 0
m + 1 + V6m - 4 > 0

- ( m + 1 ) < V 6 m - 4 < (m + 1 )
2

m >
-3

m + l > 0 ^
\/6m - 4 <l m + 11

9

m >
-3

6 m - 4 < m ' + 2 m + l

m >
-3
m > - 1

0 < 6 m - 4 < m ' + 2m + l

2
m >

-3

m^ - 4m + 5 > 0
2

m >

-3 m > - . 2
( m - 2 ) ' + l > 0

V a y k h i m > - phu'dng trinh da cho cd hai nghiem phan biet
deu du'dng.

^^Qishu^ I) Cho A>0.T^ic6: X <A^-A<X < A. ,

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(46)<div class='page_container' data-page=46>

2) Bai toan tren c6 the giai nhiT sau:

• Phifdng trinh da cho c6 hai nghiem phan biet deu difcfng 
A > 0

jc, > 0

A , > 0

A > 0
A, .x^ > 0

A > 0
P > 0
S>0 
1 ^ 0

6m - 4 > 0
- 4m + 5 > 0
2(m + 1) > 0

2
m >

-3 2
m G <^ m > —.

3
m > - 1

Vay khi m > -J phi/dng trinh da cho c6 hai nghiem phan
biet deu dtfOng.

3) Bang each chiJng minh tufcfng tuf, ta cd ke't qua:

Cho phu'Ong trinh bac h&i: ax^ + bx + c =^ 0 (a ^ 0).

• Phu'Ong trinh c6 hai nghiem phan biet trai dau <^P<0. 
'A>0

PhtTcJng trinh c6 hai nghiem phan biet deu du'cfng <^

PhUdng trinh c6 hai nghiem phan biet deu a m <^

P > 0 .
5 > 0
A > 0
P > 0 .
S<0

Bi 5.11: Cho phu'Ong trinh: mx^ - 2(m + 2)A + m = 0 (c6 an x). 
D i n h m d e :

1) Phu'dng trinh cd nghiem.

2) Phu-dng trinh cd hai nghiem trai daii.
90

3) Phu'dng trinh cd hai nghiem phan biet deu am.
4) Phu'dng trinh cd hai nghiem phan biet deu diTdng.

GiSi

1) X e t hai trrfdng hdp:

• m = 0: phu'dng trinh ^ O.x^ - 4^ + 0 = 0.

<^ A — 0 : phu'dng trinh cd nghiem. 
• mr;^0: phu'dng trinh cd nghiem<^ A ' > 0

<»(m + 2 ) ^ - m . m > 0

<^4m + 4 > 0 - ^ m > - l (m^O). 
Tit hai tru'dng hdp tren, ta cd:

Phufdng trinh cd nghiem • o m > - 1 .

2) • Phu'dng trinh cd hai nghiem trai dau <^ P < 0
<^ 1 < 0 (v6 nghi6m).

Vay khong cd gia tri m nao thoa yen cau bai toan.
• Phu'dng trinh cd hai nghiem phan biet deu a m

A > 0
P > 0 ^
S<0

4m + 4 > 0
^ > 0
m . 
m + 2

< 0

wi > - 1

m + 2<0 vci m>0 
m + 2>0 va m<0 
m > - l

m > - 2 va m > 0 < ^ 
m > —2 va m < 0

m > - 1

me0

< ^ - l < m < 0
- 2 < m < 0

Vay khi: - 1 < m < 0 phu'dng trinh da cho cd hai nghiem phan biet
deu am.

4) • Phu'dng trinh cd hai nghiem phan biet deu du'dng

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A > 0
P>0 <^ 
S>0

4m + 4 > 0

^ > 0
m 
m + 2

> 0

m>-l

m + 2>0 va m>0 
m + 2<0 va m<0 
m

m>-\

m>-2 va m>0<^ 
m<-2 va m<0

m>-\

m > 0 <^m>0. 
m<-2

Vay k h i : m > 0 phifdng trinh da cho cd hai nghiem phan biet deu
diTcfng.

Bi 5.12: Cho phiTdng trinh: (m +1)^ - 2(m + 2)^: + m - 3 = 0 (cd an
so l a x ) .

1) Dinh m d^ phiTcfng trinh cd nghiem.

2) Dinh m de phuTdng trinh cd hai nghiem x^,x^ thda:
(4x, +1)(4X2 +1) = 18.

Giii

1) * V d i m + l = 0 ^ m = - l

PhiTdng trinh ird thanh : -2x - 4 = 0>^2;c = -4<i^;c = - 2
Vay khi m = - 1 phiTdng trinh cd nghiem x = - 2.

• V d i m + 1 0 <^ m ^ - 1
A ' = (m + 2 ) ' - ( m + l ) ( m - 3 )

= m^ + 4m + 4 - w ^ + 3 m - m + 3 = 6m + 7.
Phirdng trinh cd nghiem 4=^ A ' > 0

^6m + l>0^6m>-l ^m> - ^ ( m ^ - l ) .
K^'t luan: PhiTcfng trinh cd nghiem <=> w > - - .

92

2) D i e u k i e n O T > —- m ^ - l . 
6

Theo dinhly Viet ta cd:

5 = + = 2(m + 2)

P = JC, V =

m + 1
m - 3

Do dd: (4^1 + l)(4;c, + 1) = 18 •'-> \6x,x^ + 4;Cj + 4x, + 1 = 18

«!=J> 16 A:, A:^ + 4(x, + J C j) - 1 7 = 0

1 6 ( m - 3 ) , 8(m + 2)

17 = 0

m + 1 m + 1 •
16(m - 3) + 8(m + 2) - 1 7 ( m +1) = 0

<^ 16m - 48 + 8m +16 - 17m - 1 7 = 0
7m - 49 = 0 7m = 49 <i=^ m = 7.

m = 7 thoa dieu kien

m —1

m > - Z . 
~ 6

Vay khi m = 7 thi phiidng trinh cd hai nghiem x^,x^ thoa :
(4x, +1)(4X2 +1) = 18.

1)6 5.13: Cho phu-dng trinh: x^ - 2(m - l)x + 2m - 4 = 0
^ (cd an so' la x).

p 1) Chitng minh rang phiTdng trinh cd hai nghiem phan biet.

^ I l ^ 2) Goi x^,x^ la hai nghiem cua phifdng trinh. Tim gia tri nho
nhatcua y = x^ + x^.

Giii

1) Tacd : A ' = ( m - 1 ) ' - ( 2 m - 4 ) = m ' - 2 m + l - 2 m + 4
= m ^ - 4 m + 4 + l = ( m - 2 ) ^ + l > 0 .

Vay phu-cfng trinh cd hai nghiem phan biet vdi moi m.

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2) Theo dinh ly Viet ta c6: x ^ + x ^ = 2{m - 1) = 2m - 2

X j = 2m — 4

y + xl = (jc, + ;c, )^ - 2XjX, = (2m - 2)' - 2(2m - 4)
= 4 m ^ - 8 m + 4 - 4 m + 8 = 4m^-12m + 12

= ( 2 m ) ' - 2.2m.3 + 3 ' + 3 = (2m - 3 / + 3 > 3

Dau " = " xay ra ^ 2m - 3 = 0 4=^ 2m - 3 ^ m = - .
2 
3

Vay min y = 3 dat difdc khi m = —.

5.14: Cho phufdng trinh - (2m + 3)x + m - 3 = 0 (cd an la x).
1) ChiJng to rang phifOng trinh luon c6 nghiem.

2) Goi la cac nghiem cua phu'dng trinh tren. Tim m de

-C[ X2 dat gia tri nho nha't. Tinh gia tri nho nha't ay.

Gi^i

1) A = (2m + 3)' - 4 ( m - 3 ) = 4m- +12m + 9 - 4 m + 12.
= 4 m ' + 8m + 4 + 17 = ( 2 m + 2 ) ' + 1 7 > 0 .

Do do phu'dng trinh c6 hai nghiem phan biet vdi moi m.
\x, + x^ =2m + 3

2) Theo dinh ly Viet ta cd •

[x^X2 = m — 3.

Mat khac ta cd: (x^ - x^ f = (;c, -\- x^f -Ax^ x^

= (2m + 3)' - 4(m - 3) = (2m + 2)" +17 > 17

Do dd \x, - X 2 \ yl(2m + 2f +17 > Vl7 .

Dau "=" xay ra <^ 2m + 2 = 0 <s=^ 2m = - 2 <^ m = - 1 .

Vay min x^-x^ = Vl7 dat du'dc khi m = - 1 .

5.15: Cho phufdng trinh x^ - 2mx + 2m - 1 = 0 vdi m la tham s6'.

94

1) ChiJng minh phu'dng trinh luon c6 nghiem.

2) Tinh bieu thitc y=— 3 5 theo m (vdi x,,x, 
x; +xl+2(\ x^x2) . " ^ 
la nghiem cua phu'dng trinh).

3) Tim gia tri Idn nhat va nho nhat cila y.

1) Tacd A ' = m ' - ( 2 m - l ) = ( m - l ) ' > 0 vdi moi m nen phifdng
trinh luon cd nghiem.

2) T a c d : S = x^ + x^ =2m,P = x^x^ =2m-\ 
D o d d : y = ^ ^ • ^ ^ + ^ ^^x^x^ + 3_

x^ +xl+2x,X2+2 (x^+X2f+2

_ 2(2m - 1 ) + 3 _ 4m + 1
4 m ' + 2 ~ 4 m , ' + 2
4 m + 1

3) Tu^ y = suy ra:
4 m ' + 2

4 y m ' + 2 y = 4m + l < ^ 4 > ' m ' - 4 m + 2 y - l = 0 (*)
Ta dijng dieu kien cd nghiem ci\ phu'dng trinh de tim min y va
max y

• N e u y ^ O : (*) <i=^-4m - 1 = 0 <^ m = - - ( p t (*) cd
4

nghiem).

• Neu y^0:{*)c6 nghiem khi

A ' = 4 - 4 y ( 2 y - l ) > 0 ^ 4 y ' - y - l < 0

. ^ - | < y < i ( y - ^ 0 )

Tiir hai tru'dng hdp tren ta cd: pt (*) cd nghiem

V a y m i n > ' = - — va max>' = l .

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5.16: Cho phiTdng trinh : - 2mx r = 0 •

m

1) T i m m de phifcfng trinh CO nghiem.

2) Dat A = x," + xl vdi x^,x^ la nghiem cua phiTdng trinh neu 
tren. T i m gia tri nho nhat ciia A.

Gidi

1) Ta cd A ' = + — > 0 vdi moi m ^ 0 n^n phifdng trinh cd

m

nghiem khi m 0 .

4
2) Theo dinh ly Viet, ta cd: x^+x^= 2m,x^x^ = ^

J m

A^ixlf+{xlf={x]^xlf-2xlxl

{x, + x,f-Ax,x, -l{x,xS = f 1

2 
2

- 2 4 ] 2

4 m ' + • + ^ = 16

m m

+ •

32
m
= 16 4 I 32

m ) m m ^ + - ^ + 4 m

Mat khac theo BDT Cosi ta cd:

m \ m

Dau " = " xay ra khi = ^ = 6 ^ m = ± ^ 6 .
m

Va luc do A dat gia tri nhd nhat la: min A = 16(2V6 + 4) = 32(V6 + 2).

D(! 5.17: Cho phufdng trinh : x ' - 2(m + 1 ) + 2m = 0. (1)
1) ChiJng to rang phu-dng trinh (1) luon lu6n cd hai nghiem phan

biet vdi moi gia tri cua m.
(

96

2) Goi x^,x^ la hai nghiem ciia phUdng trinh (1). Chifng to rang 
gia tri ciia bieu thitc x,-]r x^-x^.x^ kh6ng phu thudc gid tri 
ci5a m.

1) Ta cd: A ' = ^•' - = [-(m +

l)f -

2m
= m ' + 2 m + l - 2 m = m H l .
V i m ' > 0 n e n m ' + l > O . S u y r a A ' > 0 .

Vay phiTdng trinh : x ' - 2 ( m + + 2m = 0 luon luon cd hai
nghiem phan biet vdi moi gid tri cua m.

2) Phu-dng trinh : J : ^ - 2 ( m + l ) j : + 2m = 0 luon luon cd hai 
nghiem phan biet nen theo dinh ly Viet, ta cd:

x^=2{m-\-\) va x^.x^—2m.

T a c d X, +x^ -x^.x^ = 2 ( m + l ) - 2 m = 2m + 2 - 2 m = 2

Do dd, bieu thiJc x^ x^ -x^.x^ khong phu thuoc vao gia tri 
ciia m.

5.18: Cho phtfdng trinh: (m + \)x^ - 2{m-\)x + m-2 = Q.

1) Xac dinh m de phufdng trinh cd hai nghiem phan biet.

2) Xac dinh m de phtfdng trinh cd mot nghiem bang 2 va tinh
nghiem kia.

3) Xac dinh m de ph^dng trinh co hai nghigm phSn bid! x^,x^ 
thoa man he thitc

x^ x^

Gi§i

#. Cho phutfng trinh (m + \)x'^ - 2(m - + m - 2 = 0 .

1) Phu-dng trinh : ( m + - 2 ( m - 1)A:+ m - 2 = Ocd hai nghiem 
phan biet khi va chi khi

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m + 1 ^ 0 
A ' > 0

m ^ —I

- ( m - l ) f -(OT + l ) ( m - 2 ) > 0

OT ^ — 1

m - 2 m + l - m ^ + 2 w - m + 2 > 0 
- m > - 3

m ^ —1

m < 3

Vay k h i m < 3 va m ^ - 1 thi phiTcfng trinh trSn c6 hai nghiemphan
biet.

2) Phtfdng trinh (m +1)^:^ -l{m-X)x + m-l = 0c6 mot nghiem 
bkng 2 k h i chi k h i

(^4-1)22 - 2 ( m - l ) 2 + m - 2 = 0 ^ 4 m + 4 - 4 m + 4 + m - 2 = 0

Vay k h i m = - 6 thl phifdng trinh (m + \)x^ - 2(m - l)x + m - 2 = 0 
c6 mot nghiem bkng 2.

Khi m = - 6 phiTdng trinh (m + - 2(m - ^ m - 2 = 0 c6 hai
nghiem phan biet x^,x^ nen theo dinh Viet ta c6:

2 ( m - l ) 2 ( - 6 - l ) _ 1 4
- 6 + 1 5
m + 1

14 14

Vay nghiem con lai cua phiTdng trinh tren \k:x^= —.

3) Ta x6t trirdng hdp phifdng trinh c6 hai nghiem phan biet x^,x^ 
tiJclk: m < 3 va m ^ - 1 .

2 ( m - I ) . m - 2
Theo dinh ly Viet ta c6 x.+x^= va x, .x^ = .

m + 1 /n + 1

Phifdng trinh (m + l ) ; c ^ - 2 ( m - l ) x + m - 2 = 0c6 hai nghiem
x^,x^ th6a man he thiJc

^ 2 ( m - l ) _ m-2 ^ 7 ^ 2 ( m - l ) _ 7 
m + 1 m + 1 4 m - 2 ~ 4

m vt2

OT = —6
m:;^2

8 m - 8 = 7 m - 1 4
4=^ /n = - 6 .

Do do khi m = - 6 thi phUdng trinh

(m + l ) j r ' - 2 ( m - l);c + m - 2 = 0c6 hai nghiem phan biet x^,x^ 
thoa man he thitc — + — =: — .

D d 5 . 1 9 : Cho phifdng trinh : - 2 ( m - l)x + m - 3 = 0 .

1) Chitng minh rang, vdi moi gia tri ciia m phiTdng trinh luon c6
hai nghiem phan biet.

2) Goi x^,x^ la hai nghiem ci5a phUdng trinh da cho. T i m he 
thiJc lien he giffa x^,x^ doc lap do'i vdi m.

Gi^i

1) Ta CO A ' = b''~ ac

- ( m - l ) f - ( m - 3 ) = m ' - 2 m + l - m + 3 
= m^ - 3 m + 4 = m

Do dd, phu-dng trinh x^ - 2 ( m - l ) x + m - 3 = 0 luon luon cd hai 
nghiem phan biet v d i moi gia tri ciia m.

2) PhUdng trinh - 2 ( m - l ) x + m - 3 = 01u6n luon cd hai
nghiem phan biet x^,x^ nen theo dinh ly Viet ta c6

x^+x^=^2{m-\) 
x^x^ = m — 3

(1)
(2)

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Tu" ( 2 ) t a c 6 : m = X i X 2 + 3 . T h a y v a o ( l ) :

x,+x^^ lx,x^ + 4 =^ lx,x^ -x,-x^+A = Q.

He thiJc lien he giffa

x,,x^

doc lap doi vdi m la :

l x ^ x ^ - x ^ - x ^ ^ A = Q.

5.20: PhiTcfng trinh : 2J:^ - 6x + m = 0.

1) Vdi gia tri nao cua m thi phu-dng trinh c6 hai nghiem deu

du'dng?

2) Vdi gia tri nio cua m thi phtfdng trinh c6 hai nghiem

x^,x^

sao cho :

+ — = 3

x^ X, . •

GiSi

1) Phu-dng trinh

2x^ -6x^m = Q c6

nghiem

<^ A ' > 0 < ^ ( - 3 ) ' - 2 m > 0 < ^ 9 - 2 m > 0

9

2 Khi phu-dng trinh tr^n cd hai nghiem x, va

x,,

theo dinh ly Viet ta c6 :

X,

= 3 va

x.Xj = - J .

Phifdng trinh 2x^ - 6x + m = 0 c6 hai nghiem dufdng khi:

[ A ' > 0

5

= X, + Xj >

0 <s=^

P =

x,.X2

> 0

~2

3 > 0 4=>

^ > 0

2

9 m < - ^ ^ 9

~ 2 < ^ 0 < m < - .

m > 0 ^

Do dd khi 0 < m <

I

thi phtfdng trinh 2x^ - 6x + m = 0 c6 hai

nghiem dufdng.

100 2) Khi w < - .

~ 2

Phu-dng trinh cd hai nghiem Xj .Xj thda ^ + ^ = 3.

Xj Xj

x f + x ^ ^ (x,

+ X 2 ) ' - 2 x , X 2

= 3

Xj

.x^

x,.x,

( 3 ) ' - m ,

^ — = 3 4=^

m

18 — 2m = 3m

m :;^0

18 9

l ^ ^ m = — ( t h 6 a d k m < - )

m = — 5 2

18 Do dd khi '^ = — ^hi phufdng trinh 2x^ - 6x + m = 0 cd hai

nghiemx, .x, thoa :

^ + ^ = 3. .

Xj Xj

DC' 5.21: Cho phu-dng trinh bac hai do'i vdi x :

(m + l ) x ' - 2 ( m - l ) x + m - 3 = 0 ( m ^ - 1 ) (1)

1) Chitng minh r^ng phifdng trinh (1) luon luon cd hai nghiem

phan biet vdi moi gia tri ciia m.

2) Goi

Xi,X2

la nghiem cua (1), tim m de

x^x^

> 0 va

Xj =

2x2

1) Tacd A' = [ - ( m - l ) f - ( m + l)(m-3)

= m^ - 2 m + l-(m.^ - 3 m + m - 3 )

= - 2m +1 - m^ + 2m + 3 = 4 > 0

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Vay phu-cfng trinh (m + l)x^ -2(m -l)x + m-3 = 0 luon luon c6 
hai nghiem phan bi6t vdi moi g i i tri cua m ^ - 1 .

2) Do phu-dng trinh (m + \)x^ - 2(m - l)x + m - 3 = 0 luon luon c6 
hai nghiem phan bidt v d i moi gia tri cila m ^ - 1 nen theo dinh
1^ Viet ta cd :

2 ( m - l ) . m - 3

x,+x.= — va X. .X, = . 
' ' m + 1 ' ' m + 1

Phu-cfng trinh (1) cd hai nghiem x^ ,x^ thoa man = 2x^ khi

2x^ +x^ = 2 ( m - l )

m + 1

2x^.x^ = m - 3

m + 1

•^2 = 2 ( m - l ) 3(m + l ) 
m —3
2(m + l )

(1)
(2)
Binh phtfdng (1) va so sanh vdi (2) ta diTdc:

4 ( m - 1 ) ^ ^ _ m - 3 _ ^ , _ 2 m + 1 ) = 9(m^ - 3 m + m - 3 )
9(m + l ) ' 2(m + l )

8 m ' - 16m + 8 = 9 m ' - 1 8 m - 27.

K h i m = 7 ta cd: x,.x^ =

- 2 m - 3 5 = 0
m - 3 7 - 3 I

mj = 7
= —5

K h i m = - 5 ta cd: x^.x^ =

' m + 1 7 + 1 2
m - 3 - 5 - 3

= - > 0 .
= 2 > 0 .
m + 1 - 5 + 1

Do dd khi m = 7 hoac m = - 5 thi phu-dng trinh (1) cd hai nghiem
thoa man x^ = 2x^ vk x^x^ > 0.

Bi 5 . 2 2 : Cho phu-dng trinh x^ -ax + a + \ 0. ChiJng minh r^ng 
n6'u •.a + b>2 thi it nhat mot trong hai phu-dng trinh sau day cd 
nghidm : x^ + 2ax + b = 0,x^ + 2bx + a = 0.

102

Giii

Phu-dng trinh : x^ + 2ax + b = 0 c6 biet so' A\=a^ -b. 
Phu-dng trinh -.x^ + 2bx + a = 0 cd biet so' A\=b^ -a. 
Suyra A' l + A ' j = a ' + Z J ' - ( a + Z?)

= ( a ' - 2 a + l) + ( ^ ' - 2 Z j + l) + (a + ^ - 2 )

= (a-lf +(b-lf +(a + b-2)>0

( V i (a-\f > 0 , ( ^ - 1 ) ' > 0 v a a + ^ > 2 (gt) nen a + b-2>0) 
=^ ft nhat mot trong hai biet so' A ' j , A cd mot so' khong am.

That vay ne'u ca hai am nghla la A ' , < 0 va A'^ < 0

A'l + A'^ < 0 (vd ly)

Vay it nhat mot trong hai phu-dng trinh da cho cd nghiem.

5.23: Chii-ng minh trong hai phufdng trinh sau cd it nhat mdt
phu-dng trinh cd nghiem:

ax^ +bx + c^O (1)

ax^+cx + b-c-a = 0 (2) (vdi a * 0,^ e/?,c G/?).

Giii

(1) c d b i e t s o l a A , = Z ? ' - 4 a c .

(2) cd biet so la A^ = c' - 4a(b -^c -a) = c^ - 4ab + 4 a ' + 4ac 
Tacd A , + A , =Z>' + c ' - 4 a i + 4 a '

= (b^-4ab + 4a^) + c^ = ( b - 2 a f > 0 .

V i A j + A j > 0 nen suy ra A , > 0 hay A j > 0.
Vay (1) cd nghiem hay (2) cd nghiem.

I>6' 5 . 2 4 : T i m k de hai phu-dng trinh sau cd nghiem chung:

x^+kx + \=0 (1)

x^+x + k = 0 (2)

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-Gidi

Goi la ng^^iem so' chung thi ta c6:

xl+kx,+l = 0 (1')

xl+x,+k = 0 (2')
Tru" ve'(1') cho (2') thi c6: (k-l)x,+ l-k = 0

^(k-l){x,--l)=^0

<^ k — I hay = ^

• V d i k = 1 thi (2) trd thanh + A; + 1 = 0 . Day Ik
phuTdng trinh v6 nghiem.

• V d i = 1: the vao (2') ta c6 k = - 2. Liic d6, ta cd:

(1) ^ x^ -2x + l = 0^ x = \

(2) ^ x''.+x-2 = 0^ x = l,x^-2.

Do do hai phifdng trinh cd nghiem chung la x = 1.

5.25: Cho a, b, c la do dai ba canh cua mot tam giac. QnJng minh
phu'dng tiinh sau la v6 nghiem: a^x^ + (a^ + — c^)x + = 0.

Gi§i

Taco A = (a'+ b'-cy-4a'b'= (a'+b'-c')^-{2abf 
= (a^+b^~c^-2db)ia^+b^-c^+2ab)

= \a-bf-c^].[{a + b)''-c^

= {a-b-c)ia-b + c)(a + b + c){a + b-c)

V i a

b<:a + c^a + c-b>G 
c<a+b^a+b-c>0

va a + b + c>0 nen A < 0 va do do phu'dng trinh tren v6 
nghiem-De 5.26: T i m m de 2x^ - 3x + 2m = 0 c6 mot nghiem khac 0 va 
gap ba Ian mot nghiem cua 2x^ — x + 2m — 0.

104

Gi^i

X6t 2x^-x + 2m = 0 (1)

2 x ' - 3 j c + 2m = 0 (2)
Goi XQ la nghiem cua (1) sao cho 3XQ la nghiem cua (2) thi ta c6:

2xl-x,+2m = 0 (1')

2 ( 3; C o/ - 3 ( 3 X o ) + 2m = 0 (2')

Trir ve (2') cho (1') thi: 16x1 -^x^ = 0 ^ x^ = 0,Xo

V i dieu kien cua de bai \& x^ ^0 nen ta chi xet -^o = ^

1 1 1

The = - vao ( l ' ) t a du-dc: + 2m = 0 ^ m = 0.

Liic d6 (I) ^2x^-x = 0^ x = 0,x = ^

(2) < ^ 2 x ^ - 3 x = 0 ^ x = 0 , x - |

nghiSm = ^ cua (2) la gap ba Ian nghiem = ^ cua (1)

Vay m = 0 la gia tri can tim.

b e 5.27: T i n a va b de hai phu'dng trinh sau la tu-dng du'dng

x''-(4a + b)x-6a = 0 (1) •

x^ -(2a + 3b)x-6 = 0 (2)

Giii

Neu (1) va (2) la tu-dng diTdng thi (1) va (2) c6 nghiem trung nhau

Do do 5 , = 5 , 4a + ^ = 2a + 3Zj
- 6 a = - 6
4^ a = \,h = \.

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L i l c ay, ta c6: (1) - 5 x - 6 = 0 <^ ;c =

V a y a = b = 1 la dap so.

-\,x = 6 
- \ , x ^ 6 .

D e 5 . 2 8 : Cho hai phiTdng trinh :

x^ -{2m-3)x + 6 = 0.

+ X + m - 5 = 0 (x la an, m la tham so').
T i m m de hai phifcfng trinh da cho c6 diing mot n g h i f m chung.

Gi§i

Gia su" XQ la nghiem chung cua hai phifdng trinh

Ta CO

xl - (2m - 3)Xo + 6 = 0 xl-(2m-3)x,+6 = 0 (1)

m = —2XQ — + 5

2x1 + ^ 0 +m — 5 — 0 
Thay m t i r ( 2 ) va (1), ta c6:

JCQ^ - [2(-2;Co^ - + 5) - 3] + 6 = 0

<^ xl - 4^ 0 + 2^ 0 - 7^ 0 + 6 = 0

^4x1+3x1-1x^+6 = 0

^ + 8^ 0 - 5^ 0 - I O X Q + 3^ 0 + 6 = 0

<^ 4xl(x, + 2) - {x, +2) + 3{x, + 2) = 0

^ix,+2)(4xl-5x^+3) = 0 
j C o+ 2 = 0

4x1-5x^+3 = 0

(2)

44>

X =—2

XQ — L

x^e0 ( v i A = 2 5 - 4 8 = - 2 3 < 0 )
V d i : x^ = -2 thay vao (2), ta dufdc:

m = - 2 ( - 2 f - ( - 2 ) + 5 = - l .

vay hai phtfcfng trinh c6 mot nghiem chung la -2 k h i m = - 1 .

106

^^fs^- G i a i cac phu-dng trinh bac ba sau :

\) +2x^-lx + A = 0. 2)x'-2x^-x + 2 = 0.

Gi§i

1) x' +2x^-lx + 4 = 0 (1)

De thay phu'dng trinh c6 dang: a + b + c + d = 0 nen c6 mot
nghiem: x = 1.

Chia: x^ + 2x^ - lx + 4 cho x - 1, ta du-cfc:

{\)^{x-\){x'' + 3A: - 4 ) = 0

U

= l

44-x - \ 0

; C 2+ 3J C- 4 = 0

x = \ x = \y x = -4.

x = - 4
2) x' -2x^ - x + 2 = 0 (2)

De thay phu'dng trinh c6 dang: a - b + c - d = 0 nen c6 nghiem: x =
Chia: x^ - 2x'^ - x + 2 = 0 cho x + 1, ta drfdc:

( 2 ) < ^ (A : + 1 ) (X '- 3 x + 2) = 0

U

= - i ,

- 1 .

x + \ 0

x^-3x + 2 = 0

x = \ 
x = 2.

5 . 3 0 : Cho so x^ = ^ 9 + 4^5 + ^ / 9- W 5 .

1) ChiJng to XQ la nghiem ciia phu'dng trinh x^ - 3x-\% = 0.

2) T i n h .

1) T a c d :

</div>
(55)<div class='page_container' data-page=55>

= 9 + 4 ^ / 5 + 3 ^ 9 + 4 7 5 . ^ 9 - 4 7 5 (V9 + 4V5 + ^ 9 - 4 7 ? '
+ 9 - 4 7 5 - 3 ( ^ 9 + 475 + V 9 - 4 7 5] - 1 8

= 3781-80(^9 +

4 7 5+ ^ 9- 4 7 5 ] -3

[^9 +

475

+ ^ 9 -

4 7 5 ]

= 0

Vay =

V9 + 475 + ^ / 9 - 4 7 5

la nghiem cua phiTdng trinh:

A

; ' - 3

A

; - 1 8 - 0 .

2) X6t phircfng trinh: x ' - 3 x - 1 8 = 0 (*). Bkng each nham

nghiem ta thay phu'dng trinh (*) c6 mot nghiem x = 3.

Chia:

A

:^ - 3;c -18 cho x - 3. Ta dtfcJc:

(*)^(x-3)(;c^+3x + 6) = 0

\x = 3

+3x + 6 = 0(VN)

-ip^

x = 3.

Vi phu'dng trinh (*) cd mot nghiem duy nha't x = 3 nen: x^ = 3.

D^5.31: Chophu'dngtrinh:

( ^ - 2) ( x ' - A : ) - ( x- 2 ) ( 2 A : - m )

= 0 (1)

1) Giai phu'dng trinh (1) khi m = 1.

2) Vdi gia tri nao cua m, phu'dng trinh (1) cd ba nghiem phan

biet?

= 0

Ta cd :

(x -2)ix^

-x)-(x-2)(2vC-m)

=

0^{x-2)

[(x'

-x)-(Ix-m)

4=>

(;c - 2)(x^ - ;c -

2JC

+ m) =

0 ^{x-2){x'-3x + m) = Q

1) Khi m = 1 phu'dng trinh (1) trdjhanh:. ^ . . ^ , ,

108 {x-Dix" -3x^\) = 0<^

x

;c2_3x

- 2

= 0

+ l = 0

^2_3;c + l = d

x,=2

^2 =

X3 =

3 + 75

2

3 - 7 5

Vay khi m = 1 thi phu'dng trinh (1) cd ba nghiSm phan biet

. 3 + 75 . 3 - 7 5

X. =

2;x. = va x. = .

^x-2 = 0

x^ - 3x + m = 0 >

x = 2

x^-3x + m = 0

2) Tacd:

( A

:-2)(x'-3x + m) = 0 ^

44>

Tir dd suy ra phifdng trinh (1) cd ba nghiem phan biet khi va chi khi

phifdng trinh ;c' - 3x + m = 0 cd hai nghiem phan biet khac 2

A > 0

2'-3.2 + m ^ 0

- 4 m

> - 9

9 -

> 0

- 2 + m ^ O

m ^ 2

9 m < -

4

m ^ 2 .

Vay khi m < - va m ^ 2 thi phtfdng trinh (1) cd ba nghiem phan

biet.

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(56)<div class='page_container' data-page=56>

§ 2.

PHUdNG

TRINH Q U I

VE

B A C HAI

I. Phifong trinh trung phiforng: ax" + bx^ + c = 0 (a ^ 0).

Cach giai: Dat t = x^ (dieu kien: ; > 0) phifdng trinh trd thanh:

at^ + bt + c = 0.

• Giai va kiem tra dieu kien < > 0.

II. PhiTorng trinh: {x + a)ix + b)ix + c){x + d) + e = 0

Vdi: a + b = c + d.

Cach giai: Dat: t = {x + a)(x + b) = x^+(a +b)x + ab.

Khi do: (x + c)(x + d)^x^+(c + d)x + cd

^x^ -\-{a + b)x + ab-ab + cd

= t-ab + cd.

nen phu'cfng trinh trd thanh:

t(t-ab + cd) + e = 0^t^ + (cd-ab)t + e = 0 .

III. PhiTrfng trinh: {x +af +{x +b)' + c =^Q.

Cach giai: Dat: t = x + ^ - i ^ rdi du^a phu'cfng trinh da cho ve

pJlu-dng trinh triing phu'cfng.

IV. PhtfcTng trinh: ax' + bx^ + cx' +dx + e=-0 (a ^ 0,e ^ 0) (vdi

'd\'

e

a

)

Cach giai: Chia hai

va'

cho x^, phu'cfng trinh trd thanh:

+ c = 0.

2 ^

{ d\

^b

;c + —

bx]

•> CI

ex

+

Id

va phufcfng trinh trd thanh: at^ + bt ^ c - ~ = 0 .B&y la phiTcJng

b

trinh bac hai theo t.

110 5.32: Giai cac phu'cfng trinh:

1) x' - 4 x ^ + 3 = 0

2) x ' - 3 x ' - 4 = 0

3)

(X

- l)(x - 5)(x - 3)(x - 7) - 20 = 0

4) {x + \){x + l){x + 3)(x + 4) - 35 = 0

5) x'+{x-A)'=%2

6) / + 5 x ' - 1 2 x ^ + 5 ^ + 1 = 0

7) x ' - 3 x ' - 6 x ' + 3 x + l = 0.

Gl§i

1) x ' - 4 x ' + 3 = 0

ãDat: ô = x' (di^ukien r > 0 ) .

Khidd: -At + 3 = Q

t = \

ô|, = 3.

ã Vdi: ? = l < ^ x ' = l<^x = ±l.

(1)

• Vdi: f = 3<s=>x'=3<^x =

± V 3 .

Vay phu'cfng t inh cd bon nghiem: x = ±l;x =

± A ^ .

2) x' - 3 x ' - 4 - 0

Dat: i = x' (dieu kie i

t

> 0).

Khidd: (2) ^ - 3 r - 4 = 0

t = -\{loai)

t = A.

• Vdi: i = 4 <i=^ x' = 4 ^

X

= ±2.

Vay phu'cfng trinh cd hai nghiem: x = ±2.

3) ( x - l X x - 5 ) ( x - 3 ) ( x - 7 ) - 2 0 = 0

, Ta cc': (3) ^{x- l)(x - 7)(x - 3)(x - 5) - 20 = 0

(2)

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(57)<div class='page_container' data-page=57>

Dat: f = (;c-l)(;c-7) = ;c'-8x + 7.

Khido: (x-3)(x-5) = x^ ~Sx + \5 = t + 8

Dod6: (3)^r(r + 8)-20 = 0

^ +St-20 = 0

r - 2

; = - l 0 .

• Y6i t ^ 2 ^ x^ - Sx + l = 2

^x^-Sx + 5 = 0^x = 4±^.

• vdi f = - 1 0 ^ x ^ - 8 j ; + 7 = -10.

<^x^-8;c + 7 = -10

<^;c^-8;c + 17 = 0 (pt v6 nghiem).

Vay phifdng tfmh c6 hai nghiem: x = 4± VlT.

4)

(;

C

+

1)(

X

+

2 X;C

+

3)(

A

;

+ 4)-35 = 0 (4)

Tacd: (4) ^ (;c + l)(;c + 4)(x+ 2)(x + 3) = 0

Dat: t = (x + l)ix + 4) = x^+5x + 4

Khidd: (x + 2)ix + 3) = x^ +5jc + 6 = r + 2

Dodo:

(4

)<i

=^;(i

+ 2)-35 = 0

<^f' + 2 ; - 3 5 = 0

t = -l

t = 5.

• V6i t = -l ^ x^ +5x + l\ 0 (ptv6 nghiem)

• Y6i t = 5 <^ x^+5x-l = 0.

<^ X —

-5±^/29

Vay phiTdng trinh cd hai nghiem: x =

5) x' +ix-4y =82

0 - 4

Dat: t = x + = x-2

-5±yf29

(5)

Khi

dd

x = t + 2

2 x-^4 = t-2

112 nen (5) <^ + 2)'+ (/- 2)' - 82

^ + 4 / . 2 + 6r\ + 4.r.8 + 16

+ - 4 / . 2 + 6r'.4-4./.8 + 16 = 82

^2t' +48?' +32 = 82

^ t ' +24f'-25 = 0 (*)

Dat X = t^ (X> 0)

thi ( * ) ^ ^ ' + 2 4 Z - 2 5 = 0

\X = l

^\x

^-25(loqi)

Vdi x = i<^f' =i<^f = ±i.

• Vdi f = l<;=>;c = l + 2 = 3.

• Vdi f = - 1

A

:

= -1 + 2 = 1.

Vay phiTdng trinh cd hai nghiem: x = 3 ; x = I.

.) JC

'+5;

C

'-12

A

:'+5

A

; + 1 = 0 (6)

De thay x = 0 khdng la nghiem ciia phiTcfng trinh (6).

Vdi x^O, chia hai ve cua phifcfng trinh (6) cho x^, ta diTdc:

I

(6)^x^ +5;c-12 + - + — = 0

X X

n

+ 5 x +

-X

-12 = 0

Dat: t = x + -.Th\: =x' + \ 2<^x' + \ e-2.

X X X

Dodd:

(6

)<i^a

'-2)

+ 5r-12 = 0

<^<'+5r-14 = 0

t = -l

f = 2.

• Y^i t = -l ^ x + -=-1

X

<^ A:'

+ 7 x - l = 0

<!4> J: = •

-7±3

N

/5

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(58)<div class='page_container' data-page=58>

• Vdi t = 2^ x + - = 2

X

<^x^-2X + 1 = 0<!F^ x = \.

Vay phifcfng trinh cd ba nghiem: x-\ x = - 7 ± 3 ^ 5

7) x'-3x'-6x'+3x + \^0

• De thay x = 0 khdng la nghiem ciia phiTdng trinh .
• V d i x^^O, chia hai ve ciia phuTdng trinh cho x^ ta diTdc:

; c ' - 3 ; c - 6 + 3 . - + -!- = 0

X X

X ' + \ - 2

X 
\2 
X

X

- 3

1
- 3 X

x^

r

X — - - 4 =

- 4 = 0

Dat: t = x phiTdng trinh trd thanh - 3; - 4 = 0

X

=-\\t^ = 4 v i a - b + c = 0

• t = - 1 tacd ; c - - = - l <i4> A : ^ - l = -;c,

X

^ x'^ +x-\ 0

- i +

Vs

- 1 - V 5
= ,

• t = 4 t a c d j : = 4 - ^ x ^ - l = 4 x < ( ^ x ^ - 4 ; c - l = : 0

X

^x,=2^S\x,=2-S.

Phifdng trinh cd bon nghiem la

z l ± ^ . z i ^ ; 2 + 75;2-V5.

De 5.33: Giai cac phifdng trinh sau:
' x _ 4 l

.3 X

1) l ^ ^ + ^ - i o
114

2) (4A: + l)(3x + 2){\2x -\){2x + 2) - 8 = 0.

3) 4(JC + 5)(;C + 6)(A: + 10)(X + 1 2 ) - 3 X ' = 0 .

Giai

1) Dieu kien x ^ 0 .
T a c d : 1 x ^ + ^ = 1 0

3 x^

. x 4 .. 0 
3 X

Phirdng trinh trd thanh: 3

X 4 '

^ 3 ix' 16' = 10 ' X A 
— ^ 3 + - T = 10

13 x^ 9 x\ >3 X,

x' 16 8 • x' 16 2 8

+ / + - .

9 3 9 x' 3

3 = 1 0 / ^ 3 ^ ' - 1 0 / + 8 = 0
4^t = 2,t =

-3
* V d i ; = 2 < ^ - - - = 2 < ^ x ' - 1 2 = 6x

. 3 X

x ' - 6 x - 1 2 = 0 < ^ X = 3 ± >/2T 
* V d i / =

-3

X 4 4

4^ =

-3 x -3

^ x^ - 1 2 = 4x <^ x^ - 4x - 1 2 = 0

<i=^A: = 6,x = - 2 .

Vay nghiem ciia phuTdng trinh l a : x = 6,x = - 2 , x = 3 ± ^ITA. 
2) Dat / = (4x + l)(3x + 2) = 12x' +11X + 2

thi (12x - l)(2x + 2) = 2 4 x ' + 22x - 2 = 2f - 6

nen phufdng trinh (4x + l)(3x + 2)(12x - l)(2x + 2) - 8 = 0

2/' - 6 / - 8 = 0 r = - l , f = 4
* V d i t = -1 <^ 12x' + 1 I x + 3 = 0 (v6 nghiem).

V d i t = 4 <^ 1 2 x ' + l l x - 2 = Q < ^ x = ' ^ .
24

-\\±yl2ri

Vay nghiem cua phiTdng trinh la x =

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(59)<div class='page_container' data-page=59>

^ 4(x + 5)(x + I2)(x + 6)(x +

W

)-3x^ =0

^4(x^ +nx + 60)(x'' +l6x + 60)-3x^ = 0

Dat t = x^ + l6x + 60^ x^ + nx + 60 = t + x

nen phiTcfng tfinh trd thanh: 4(t + x)-3x^ =0

^4t^+4xt-3x^

=0

^(41^

+4xt

+ x'^)-4x^ = 0

^ {It4-xf -{2xf =0^{2t + 3x)(2t-x) = 0

3

X

<^ t = x,t = —

2 2

*\6\t = - ^ x'+16;c + 60 = - 4 ^ 2 ; c ' + 3 1 ^ + 160 = 0 v6

2 2

nghiem vi A < 0 .

> A:'+16X + 60 =

-—

2

^2JCN

-35±N/265

*Vdi / = - - ; c ^ A:'+16X + 60 = - — ^ 2 J C ' + 3 5 ^ + 120 = 0

2 2 •

Vay nghiem ciia phu'dng tnnh la x =

—

^ V265^

5.34: Giai phifdng trinh: {x^ +lx + \ +1 IA: + 28)-72 = 0.

Giii

Tacd: (x^ + 7A:+ 10)(A:^ +1 IJC+ 28) - 72 = 0 (*)

^{x" +2X + 5A: + 10)(X' +4JC + 7X + 28)-72 = 0

x{x + 2) + 5{x + 2)][;c(jc + 4) + l{x + 4)] - 72 = 0

^{x^ 5){x + 2){x + l){x + 4) - 72 = 0

^ (x + 2)(A: 4- 7)(x + 4)(x + 5) - 72 = 0

Dat: < = + 2){x + 7) = + 9x +14.

Khidd: (;c + 4)(jc + 5 ) - + 9 x + 20 = / + 6

Dod6: (*)<^f(/ + 6)-72 = 0 +6/-72 = 0

/ = -12

116 • Vdi f = -12-4^;c^+9;c + 14 = -12

4 4 - +

9x + 26 = 0.

Phifdng trinh v6 nghiem vi: A = 81 -104 = -23 < 0,

ã Vdi ô = 6 44>x^+9A: + 14 = 6

A:^

+9x + 8'=0

A:

= - 1

Vay phu'dng tnnh cd hai nghiem: x = -1 ; x = - 8.

m 5.35: Giai phu'dng trinh: x' - 4;c' -19x' +1 06A: -120 = 0.

Gi§i

J Tacd: jc'-4A:'-19A:'+106x-120 = 0

^x' - Sx"

+ 6 A r '

+

A:' - 5A:' + 6A: - 20A:' + 100A: -120 = 0

A:'(JC' -5A: + 6) + A:(A:' -5X + 6)-20(A:' -5A; + 6) = 0

(A:'+ A: - 20)(x'- 5A: + 6) = 0

^{x'' -4A: + 5A:-20)(X' -2X-3A: + 6) = 0

^ \x{x - 4) + 5(X - 4)][A:(A: - 2) - 3(A: - 2)] = 0

(AT

+ 5)(x - 4)(A: - 3)(A: - 2) = 0

x + 5 = 0 [A: = - 5

jc-4 = 0

A: =

4 x - 3 = 0

A:

= 3

x - 2 = 0

A:

= 2.

Vay phu'dng trinh cd bdn nghiem: x = -5;x = 4 ; x = 3;x = 2.

5.36: Phan tich thanh nhan tiJ : A = x' - 5A:' +1 OA: + 4.

x" + 4

Ap du ng: Giai phu'dng trinh — = 5A:,

Giii

Tacd: A = x' -5A:' +10A: + 4.

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(60)<div class='page_container' data-page=60>

= x' -x'-2x^-4x' + 4A:' + 8;c -2;c" +2x + 4

^x^(x^-x-'2)-4x{x^ ~x-2)-2ix^-x-2) 
• ={x^ -x-2)(x^ -4x-2)

^(x^-2x + x-2)ix^ - 4 J C - 2 )

= x(x-2) + (x-2)](x^-4x-2) 
= (x-2)(x + \)ix' -4x-2)

A p dung: D i e u k i e n x^ —2 ^ 0 x ^ ±yj2 . 
x' + 4

Ta c6:

x'-2

^5x ^ x'+4 = 5x{x'-2)

^x' - 5 ; c ' + 1 0 x + 4 = 0

^ (jc - 2){x + l){x^ - 4x - 2) - 0

x-2 = 0

<^ ;c + l = 0 4=^

x^-4x'2 = 0

Cac gia t r i ciia x t i m dUdc deu khac ± 2 .
V a y nghiem cua phUdng trinh da cho la:

x = 2

x = -l 
x = 2 ± S

= 2\x^ = -\;x^ - 2 + ^/6;x, - 2 - V 6 .

5.37: Cho biet phUdng trinh: {x + \){x + 2){x + 3)ix + 4) + m = 0
CO nghiem la x^,x^_,x^,x^. Hay tinh M — — + — + — + — theo m.

Gi§i

T a c 6 : M = : ^ + ^ l ± A

x^x,

PhUdng trinh dUdc viet thanh:

{x + l)(;c + 4).(;c + 2){x + 3) + m = 0 .
Dat t^{x + \){x + 4) = + 5x + 4 thi
(;c + 2)(;c + 3) = x ' + 5 ; c + 6 = / + 2

PhUdng trinh trd thanh: / ( / + 2) + m = 0 ^ + 2/ + m = 0 (*)
118

G o i t^,t^ la nghiem cua (*) thi ta c6: t^+t^= - 2 , / / ^ = m

* V d i / = t, thi jc' + 5;c + 4 = <^ + 5JC + 4 - = 0

Theo dinh l i V i e t , ta cd x^+ x^= —^,x^x^ =4 — t^

* V d i r = thi x ' + 5JC + 4 = ^2 A;^ + 5x + 4 - ^2 = 0
Theo dinh \i Viet, ta c6 x^+ x^ =^-5,x^x^ =:4 — t^

Do do: M = — — = ^ = - 5
4 - t , 4 - t ,

%-{t,+t,)

+

4 - t , 4 - t 
•2 J

= -5.

= - 5 . 8 - ( - 2 ) - 5 0
1 6 - 4 ( - 2 ) + m m + 24

f)i 5.38: Cho phu-dng trinh: x' - 2(m + 2)x^ = 0. T i m m de
phUdng trinh cd bon nghiem phan biet.

Gi^i

x' -2im + 2)x^ =0 (1)

D a t : / = A ; ' ( ? > 0 ) thi (1) < ^ - 2 ( m + 2)r + m ' = 0 (2)
IjPhu'dng trinh (1) c6 bon nghiem phan b i ^ t

<^ phUdng trinh (2) c6 hai nghiem phan biet diu dUdng
A ' > 0 [(m + 2 ) ' - m ' > 0

S>0 2(m + 2 ) > 0
4/n + 4 > 0 [ m > - l

m + 2>0 [m>-2.

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(61)<div class='page_container' data-page=61>

§ 3. PHLfONG TRINH CHLfA GIA TR! TUYET D

6\

PHJCNG TRiNH CHLfA CAN THLfC.

I. Phifrfng trinh chiJa gia trj tuy^t do'i:

1) Dinh nghia va tinh chS't:

X

ne'ux

>0

—X ne'uX <0

.

V7=

2) PhiTcfng trinh chiTa gia tri tuyet do'i:

f(x) = g{x)

m =

g

(x)

fix) =gix)^

[fix) = -gix) 
'gix)>Q

f(x) = gix) 
fix) =^-gix)

Nhan xet: Cac phiTdng trinh tren cd the bien doi nhuf sau:

• \f(x)\ \gix)\^fix) = g\x).

\gix

)>0

fix) = gix) <^

fix) = g\x)

II. Phufofng trinh chiJa can thiJc:

1) PhiTdng trinh chiJa mot can:

r-

\B>0

A = B^

A = B

A > 0 V i 5 > 0

Ghi Chu: Neu phifcfng trinh cd dang:

afix) + b^fix)

+ c = 0 .

Dat:

t = 4fix)

; (/>0).

2) Phifdng trinh chiJa nhieu can:

Phifdng phap:

4A

=

4B

^

120 • Dat dieu kien.

• Bia'n doi de hai w€ deu khong am.

• Binh phiTdng hai ve de khiJ bdt can.

• Riit gon de diTa ve dang cd ban.

• Giai va ket hdp vdi dieu kien de chon nghiem.

Ghi chii: * Neu phifdng trinh cd dang:

g(x)

fix)

+ c = 0.

Dat: / = / ^ ; ( / > 0 )

\gix)

* Neu phu'dng trinh cd dang:

V A + 2 V A ^ + V A - 2 V A ^ =

B .

Ta dung phep bien doi:

* A + 2 V A - 1, = (/\ 1) + 2 V A - 1 + 1

= ( V A ^ + I ) ' .

* A - 2 V A - I = ( A - 1 ) - 2 V A - 1 + 1

.

=(vr

1 - 1

5.39: Giai cac phu'dng trinh:

1) jc-5 = 3

•

2)

2 X - 3

-

x + \

= 0.

3)

x-l =2x + l.

:) Tacd:

x

-5 = 3^

Giai

;c-5 = 3

x

-5^-3

x = 2

</div>
(62)<div class='page_container' data-page=62>

2x-3 = x + \

2x-3 = - x - \

4^

x = 4

3x = 2

x^4

2

X = —

Vay phifdng trinh c6 hai nghiem: x = 4 va — j •

3) Tacd: x-l =2x + 7 ^

2x + l>0

x-l = 2x + 7

x - \ -2x-l

4=>

3x = -6

Vay phiTcfng trinh c6 nghiem: x- - 2.

~ 2

x = - 8

3x = -2

^ x = -2.

De 5.40: Giai cac phiTdng trinh:

1) x - x - 2 = 2 . 2) 2 x ^ - 7 A : - 5 A : - 2 = 0 .

3) I J C- 2 1 - 1 - 2 = 0. 4) x^ -5 x +6 = 0..

1) * Xet ;c < 0 phu-dng trinh trd thanh:

~x + x-2 = 2^Ox = 4^xe0.

* Xet 0 < jc < 2 phiTdng trinh trd thanh:

x + x - 2 = 2<;=>2;c = 4<^;c = 2 (loai)

* X6tx > 2 phiTdng trinh trd thanh:

x-x + 2 = 2^0x = 0^ xeR.

Do do x>2.

Ke't luan: Nghiem ctia phu^dng trinh la : ;c > 2.

2) * Vdi ;c < 0 thi

= -X,

x-2=2~x.

Phu-dng trinh trd thanh: 2x' + 7x - 5(2 - jc) = 0

<^ 2 x ' + 12x -10 = 0 <;=> ;c = - 3 ±

Vi4

Vi JC < 0 nen chi nhan A; = - 3 - Vi4 .

122 * Vdi 0

< A: <

2 thi |x| = 2| = 2

-Phu-dng trinh trot thanh: 2x^ -lx-5(2-x) = 0

<^ 2x^ -12x - 1 0 = 0

-

X

- 5

= 0

1

^X = -(l±yf2\^

1 \

Vi - ( l -

V2T)

< 0 <

2 < - ( 1 +

ylri^

nen khong nhan

\±yl2\]

1 x = — 2

* Vdi

JC >

2: thi

x-2=x-2.

Phu-dng trinh trd thanh: 2

JC^ -

7 X -

5(

X -

2) = 0

<i:^2x'-12jc + 10 = 0 ^

J: =

1, jc = 5.

Vi

JC >

2 nen ta chi nhan x = 5. ;

Ke

't luan: Nghiem la x = 5, jc = - 3 - Vl4.

3) Taco:

l x

- 2 1 - 1

- 2

= 0 ^

= 2

x-2

- 1 = 2 hay x-2

= 3(v« x-2

x-2 - I

- l = - 2

— 1 ——2<;4-Jc—2=—1 vd nghiem

< ^

X

- 2

= ±3 4=^

X =

5,jc = - 1 .

Vay nghiemla: x = 5, x = -1.

4) Dat/ = |x| (/>0)thi: jc^ - 5|

JC

| + 6 = 0 <^ - 5/+ 6 = 0

^ t =

2

,t

=

3

* Vdi t = 2 thi |jc| = 2

^ JC =

±2.

* Vdit = 3thi |x| = 3<^jc = ±3.

Vay phu-dng trinh c6 nghiem la: x = ±2,x = ±3.

De 5.41: Tim m de phiWng trinh x + •2x + m = 0 CO nghiem.

Gi^i

</div>
(63)<div class='page_container' data-page=63>

-x>0

(x^ -3x + m)(x'^ -x + m) = 0 
x<0

x^ -3x + m = 0 
x<0

(x^ - 2x + mf -x"^ =0

x<0

(1) hay x^ -x + m = 0 
x<0 (2)
* X a t m > 0: (1) c6 S = 3 > 0, P = m > 0

(2) C O S = 1 > 0, P = m > 0

nen chiing c6 nghiem thi cac nghiem deu du'dng va do do ta't ca
deu bi loai vi dieu kien x <0 .

* X e t m = 0 : thi (I) ^ x^ -3x =^0 ^ x = 0,x = 3.

(2) ^ x''-x^O^ x^O,x^l 
V i dieu kien x<0 nen nhan nghiem x = 0.

* X6t m < 0: (1) C O P = x^x, = m <-0 nen < 0 <

Va do do phu'dng trinh c6 nghiem la x^.

Ket luan: Phufdng trinh c6 nghiem khi m < 0 . 
5.42: Giai cac phu'dng trinh:

1) i x^-Ax + A= A9 
2) 4x^A^x-\ 
3) Vl - 2JC' = ;c - 1 .

Gi^i

1) T a c o : ^x--Ax + A =A9 ^

4^x-2f

= 4 9 ^ x-2 = 4 9

A ; - 2 = 4 9

2) T a c 6 : ylx + \=x-\^

X- 2 = - 4 9
x-\>0

x^-\ {x~\f

x = 5\ 
x = -A7.

124

<^ x = 3.

•3) V l - 2 x ' = j c - 1 <^

x > l

x + 1 = - 2 x + l
x>\

x = 0 V x = 3
x - l > 0

I - 2 x ' = ( x - l )

x > l

1-2A:' - 2X + 1
x > l

X = O V J C =

-3

x>l

x^ -3x = 0

J C> 1

3X' - 2A: = 0

Vay phufdng trinh v6 nghiem.

5.43: Cho phiTdng trinh c6 an x : yjx'^ - 4 = x - a. 
1) Giai phu'dng trinh vdi a = 2.

2) Giai va bien luan phu'dng trinh theo tham s6 a.

GiSi

1) K h i a = 2 : phu'dng trinh trd thanh 4x^-A =x-2. 
\x-2>0

V x ' - 4 = x - 2 4=^

x''-A = {x-2f 
x>2

x^-A = X^ - 4A : + 4
x>2 \x>2 
4x = 8 \ix^2 
Phu'dng trinh c6 nghiem la x = 2.

» : - a > 0

-A = (x-af

<^ jc = 2

2) ^lx^-A=x-a

</div>
(64)<div class='page_container' data-page=64>

x> a

• a = 0 ta CO (*) 4 ^

• a ^ 0 ta CO (*)

x> a

lax = a^ +4 
x>0

Ox = 4

X > a

xe0

X = — a' +4

la

G i a i d i e u k i e n ^^—^> a ^ a> 0

la la

4-a^

la ~

a > 0 ; 4-cr>0

a<0 • 4-a" <0 
a>0

;

a^ <4 0<a<l

a<-l. 
a<Q ; a ' > 4

T o m l a i :

• N e u 0 < a < 2 hoac a < - l : PhiWng t r i n h c6 nghiem 
a ' + 4

X =

la

N e u - 2 < a < 0 hoac a > 2: Phtfcfng t r i n h v 6 n g h i e m .

B e 5.44: G i a i phu-dng t r i n h : + 4 x + 5 = l^Jlx^~T^.

D i e u k i e n :lx^-3>0^1x>-3^x>--. 
~ 2

T a c d A:'+4Jc + 5 =

2^y2IT3.

^ AT'+ 4A: + 5 -2 V 2^ + 3 = 0

^ + 2A: +1) + |(2A: +

3) -

2V2

^T3

+1

^ U +1)' + (V2;c +

-

1)'

= 0

126

4=>

44>

(;C + 1 ) ' = 0 '

V2A: + 3 - 1 = 0

x + 1 = 0

V2A: + 3 - 1 = 0

A: = - 1

V2A: + 3 = 1 4=^

A: = - 1
2A: + 3 = 1

< ^ Ar = - 1 .

N g h i e m cua phufdng trinh la x = - 1.
p 6 5.45: G i a i cac phu'cfng t r i n h :

1) A: - 5V X- 6 = 0.

2) 3A:^ + 2A: = lyjx^ +x +\-x.

3) A : '- 4 x - 6 = yjlx^ - 8 x + 12.

1) A: - 5V^ - 6 = 0. ( 1 )

D a t : t = ylx ; (t>0). • 
K h i d d : ( l ) ^ r ' - 5 / - 6 = 0 4=!>

[/ = 6

Ydi l = 6^4x = 6^ x = 36.

2) 3A:' + 2 A : - 2 V ? T^ + 1 -A: ^ 3U ' + X ) - 2 7 ^ ^ ^ T^ - 1 = 0(2)

D a t : t = y[x^~+x ; ( / > 0 ) . 
K h i d o : ( 2 ) 4^ 3/' - 2? - 1 = 0

t = l

' - 3 «
V d i / = l< ^ V ^ + A: = l<;=>A:'+x = l

- 1 - V 5 
2
•i +

Vs

<^A:'+A: - l = - 0

A: =

x —

V a y p h i cfng t r i n h c6 h a i n g h i e m l a : x =

1 
- 1± 7 5

2) T a c 6 : x'-4x-6 = ^lx'' - 8 x + 12.

</div>
(65)<div class='page_container' data-page=65>

-4;c + 6)-12-V2(x^ -4

A

: + 6) = 0 (3)

Dat: t = ^2{x' -Ax+ 6) ; (i>0).

Khido: ( 3 ) 4 ^ y

- 1 2

- r = 0 ^ f ' - 2 f - 2 4 = 0

r = 6

' = - 4 (/)

* Vdi f =

6 ^ 7 2 ( X

^ - 4

A

:

+ 6) =6<^2(

X

'-4

A

; + 6) = 36

<^;c^-4x-12 = 0<(=> '^"'"^

l;c=:6.

Vay phiTdng trinh cd hai nghiem la: x = -

2 va x = 6.

De 5.46: Giai cac phifdng trinh:

1) ylU-x-Jj^=^2

2)

V

5 A; -

1

- V3x - 2

V

;c

- 1 = 0.

1) Tacd: V l l - x - V ^ = 2.

Dieu kien: l l - x > 0

x-l>0

x<\\

x>l < ^ l < ; c < l l

( 1 )
(*)

Khi dd: (1) Vl 1 - ^ =

2

+ VA:

-

1 <^ 11 - ^ = 4 + 4 + (;c -1) <^ 4 = 8 - 2 ;c

4

- X >

0 2 V I ^

= 4 -

X ^

4 U - l ) = ( 4 - x f

x<4

x<4

4x-4 = l6-Sx + x^

x<4

x = 10\/x = 2.

^x = 2 (thoa dieu (*)).

Vay phiTdng trinh da cho cd mot nghiem x = 2.

2) ^5x-l-fix--2-^f^ = 0.

x" -12

A:

20 = 0

(2)

128

Di^u kien:

5x-\>0

3x-2>0^

; c - l > 0

x>-~ 5

2 x> — <^x>\

~ 3

x>\

Khi dd: (2) ^ yl5x-\ y/3x-2 + ^x-l

<!^5

A

:-1 = (3

A

:-2) + (;

C

-1) + 2

V

(3

A

:-2)(

X

-1)

x + 2>0

^2yji3x~2Xx-\)=x + 2 4:^

44>

x>-2

12 X'

12;c -

8

A:

8 = + 4x + 4

2

A: =

2

V X

= —.

A; =

2 V;C

= —

11 4(3;

C

-2)(;

C

-1) = (

X

+ 2)'

x>-2

l l x ' - 2 4 ^ + 4 = 0

44>

So vdi dieu kien (* *) ta chi nhan nghiem x = 2.

Vay phiTdng trinh da cho cd mot nghiem x = 2.

Bi 5.47: Giai cac phu-dng trinh:

y/x-\ + ylx-5

Gi§i

</div>
(66)<div class='page_container' data-page=66>

ix

+

\f

>4

x > \

x + l>2 hoqc x + \ < - 2 
x>l

_ sJx^+2x-3

W

V d i d i l u kien do, ta c6:

y/x-l 
ylix+3)(x-l)

^ x > \

= 3 + x.

= 3 + x

• = x + 3<=^^Jx + 3=x + 3

4=^

'x + 3>0

x + 3 = (x +

3Y

'x>-3

x^ +5x + 6 = 0

^ x = -3Vx = -2

So vdi dieu kien (*) =^ phu'dng trinh da cho v6 ng liem.

2) Dieu kien:

'x>-3

x +

3

= ;c^ +

6 i + 9

;c

>-3

x = -3\/ x ^ - 2

x - 5 > 0

3 + ylx-5^0 
<^ x>5

Ta cd:

^ 5 - ^ ^ = 3 ^ ^ 5 - ^ ' - ' ' ^ ^ ' - ^ ' ^ ^ 3

^ _ ( . - 1 4 ) ( 3 - V I 3 5 ) ^ ^

9-

A: +

5

4=^

VJ

:-5+3-

V;C

-5 = 3

OyJx-5 = 0 pt dung vdi moi A: > 5.
Vay phu-dng trinh cd nghiem tuy y x>5.

De 5.48: Giai cac phtfdng trinh:

1) ^Jx + 3 + 4yfI^ +^Jx + S-6^JJ^-5 = 0. 
130

2)

^x +

2

^x-[ +-ix

-2

^x-\

x-\-3

GiSi

1)

Tacd

\lx +

3 +

4

^x

-l

+^jx

+

S-6s/

x-1-5

= 0

^ V

(-^-1) + 4

V ^ T

4

+ V

(

A

: - 1 ) - 6

V ^

9 - < = 0

f V ^ + 2f + j f V I

^ - 3 V - 5

= 0

<i=^ V x

- H

- 2 +

- 1 -

3

• Dieu kien Jc > 1.

- 5

= 0

(1)

• Vdi A;> 10(a) : thi - 1 > N/9 = 3 n e n V x - l - 3> 0 va

(1)

<!=^VA:-1+2

+ V A

: - 1

- 3 - 5

= 0

<^ V x - 1 = 3< ^ x - l = 9< ^ j ; = 10 (thda dk (a))
• V d i l < j c < 1 0 : t h i V ^ < 3 n e n V ^ - 3< 0 v k

(1) V ^ ^ +

2

- ( ^ / ^ - 3 ) - 5

= 0

< ^ 5 - 5 = 0<i^0 = 0 ( luon diing)

Dieu nay chiJng to moi x E I,10) deu la nghiem.

Ket ludn: 1 <

x

< 10 deu la nghiem ciia phtfdng trinh da cho.
*' 2) Tacd: + 2 ^ / ^ + - 2 V ^ ^ = ^ ^ i ^ .

^ 7

(^-1)

+-2V^^

+ 1 +

VU

-1)-

2

N

/7^

+ 1

= ^ i ^

- i + i ) ' + ^ | ( 7 ^ ^ T ^ =

- 1 + 1 + - 1 - 1

A: +

3

x +

3

(2)

• Dieu kien: x > 1.

• V d i

A; >

2

(a) : thi

V ; c

-1 > 1 =^ - 1

> 0

(2)

^ ^ / I ^ + l + V I ^ - l =

</div>
(67)<div class='page_container' data-page=67>

2VA: - 1 - x + 3

4=>

\e{x-\) = {x + 3f 
x>-3

-10A: + 2 5 = : 0

x > - 3

1 6 X - 1 6 = A:^ +6;c + 9
; c > - 3

x = 5 
^x = 5 (th6a dk (a))

V d i l < 0 < 2 {b) thi V ^ < 1 nen V ^ - K O
d o d d ( 2 ) ^ + + l =

^ 2 = ^ ^ j i < ^ A ; = l ( t h 6 a d k ( b ) )

T d m lai phtfdng trmh da cho cd hai nghiem: x = 1 va x = 5

5.49: Cho phiTcfng trinh cd an x :

^x^-lx + \=yj(i + A^-4^-A4i .

1) Rut gon v6' phai cu a phifdng trinh.

2) Giai phu'dng trinh.

GiSi

1) T a c d :

6 +

4 N

^ = ' 4

+ 2 . 2 ^

+ ( N ^ ) '= ( 2

+ .^'

Tu'cfngtif 6-A-J2 =(l-yj2f 
Ta cd va'phai

= 2 + > y 2 | - | 2 - N ^ = 2 + V 2 - 2 + >y2 {vil> y[i)

2) yjx^-2x + l=yl6 + 4.j2 -yl6-4yf2

<^^[x-\f =2yf2

132

- 1 = 2 ^ 2 ^ x - 1 + 2 ^
X = \-2yj2

Phu'dng trinh cd hai nghiem la: x = \ 2^2 \x = \- 2yf2 .

5.50: Giai cac phu'dng trinh sau:

\^ +4x + 2

1) c ' - 2 ; c - 2 + ,

+ 4x + 2 V A : ' - 2 ^ - 2 = 2.

2) + = 5^{x-2){x-3).

1) Dat f = - 2 A ; - 2
V x '

+4

;c

+ 2

G i i i

thi t > 0 va phu-dng trinh trd thanh:

/ + - = 2<i=>/^-2/ + l = 0 < ^ f = l

t 
x^~2x — 2

^ —z = 1 ^ 3 x ' + 6x + 4 = 0 (v6 nghiem)

4x^+Ax + 2

Vay phu'dng trinh v6 nghiem.

) Nhan xdt x = 2, x = 3 khong la nghiem.

jgChia hai va' ciia phu'dng trinh cho ^[x - 2)(,x - 3) thi cd:

6.6 + .6 ( . - 2 f

^ ( . - 2 ) ( . - 3 ) ^ ( x - 2 ) ( . - 3 ) = 5

x-3 x-2 , 
x-2

'x-3

<^ 6/ + - = 5 (vdi t = 6

t V 
x-3

x-2 ,t>0)

1
> ^ 6 r - 5 / + l = 0 < ^ / = - , / = - .

... x-3 6 x-3

U-2 x-2

</div>
(68)<div class='page_container' data-page=68>

It"-3

x =

Vay nghiem 1^: * x =

* V — X = —^

3 - 1

- 3

- 1

2-3.2^ _ 190

1-2' " 63

2 - 3 ' _ 2185

1-3' ~ 728

Bi 5.51: Giai phiTdng trinh: ^(x + 1)' +4^(x-\f = 6 ^ x ' - l .

- _

Nhan xdt ring x = ±\g la nghiem.

Chia hai va'cua phiTcJng trinh cho ^x^ - 1 ta diTcJc:

{x-\f ^

JC +

1 , ,

x^-\

x-\

x + \

-6

= 0.

<^/ + l - 6 = 0 (vdi/=3pi-!-)

t. \ x - \

<^f'-6r + 4 = 0<^r = 3±V5

Vdi

r =

3^

X-\-\ X + \ 3

<^ r = ^ r x - t = x + \

x-\

^(t'-\\x =

e

+\^x = ^^

(3±V5)'+1

<^ x =

134 5.52: Giai phifdng trinh:

^ji^-Ax + n + V3;c' - 6 x ' +28 = -3;c' +-6x + 5.

pjTiffVng

phap: X^t phifcfng trinh: VT = VP (*)

VT > K

Neu tim diTcic so' K sao cho: ~

Khi do : (*) ^

vr = K

VP = K

VP<K

(tu-c cac dau "=" xay ra )

GiSi

Ta cd:

Dau xay ra: ^

^Ix" -4;c + ll + V 3 / - 6 x ^ +28

= ^l[x^ -2x^\) + 9 +

p

[x'

-2

A : ' +

I)

+ 25

= ^2(.

X

-1)'+9 + ^3(

A

;'-1)'+25 >79 + V25

= 3 + 5 = 8.

( x - l f = 0

br^-l = 0

Matkhac: -3

A

;^+6

JC

+ 5 = -3(

A

:^-2x + l) + 8

= - 3 ( ; c - l ) ' + 8 < 8 .

Da'u "=" xay ra: <^

X

- 1 = 0

<^ X

= 1.

E)od6: V2;c' - 4 x +11 + ^3.^' -6

JC

' + 28 = -3

A

:' + 6;

C

+ 5

V

2

A

;'-4;

C

+ 11+

V

3

A

:'-6

A

:'+28 = 8

</div>
(69)<div class='page_container' data-page=69>

x = l

<=> x = \. 
PhiTdng tnnh c6 nghieir duy nha't x = 1.

136

§ 4.

PHLfdNG

TRINH

V

6 |

NGHIEM

SO

NGUYEN

pi 5.53: H m so' nguyen a sao cho phifcfng tnnh:

[x-a)[x-lO) + l = 0 CO nghiemdeu la so'nguyan.

T a c 6 : {x - a){x-\0) + l = 0 ^ {x - a){x-10)=-I (*) 
Neu xEZ,aeZ thi ta c6 x - a e Z,x-10 E Z .

x — a — \ — a = —l 
hay 
x-\0 = 
Do do tiy (*) ta suy ra rang: hay

-1 [ ; c - 1 0 = l
a = x — \

x = 9 hay

a x +1

X

= n

nan : a = 8, a = 12.

* V d i a = 8 thi phuTdng trinh troi thanh:

(x-S)(x-\0) + l = 0^ x^ -\Sx + Sl = 0 • x = 9 . 
* V d i a = 12 thi phiTdng trinh trd thanh:

{x-l2){x-\0) + \ 0^x^ - 2 2A: + 121 = 0 <^ A: = 11
Vay a = 8, a = 12 la gia tri can tim.

5.54: T i m nghidm nguyen cila phifdng trinh:
4xy + 2x + 2y = 2S.

Giii

Ta cd: 4xy+ 2x+ 2y = 2^ ^ 4xy+ 2x+ 2y+ \ 29 
^ ( 2 x + l)(2>' + l ) = 29

V i xeZ,yeZ nen 2;c + 1 G Z, 2^ + 1 € Z do d6 phiTdng trinh 
trd thanh:

2x + l = l \2x + \ 29 
hay

2 j + l = 29 [2>' + l = l

2x + l = - l f2v + l = - 2 9
hay '

2)^ + 1 = - 2 9 ;?.;y + l = - l

</div>
(70)<div class='page_container' data-page=70>

F)e 5.55: T i m nghiem nguyen cua phu'dng trinh:

2xy + x + y = S3.

4^

hay

T a c d : 2xy + x + yS3 ^ 4xy+ 2x+ 2y = 166 
^ 4^;}^ + 2A: + 2>'+1 = 167 ^ (2x + 1 ) (2^ + 1 ) = 167

'2x + \ \A: + 1 = 167

hay

2>' + l = 167 [2>' + l = l

2x + \ -l , f2x + l = - 1 6 7 
hay

2 j + l = - 1 6 7 |23; + 1 = - 1
^ ( x = 0,>'=:83) hay [x = S3,y = 0)

hay{x = -\,y = -U) hay (x = - 8 4 , ) ' = - l )

D e 5.56: T i m nghiem nguyen dtfcfng cua phu'dng trinh:

[x^+4){y^+\){z'+25) = S0xyz (*)

Gi^i

Ta bie't rang v d i a > 0, b > 0 thi a + 6 > yfcib va d i n g thitc chi
xay ra khi a = b.

Ap dung ket qua tren, ta suy ra:

+4>2yj4x' =4x

/ + l

>2V7 = 2x

( V I X > 0)

(vi y > 0)

z^+25>

2^/257

= lOz (vi z >

0)

Nhan ve'theo ve ba bat d i n g thitc vuTa neu thi c6:

(x' +4)(y' +l)(z' +25)>S0xyz

Do dd:

; c ' + 4 = 4x

y^ +\ 2y ^

z ' + 2 5 = 10z

x' =4

/ = 1

4=>

z' =25

x = 2 
y = l 
z = 5.

Vay X = 2, y = 1, z = 5 la nghiem can tim.

f)i 8.57: T i m nghiem nguyen cua phu'dng trinh: 
4x' +Sx^y-4y + 3y^-n = 0.

GJii

Phu'dng trinh du'dc viet

thanh-4(x' +2x^y + y^)-[y^ +4> ' + 4 ) - 1 3 = 0 
^4[x^+yf-{y + 2f =13

^[2x^+3y + 2)[2x^+y-2) = \3

hay

2x^+3y + 2 = l 
hay

2x^ + y-2 = l3

2x^+3y + 2 = -\ 
2x'+y-2 = -l3

2 x ' + 3 y + 2 = 13

2x^ + y-2 = l 
.2

hay

2x'=23

y^-S hay

2x'=-\ 
y = 4

2 A : ' + 3y + 2 = - 1 3

2x' + y-2 = -\ 
2x'=-\5

y = 4 5

hay hay 2x^=9

y = -S.

Vay phu'dng trinh khong cd nghiem nguyen.

De 5.58; T i m tat ca cac so' nguyen x thoa: x^ + S = l^Sx +1.

D i ^ u k i e n Sx+ 1> 0 ^ Sx >-I ^ x > 
~ . ~ ~ 8

Do xeZ n e n j c > O . T a c d :

T a c d : x^ +S = l^lSx + \ +sf =hJSx+ \]^ 
^ A: ' + 16A;'+64 = 49(8A: + 1)

<i=>x' + 1 6 x ' - 3 9 2 x + 15=:0

^ x^ -3x' +3x' -9x' + 9x' -21 x^ + 43JC' 
- 1 2 9A: ' + 1 2 9 ;C' - 3 8 7X- 5J C + 15 = 0

<^ x'(x-3) + 3x^ ( x - 3 ) + 9 x ' ( j c - 3 ) + 4 3 x ^ { x - 3 )

+ 1 2 9A : ( X- 3 ) - 5 (X- 3 ) = 0

4^(x- 3){x' + 3x' + 9x' + 43;c' + 129 A: - 5) = 0

</div>
(71)<div class='page_container' data-page=71>

x-3 = 0

x' + 3x' + 9x' + 43x^ + 1 29JC - 5 = 0
x = 3

A: ' + 3 ;C' + 9A: ' + 4 3X' + 1 2 9J C- 5 = 0 (*)
X e t phufdng trinh (*):

* x = 0 khong la nghiem ciia (*) V I- 5 0
* x ^ O t a c o ; c > O v a x e Z n e n x > l .

D o d d : X' + 3A: ' + 9A; ' + 4 3A: ' + 1 2 9J C> 5 .

=^ (*) khong c6 nghiem nguyen khac 0.
Vay phufcfng tnnh chi c6 mot nghiem nguyen la x = 3.

Bi 5 . 5 9 : T i m ta't ca cac so' nguyen x, y, z thoa phu-dng trinh
3x^ + 6y^ + 2z' + 3yh^ - 1 8 x - 6 = 0 .

Gi^i

Ta c6: phu'dng tnnh da cho

^ 3(;c' + 2y' + yh'-6x-2) + 2z' = 0.

=^ chia he't cho 3 =^ z chia he't cho 3 ^ z^ chia he't cho 9.
• X ^ t z' = 0 : Phu'dng tnnh trd thanh :

3A: ' + 6 / - 1 8 x - 6 = 0

^ 3 ( ; c - 3 ) ' + 6 / = 3 3

< ^ ( A: - 3 ) ' + 2 r = 1 1

D o d d : 2 / < 1 1 = > / < 2 ' = 0 ' ; 1 ^ 2 '
* / = 0 ' = » ( j c - 3 ) ' = l l V 6 ly !

* y' = l'=^[x-3f=3'=^x-3 = ±3

^ X = 6 hoac x = 0.

Phu'dng tnnh cd cic nghidm

(J: = 6;y = l;z = 0);(x = 6;y = - l;z = 0)

(x = O;}-= l;z = 0);(x = 0; J = - l ; z = 0)
140

* / = 2 ' - ^ (A; - 3 ) = 3 V6 ly !

• X ^ t z' > 9 : Ta cd

3A: ' + 6 / + 2 z ' + 3 ) ; ' z ' - 1 8 ; c - 6 = 0

^ 3 ( ^ - 3 ) ' + 6 / + 2 z ' + 3>''z' = 33

* / > 1 t h i 2 z ' + 3 / z ' > 2.9 +3.1.9 > 33 (loai)
* / = 0 thi 3 ( x - 3 f + 2 z ' = 3 3

V d i : * z' = 9 thi 3 (x - 3)' = 15 (loai)

* =^z'>6^ =36

T a c d 3 (A: - 3 ' ) + 2Z' > 3 3 (loai)

T d m l a i nghiem nguyen cua phu'dng trinh la:
(;c = 6;); = l;z = 0);(x = 6;^ = - l ; z = 0 ) ;

(;c = 0-y = l;z = 0);{x = 0;y = - l ; z = 0)

5 . 6 0 : T i m nghiem nguyen cua phu'dng trinh:
5 x ' - 4 x y + / = 1 6 9 .

Ta c d : 5 x ' - 4; c > ' + / = 1 6 9 4 4> ( 2 x- ) ; ) ' + J c ' = 1 6 9

< ^ ( | 2; c - > ' | ) ' + | A : f = 1 6 9 .

Trong dd: | 2 - y] 6 |x| 6
Nhirng 169 = 13' + 0 ' = 1 2 ' + 5 ' .
Do dd cd cac kha nang sau:

l 2 A : - y l = 1 3

\x\0

I 2 x - y l = 0 
I A: I = 1 3
l 2 A : - y l = 1 2

IA:I = 5

;c = 0

y = 13

A: = 13

y = 26

x = 5

V

y = - 2

V

x = 0 ^

y = - 1 3

A: = - 1 3

y = - 2 6

x = 5

y = 22

V

= - 5
= - 2 2

x = - ^
v = 2

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\2x-y\5

I A I - 1 2

;c = 12 x = l2 x = -l2

y = l9 3; = 29 ^ - - I Q

De 5.61: T i m nghiem nguyen cua phu'cfng trinh:

2 " + 2 " + 2 ^ = 2 3 3 6 ( v d i x < y < z ) .

Gidi

T a c 6 : 2 " + 2 ^ + 2 ^ = 2336 ^ 2^ ( l + 2^"^+2^"^) = 2'.73 (1)
V i 1 + 2 ^ - " + 2 ^ - M a s6Me nen:

( 1 ) ^ 2 ^ = 2 '

1 + 2^-^+2^-^=73

Ta c6: (2) <^ 2^"' ( l + 2'-' ) = 2\9 
V i 1 + 2 ^ - M ^ n6n:

x = 5

( 3 ) ^ 2'-' = 2' 
1 + 2'-' = 9

y = x + 3 
z-y = 0

y — X — 3

2'-' = 8

> = 8

z^3 + y

y = \i

z =

n

Vay phufdng trinh da cho c6 nghidm Ih.: 
(x = 5;>' = 8;z = l l ) .

(2)

(3)

De 5.62: T i m nghiem nguyen dtfdng cua phu'dng trinh:

2 ^ + l - / = 0 .

Giii

Ta c6: 2^ + 1 - = 0 ^ 2 ' = - 1

^ 2 ^ = ( y - l ) ( y + l ) (*)
Tir (*) suy ra y - 1 va y + 1 la ufdc so ciia 2" (vdi xeN)

nen chiing phai CO dang: y - 1 = 2',>' + l = 2 ^ ( v d i 0 < / < 7 ' ; iJeZ) 
Ta cd: 2^ - 2' = y + 1 - (); - 1 ) = 2

= ^ 2 ' ( 2 ^ - ' - l ) = 2

142

IT

2' = 2

2^-' - 1 = 1 hay

2" = 1
- 1 = 2
=^ (/ = 1,;- = 2) (con 2'-' = 31a v6 l y )

= 3 va do do 2" = 8 = 2^ = 8 = 2'
Vay nghiem nguyen du'dng can tim la : x = y = 3.

^ x = 3.

0 6 5.63: T i m nghiem nguyen du'dng ciia phu'dng trinh:

(*)

1 1 1 1 ,

GiSi

Tuf (*) =^ ^ > l,y > l,z >l,t>l ( v i - ^ < 1 , - ^ < 1,4- < 1>4- < 1)

x^

/

z^

e

* Neu X = y = z = t = 2 thi (*) thoa man

* Neu mot trong bon so' x, y , z, t cd it nhat mot so' Idn 
hdn 2 thi:

1 1 1 1 1 1 1 1 ,

+ -^ + ^ + ^ <- + - + - + - < !

- ' / ' ' ~ 9 4 ' 4 4

(Gia svl A: > 2 = ^ X > 3 ^ ^ < - con y, z, t deu Idn hdn hay

X 9

b k n g 2 n e n 4 x^ 4 z 4 r 4 - < 7, 4 - < j )

Vay x = y = z = t = 2 1a nghiem can tim.

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CHl/dNGVI: H E PHLfOfNG T R I N H

§ 1. HE HAI PHUdNG TRJNH BAC NHAT HAI AN

HE BA PHadNG TRINH BAC NHAT BA AN

I. He hai phifrfng trinh bac nhg^t hai in:

ax + by = c 
a'x + b' y = c'

II. He ba phvfrfng trinh bac nhfi't ba an:

La he phifdng trinh c6 dang:

La he phiTdng trinh c6 dang:

ax + by + cz = d 
a'x + b'y + c'z = d' 
a"x + b"y + c"z = d"

Bi 6.1: Giai cac he phircJng trinh sau bang

[3x- y = 5

phifcfng phdp cong dai s6':

'x + 3y = \0 
x-2y = -5

1) Ta c5: • 2x + y = 5 (1) i5x = \0 ((l) + (2)) 
[3x-;; = 5 (2)^[3x-y = 5

x = 2

h = 2

Vay he da cho cd nghiem: (2, 1).

2) Tacd: < x + 3>^ = 10 (3) rx +<=> 3>^ = 10

-2y = -5 (4) [5j; = 15 ((3)-(4))

144

x = \0-3y 
x = l

y = 3.

Vay he da cho c6 nghiem: (1,3).

Bi 6.2: Giai cac he phifdng trinh sau:

'2(x + >') + 3(x->') = 4

1)

(x + ;;) + 2(x->') = 5. 2)

2(x-2) + 3(l + >^) = -2

[ 3 ( x - 2 ) - 2( l + ;;) = -3.

1) Tacd: •

Glii

2(x + >') + 3(x-;;) = 4 (Sx-y = 4

(x + y) + 2{x-y)^5

(1) 
t3x->; = 5 (2)

2x + 0 = - l ((l)-(2))

y = 3x-5

2) Tacd: •

Vay he cd mot nghiem:

2(x-2) + 3(l+>') = -2

X = —

2

13

2' 2 
3(x-2)-2(\ y) = -3

(3) 
(4) 
2x + 3;; = - l

3x-2^ = 5 
'4x + 6y = -2

9x-6y^\5

'I3x + 0 = 13 ((3)4(4))

l6>/ = 9x-15 
x = l

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<=>

6.3: Ta bie't rang: Mot da thiJc bkng da thitc 0 khi va cW khi tat

c& cac hd so'ciia n6 bang 0. Hay tim cac gia tri ciia m va n de da

thiJc sau day (vdi bien so' x) bang da thtJc 0:

P(x) = {2m + n- 5)y + {5m-n-9).

Ta c6: P{x) = 0 ( V x ) { 2 m + n-5)x + (S/w-«-9) = 0 (Vx)

'2w + «-5 = 0

[5m-n-9 = Q

'2m + n = 5 (1)

\5m-n = 9 (2)

'7/77

+ 0 = 14 ((l) + (2))

77 = 5

/77

-9.

f/77

= 2

'/77

= 2

= 1.

Vay:

6.4: Giai cac he phiTdng trinh sau bang phifcfng phap the:

^x-2y = -3 \3x + y = -5

1) 2x + 3>' = 8.

2x + 3y = -S.

1) (ly.

Gi§i

x-2y = -3 (1)

2x + 3>^ = 8 (2)

Tuf phtfcJng trmh (1), bieu dien x theo y ta dtfdc: x = 2y - 3.

Tha' vac phifcJng trinh (2) ta du'dc:

2(2>'-3) + 3>' = 8 o 7 > / = 14

<^ y = 2

[y = 2 \y = 2

Vay he (I) cd mot nghiem: (1,2).

146 2) (//>

3x + y = -5 (3)

2x + 3>' = -8 (4)

TuT phu'dng trinh (3), bieu dien y theo x ta dufdc: y =

The vao phiWng trinh (4) ta duTdc:

2x + 3(-3x-5) = -8<::>-7x = 7 ^ x = -l.

-3x - 5.

Dodo: (//) y = -3x - 5

x = - l

y = -2

x = - l

Vay he (II) cd mot nghiem: (-1,-2).

x-y 2x + y _^

6.5: Giai he phu'dng trinh: 1 17

6.5: Giai he phu'dng trinh:

^^ + ^ + ^ - ^ = 1 5 .

[ 5 19

x

—

y 2x-\-y

7 17

Ax

-\-y y — 1

= 7

= 15

5 19

31x-10> = 833

76x +24^ = 1460

93x-30^ = 2499

95x + 30y = 1825

x = 23

30)' = 1825-95x

x = 23

y = - \

4=^

17x-17>' + 14x + 7>' = 833

76x + 19>' +5)^-35 = 1425

31x-10)' = 833

19x + 6>' = 365

188x = 4324

95x + 30>' = 1825

x = 23

30y =.-360

Nghiem cua he phufdng trinh la x = 23

y =

-n

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6.6: Cho he phufcfng trinh (m la tham so'): mx — y = \ 
—x + y = -m 
1) ChiJng to khi m = 1 he phufcfng tfinh c6 v6 so'nghiem.
2) Giai he tren khi m ^ 1.

1) Khi m = 1 he phufcfng trinh trci thanh:

Ox^O xeR 
y — x — l 
x — y = l

—x + y — —\ — x — \ 
He phtfcfng trinh c6 v6 so'nghiem.

inx — y = l {mx — x = l — m 
—x + y = —m [—X -\-y — —m 
xim-\) = \-m \x = - \ 
y = -m-{-x [>' = - m - l
He phifcfng trinh cd nghiem [x = — \\y = —m —.l).
2) Khi m ^ 1:

6.7: Cho he phufcfng trinh hai an x, y: —2mx + y = 5 
mx-\-3y = \ 
1) Giai he phufdng trinh khi m = 1.

2) Giai va bien luan he phufcfng trinh theo tham so'm.

GiSi

1) Khi m = 1 he phiTcfng trinh trd thanh :
-lx + y = 5

x + ?>y = \

-6x + l,y = \5 
x + 3y = \

-Ix^U 
jc +

3y

= l
x^-2 
\3y = 3 
Nghiem cda ht phufcfng trinh Ik (jc = - 2 ; y = l ) .
2) T a c 6 : ( " ^ - ^ + ^ = 5 ^ + 3^ = 15

{mx-\-'iy = \ = \

x = -l 
- 2 +

3 j = l

x = -2 
y = \

148

4=>

-7m;c-14

mx + 3y = 1

mx = —2

\ mx

* m = 0 : Ta c6 (*) ^

Ox = -l

. = + 3

xe0

' ^ 3 
H# phufcfng trinh v6 nghiem

x = - — 
m 
* m ^ 0 : Ta CO ( * ) <^

1 - m . - 2 ^

y = m

m 
y = \

He phifdng tfinh c6 nghiem la

__2_ 
m 
y = \

(*)

6.8: Cho he phufdng trinh: 2x + 3y = m

Sx-y = \

(1)

1) Giai he ( l ) k h i m = -3.

2) Tim gia tri cua m de he (1) cd nghidm (x > 0, y < 0).

Giii

1) Khi m = -3 he phu-dng trinh (1) trd thanh :
2x + 3y = -3

5x-y^\

2x-]-3y = -3

15x-3>' = 3

;c = 0

17A: =

0

15A:-3>' = 3
4^

-3y = 3

x = 0
y = - \
Vay he phufdng trinh cd mot nghiem (x = 0, y = -1)
2) Giai he (l)tadu'dc:

2x^-3y = m 
5x-y = \

2x + 3y = m

Jl7.x = m + 3

15;c-3y = 3 ^ 1-3>' = 3-15J:

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X = m +

3

17 ^

y = 5x-l

X = m +

3

17 _ 5(m + 3)

~ 17

X =

y =

m + 3

17 5m + 15-17

17

X —

y = 
m + 3

TT

5 m - 2

17 He (1) c6 nghiem thoa (x > 0; y < 0)

m + 3

5 m - 2

X =

> 0

17 < 0

m + 3 > 0

5 m - 2 < 0

4^

m > - 3

5m < 2

m > - 3

0

3 < m <

-m <

J

5 Do dd khi - 3 < m < - thi he (1) c6 nghiem thoa (x > 0; y < 0)

5

6.9: Cho he phiTcfng trinh

mx + my =

(l-m);c + ); = 0 (2)

—3 (1)

1) Giai he phiTcfng trinh khi m = 2.

2) Tim m de he c6 nghiem (x < 0; y < 0).

Giii

1) Vdi m = 2, he phuTOng trinh trd thanh :

2;c + 2y = - 3

-x + y = 0

x-\-y =

y = x

2x = 
--2 ^ 
y = x

150 Vay vdi m = 2 he phifdng trinh cd nghiem la:

3 3^

2)

4=^

mx + my = -3 
{\-m)x + y = 0

mx + my = —3 
y = -{\-m)x

mx -m{\-m)x = -3 
y = {\-m)x

i

^m-m

+ m^^x

= - 3

y --{\-m)x

m 'x = -3

x<0 
y = -{l-m)x

Nghiem thoa dieu kien

[y <^

Vay m > 1 la gia tri can tim.

< ^ l - m < 0 4 = ^ m > l .

Bi

6.10: Cho he phifdng trinh:

mx — y = 2 3x + my = 5

(1)

1) Giai he (l)khim= 1.

2) Vdi gia tri khac 0 nao cua m thi he (1) cd nghiem thoa man :

x + y = \ m

m ' + 3

1) Khi m = 1 he phifdng trinh (1) trd thanh :

x — y — 2 
3x + y = 5

4x = l 
y = x — 2

_ 7

_ 7

^ ~ 4

1 J Vay khi m =1, he phiWng tnnh cd mot nghiem

2) Vdidi^ukien m^tO.tacd:

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mx

—

y =

2

3x

-\- my =

5

[m^ ^-?>)x = 2m + 5

2

)X-\-my

—

5

rri^x

—

my = 2m

3x + mj = 5

2 m + 5

X =

+ 3

3.(2m + 5)

m ' + 3

-\-my = 5

X =

2m + 5

+ 3

6m + 15+ m(m^ +3)y = 5m^ +15

4^

X —

2 m + 5

m' + 3

m(m^ + 3\y = 5m^

— G;??

2 m + 5

m ' + 3

(5m-6) 4=^

m(m^

+ 3 )

X =

2m+ 5

m ' + 3

5m — 6

m' + 3

He phiTdng trinh (1) c6 nghiem thoa: A: + y = 1 m^

m ' + 3

khi va chi khi:

2m + 5 ^ 5 m — m

m ^ + 3 m ^ + 3 m ^ + 3

<^ 2m + 5 + 5m - 6 = m^ + 3 - m^

44.7m = 4<^m = — 7

4

Vay khi m =

—

he phifcfng trinh (1) c6 nghiem thoa:

x+y=l-

m' + 3

6.11: Gia su" h6 phifdng trinh sau day c6 nghiem:

Chtfng minh r^ng = 3abc.

ax + by = c

bx + cy = a

cx-\-ay — b

GiSi

Goi (j^o '>o) 1^ nghidm cua hS phu'dng trinh da cho.

152 + ^Jo = c

(1)

Tac6: hx^ + cy^ = a (2)

+a>o =

Z

7 (3) . _

Nhan hai ve ciia (1), (2), (3) Ian tot vdi ,b^ ta c6

cb^x^+ab^y^=b'

=^a^+b^+c^ =[a^b + b^c + ac^)xQ+[a^c + ab^+bc^)y^ (*)

Nhan hai ve' cua (1), (2), (3) Ian liTdt vdi ab, be, ac ta c6: .

a^bx^ + ab^y^ = abc

b^cxQ + bc^y^ = abc

UC^XQ + a^cy^ = abc

=^3abc =(a^b + b''c + ac^)x^+[a^c + ab^+bc^)ya (**)

So sanh (*) va (**) ta du^dc : a' +b' +c' = 3abc.

6.12: Giai he pht/dng trinh:

'x^y + z=^\

x + 2>' + 4z = 8

x + 3y + 9z--21

Giii

x + y + z = l (1) x+y+z=\

Taco: x +

2y + 4z = S ( 2 ) ^ y + 3z = l ((2)

-(!)) (4)

x + 3y + 9z = 21 (3) y + 5z = 19 ((3)

-(2)) (5)

x+y+z=l

'x = \-(y^z) x^6

mjt ^

y + 3z = l ^

y = l-3z

y =

-n

lz = \2 ((5)-(4)) z =

e

z = 6

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Nghi6m cila he phiTdng trinh la

x = 6 
z = 6 
De

6.13:

X y z

Giai he phtfcfng trinh : 6 - 1 0 -2 
4x + 3y-2z = -\

Ta CO :

X ^ y _ z

6 - 1 0 ~ - 2
4x + 3y~2z = -l

6 - 2
y z 
- 1 0 - 2
4x + 3y-2z

X = - 3 z

y = 5z <^ 
z = - l

= - 1

'x = 3 
y = - 5 
z = -l

X = —3^

y = 5z

- 1 2 z + 5 z - 2 z = - 1

Vay he phufdng trinh c6 nghiem l a : ( x = 3 ; y = - 5 ; z = l )

x + y = Sz-l

6.14: Giai he phiTdng trinh ^ y + z = ^4x-l

Z + x = ^4y-l

dm

Dieu kien: x,y,z>-.

4

Nhan vao m6i phifdng trinh vdi 2 r d i cong lai, ta diTdc:
4x-2y/4x~l+4y-2^4y-l+4z-2.j4z-\^0

154

{4x-\) -2V4JC-1 + l] + [ ( 4} ; - l ) - 2 ^ 4 ^ + l] + [ ( 4 z- l ) - 2 ^ 4 ^ - 1 + 1
( , y 4 l ^ - i f + ( V 4 > ^ - l ) ' + ( 7 4 z ^ - i f = 0

= 0

V 4 x - 1 = 1

^4y-\=\^

V 4 z - 1 = 1

^ = 2

vay h6 CO nghiSm duy nhat 1 I 1

De 6.15: T i m m o i x, y, z trong phifdng trinh:

x + y + z + 4 = 2ylx-2 + 4 y J y - 3 + 6 V z- 5 .

GiSi

x - 2 > 0 [ ; c > 2
D i e u kien: y - 3 > 0 < ^ y > 3

z-5>0 \z>5

T a c o : x + y + z +4 = 2ylx-2 + 4^y-3+6ylz-5

x + y + z + 4-2^f7^-44y^-6^[z^ = 0

f ( A : - 2 - 2 ^ / ^ ^ ^ + l ) + ( y - 3 - 4 7 y ^ + 4) + ( z - 5 - 6 ^ / F ^ + 9) = 0 
V ^ - l f+ ( A/ r ^ - ^ 2 ) ' + ( V F= 5 - 3 ) ' = 0 •

( V ^ - 1 )
( V ^ - 2 ) ' = 0

( V ^ - 3 ) ' = 0

^ 7 ^ - 1 = 0

^ V > ^ - 2 = 0
V F^ - 3 = o

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- 2 - 1 X - 2 = 1
- 3 = 2 ^ ^ y - 3
- 5 = 3 z - 5 = 9

Nghiem ci5a he phu'cfng trinh la : [x = 3;y = 7;z = l4).

156 ,

§ 2. H E GOM MOT PHadNG TRINH BAC NHAT

V A

MOT PHLfdNG

TRINH

BAC CAO

I . Dang: 'ax + by + c = 0 (1)

fix;y) = 0 (2)

trong d6: f(x,y) = 0 la mot phu'cfng trinh bac cao theo hai an x
va y. •

I I . Phifflfng phap:

1) TO phu'cfng tnnh (1) rut mot an x hoac y theo an con lai.
2) The vao phu'cfng trinh (2) de du-a phu'cfng trinh (2) ve

phu'cfng trinh mot an.

De 6.16

1) ^

: Giai cac he phifcJng trinh sau:
j c - v + 2 = 0

+xy = 4

x-y = 0

x'^ +xy-y^ =1.

x-y + 2 = 0 (1)
x^+xy = 4 (2)

Giil

Tacd: (1)<^>' = A: + 2.

The vao (2): x'' +x{x + 2) = 4^2x^ +2x-4 = 0 <^ 
• V d i : X = 1 thi y = 1 + 2 = 3

;c = l

x = -2 
• Vdi: x = - 2 thi y = - 2 + 2 = 0.

Vay he c6 hai nghiem: (1; 3) ; ( - 2; 0).

x-y = 0 (3)

x'+xy-y^=\)

fc. Ta c6: (3) ^ y = X th6' vao (4)

x^ +x.x-x'^ = x^ =1^ x = ±\. 
2)

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• Vdi: X = - 1 I h i y = - 1.

• V d i : A ^ 1 thi y = 1

Vay he c ) hai nghidm: (-1; -1) ; (1, !).•

6.17: Cho he phifdng trinh: x + y = l

x' =m{x-y).

1) Khi m = 1: he trd thanh:
1) Giaihe khi m = 1.

V d i gia tri nao cua m thi he c6 ba nghiem phan biet ?
Giai

x ' - / = x - y (2)
Ta C O : (1). ^ >'= 1 - A; . Th6'vao (2):

x' -{\-xf =x-{\-x)<=^ x'-(\-3x + 3x^-x') = 2x 
2A: ' - 3A:^ + X = 0 <^ x(2x^ - 3A: + l ) = 0

x = 0

2A: ^ - 3X + 1 = 0 x = l ^y = 0)

2 2

Vay he da cho c6 ba nghiem: (0;1) ; (1 ;0) ; ( .
2 2

2) T a c d :
x + y = \

x' =m{x-y) 
x + y = l

{x-y)(x^+xy + y^) = m{x-y 
x + y = \

{x-y)[x^ +xy + y^-m)=^0 
y = \-x

x-y^O v ( / / )

y = l — x

x^ ^xy + y^

158

a. Tacd: (1) <^

y = \ x

x-[\-x) = 0

Vay ha (I) C O mot nghiem:

y = 1 — X

U

'2j

b. Xethe (//)

y = \ X

1 <^

X = —

2 •

(3)

X — —

2 
1

x^ +xy + y^ -m = 0 (4)
Thay: y = 1 - x tiY (3) vao ( 4 ) :

x^ +x(l-x) + (l-xf -m = 0<F^ x'^ -x + l-m = 0 (*) 
He da cho c6 ba nghiSm phan biet 4=^ (*) cd hai nghiem

phan biet khac ^

A > 0
•l1 2

2. l2,
3

m > —

-4 3
, ^ m>-.

3 4

4 4

44>

l - 4 ( l - m ) > 0
3

m ^ —
4

Vay khi m > - h6 da cho c6 ba nghiem phan bidt.

6.18: Giai he phiTdn^ tiinh:

x + y = A (1)
/ + / = 8 2 (2)

( l ) ^ y = 4- A: . T h a ' v a o ( 2 ) : x' + ( 4 - ; c ) ' = 8 2 (*)
Dat: t = X - 2 <^ A: = / + 2 . Thi:

( * ) ^ ( / + 2 y + ( / - 2 ; = 8 2

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^ t' +4.t\2 + 6.t\4 + 4.t.S + l6

+ t' ~ A

.e

.2 + 6.r' .4 - A.t.9, +16 - 82

<^2t'

+48^^ - 2 5 = 0

^t' +24r^-25 = 0

r = -25(1)

Vdit= 1, tacd: x = 3 ; y = 1.

Vdit = - l , t a c d : x = 1 ;y = 3.

Vay h6 c6 hai nghiem: (3, 1); (1, 3).

§ 3. HE DOI XUfNG - HE DANG CAP

I. d6

'i xiJng lo^i

I:

1) Binh nghia: H6 : •

duTdc goi Ik M dd'i xtfng loai

I n6u khi thay x bdi y ngifdc lai thi moi phiTdng trinh trong h6

khong thay ddi.

2)

Phifdng phap:

• Dat:S = x + y ;P = x.y.

• Dura he da cho vi he c6 hai an S; P.

• Tim S; P. Khi dd: X, y Ik nghiSm ci5a phifdng trinh:

X'-SX + P = 0.

3) Chil v:

a) Dieu kien c6 nghiem: -4P>0.

b) Neu (x, y) la mot nghiem ciia h6, thi (y, x) cung la mot

nghi6m ciia h6.

c)

*

+ =(x + yy -2xy = S^ -2P.

3P).

- (x + y){x^ -xy + y^)

= (x + yiix + yy-3xy\=S.{S'

* =(x

' + / ) ' - 2 x V

= [(x + yy-2xy\

= [S'-2PJ-2P\

II. H$ dfi'i xuTng lo^i II:

1) Bmh nghia: He:

• U(^,)') = G (2)

dtfdc gpi la h6 d6'i xiJng loai II na'u Khi thay x bdi y VJ» ngtfdc lai

thi phufdng trinh (1) bien thanh phufdng trinh

(2)

va phudng tiinh

(2)

bien thanh phifdng trinh (1)

2)

Phifdng phap:

• Lay (1) -

(2)

de difa v^ mot phiTdng trinh dang tlch s6' va

giai nghiem nay theo nghiem kia.

_ • The vao mot phifdng trinh trong he de

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III. Hf dang cfi'p:

1) Binh nghla: La he phiTdng trinh c6 dang:

ax^ + bxy + cy^ = d

a'x^ +b'xy + c'y^ = d'

2) Phifdng ph^p: X6t hai trifdng hdp:

• THl:

X =

0: Tha' trifc tiep vao he de giai.

• TH2: x^O: Dat: y = kx.

- Tha' vao he, khur x ta difdc mot phifdng trlnh theo k.

- Giai tim k, iJng vdi mSi trifdng hdp c\5a k ta tim dtfdc

nghidm (x, y) ciia he.

Ghi chu; a) C6 the xet hai trifdng hdp: y = 0 va y^O.

b) Khi he phiTdng trinh c6 bac cao hdn 2 ta van c6 th^

dung each giai tren n6u cac so'hang

d

ve trai ciia he R

ding cap.

6.19: GiSi he phtfdng trlnh sau: +y^ +x + y = S (1)

x'+y'+xy (2)

Giii

Day

Ik

mot he do'i xiJng loai I

Dat: S =

X

+ y va P = x.y

Tsic6:* {l)^{x + yf-2xy + x + y = S

4=»5'-2P + 5 - 8 (1')

* (2)^{x + yf-xy = l

<^S^~P = 1 (2')

TO (2') suy ra: P = 5' - 7. The P vao (1') ta cd:

5 ' - 2 ( 5 ' - 7 ) + 5 = 8

<=^-S^ + S + \A = S^S^-S-6 = 0^S = 3,S = -2

• Vdi S = 3 thi P = 2:

X,

y la nghiem ciia phufdng trinh:

X ' - 3 X + 2 = 0 ^ X = 1,X = 2

nencd {x = l,y = 2y,{x = 2,y = l)

• Vdi S = - 2 thi P = - 3: x, y la nghiem cua phufdng trinh:

162

+ 2X - 3 = 0 <^ X = 1, Z = - 3

nencd [x = l,y ^-3);{x =-3,y = \)

Vay h6 cd bdn nghiem gom:

{x = \,y = 2);{x = 2,y = l);{x^ l,y = -3);{x = -3,y = l),

f)i

6.20: Giai he phrfdng trinh: x^ +xy + y^ = 4 (1)

x + xy + y = 2 (2)

(!')

(2')

Gi^i

Day la h6 do'i xi^ng loai I:

Dat: S = x + y; P = xy.

Tacd:* (1) ^ {x + yf -

xy4-* {2)<:^{x + y) + xy = 2

^S+P=2

TO (2') suy ra: P = 2 - S. The vao (1'):

5 ' - ( 2 - 5 ) = 4 < ^ 5 ' + 5 - 6 = 0.

S = -3

[5 = 2

• Vdi S = - 3 thi P = 5 ;

X,

y la nghiem ciia phtfdng trinh:

+ 3X + 5 = 0 . Phtfdng trinh v6 nghiem.

• Vdi S = 2 thi P = 0:

X,

y Ik nghiem ciia phiTdng trinh:

X = 0

X = 2

n6n cd: (x = 0 ; y = 2); (x = 2; y = 0)

Vay he da cho cd hai nghiem: (x = 0 ; y = 2); (x = 2; y = 0).

, X ' - 2 X = 0<^

6.21: Giai he phufdng trinh: x'=3x + Zy (1)

/ = 3 > ' + 2x (2)

GiSi

Day

Ik

he dd'i xiJng loai II.

. (\)-(2): x'-y'={3x + 2y)-{3y + 2x)

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<^{x- y){x + y) = x-y<?^{x- y){x + y-l) = 0 
x-y = 0

[x + y-\ 0 
• Y6i: x-y = 0^y^x.Th€wa.o(\)

x^ =3x + 2x x^ -5x = 0 
\x = 0 (y = Q) 
^[x = 5 (y = 5)

• W6i: x + y-l = 0^y = l-x. Th6'w^o{l): 
x"" =3x + 2{l-x)<:^ x^ - x - 2 = 0

x^-\ = 2)

^\y = 2 {y = -l) 
Vay he c6 bon nghiem:

(x = 0,y = 0);(jc - 5,y = 5);(x = -\,y = 2);{x = 2,y = -1)

Bi 6.22: Giai he phifdng trinh: x' =3x + Sy (1)

y'=3y + Sx (2)

Giii

Day la he do'i xiJng loai II.

. il)-(2y.x'-/={3x + Sy)-{3y + Sx)

( J : -y)(x' +xy + y') = -5{x-y). 
<^[x-y)[x^ +xy + y^ +5)^0

x-y = 0

x" ^xy + y^ +5 = Q

• Vdi: ;t - >- = 0 <^ > = jc. The vao (1):
x^ = 3x + %x^ x^ - nx = Q

\x = 0 {y = 0) 
O ;c

= -VrT

{y = - ^ )

= VrT (}' = ViT)

164*

4
(v6 nghiem).

Tdm lai: He da cho c6 ba nghidm:

[x = ay = 0);(x = -^\,y = -yfu),(x = ^/^T; y = ^/^Tl

p i 6.23: Giai h6 phifdng tfrnh: ^ / = ; c ' - 3 x ' + 2 ; c (1)

. ^ ' = / - 3 / + 2 ) ; (2)

Giii

Day la he do'i xifng loai II:

. i2)-iiy. x'-y'=y'-x'-3(/-x') + 2(y-x)

^(x'-y^)-2(x'-y^) + 2{x-y) = 0 ' 
^{x-y)[x'+xy + y')-2{x-y){x + y) + 2{x-y) = 0

{x-y)[x^ + xy + y^ -2x-2y + 2) = 0 
x-y = 0

x^+xy + y^-2x-2y + 2 = 0

• Vdi: x-y = 0<=^y = x.Thivko(l): 
x\=x'-3x''+2x^x'-4x^+2x = 0

^x[x^ -4;c + 2) = 0 
x = 0 (y = 0) 
x = 2 - ^ ( y - 2 - N ^ ) 
x = 2 + yf2 [y = 2 + y/2] 
• Vdi: A:^+X>' + / - 2A: - 2 > ' + 2 = 0

2;c^ + 2xy + 2}'^ - 4x - 4}' + 4 = 0

^ x^ +y^ +(^x'^ +y^ +4 + 2xy-4x-4y) = 0 
^x' +y' +{x + y-2f =^0

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x = 0

^ y = 0 (v6 nghiem) 
x+y-2=0

Tdm lai: He da cho cd ba nghiem: .

{x = 0,y^0);(x = 2-^,y = 2-^);(x = 2 + ^,y = 2 + ^ ] 
f)i 6.24: G i i i cac hS phiTdng trinh:

1) /-3xy = 4 2) x^-3xy + y^ =-\

x^-4xy + y^=l 2) 3x^-xy + 3y^ =13

Giii

y^-3xy = 4 
x^ -4xy + y'^ =1

Day Ik h6 phifdng trinh ding cap (bac hai). 
1) (/)

• Vdi y = 0: He (I) trd thanh:
• Vdi y^O: Dat x = ky, tacd:

0 = 4

x'=l (khdng thda).

/{I-3k) = 4 (1)

/ ( ^ ' - 4 J t + l ) = l

y

-3Jk/=4

ky-4k/+y'=\ 
TO dd ta cd:

1.(1 - 3/:) = 4 ( i t ' - 4^ +1) 4=^ 4jt'-13ifc + 3 = 0 
\k = 3

4
Vdi k = 3: Thd'vao (1), ta diTdc:

(v6 nghidm).

^-4

Vdi /: = 1. The vao (1), ta diTdc: y^ = 16

166

Vay he da cho cd hai nghiem:
(x = l;y = 4);(;c = - l ; y = - 4 ) .

x^-3;cy + / = - l
3x'-;c>' + 3 y ' = 1 3

Day Ik he phiTdng trinh ding cap (bac hai) 
2) (//)

Vdi X = 0: H6 (II) trd thknh:

3 / = 1 3 (khdrg thda).

• Vdi x ^ 0: Dat y = kx, ta cd:
x'-3kx'+k'x'=-l

3x'-kx^+3k^x'=\3

TOdd tacd: l3(\-3k + e) = -\(3-k + 3e) 
k = 2

<f=^2Jfc'-5^ + 2 = 04^ 1
k = —.

2
• Vdi k = 2: The vao (2), ta duTdc:

x'(\-3k + e) = - l (2)

^2^3_k + 3e) = l3

x^=l<=¥ x = \y = 2.1 = 2)

x = - l {y = 2i-\) = -2) 
Vdi it = - : Tha' v^o (2), ta duTdc:

2
x^=4^

x = 2

x^-2

, = i

.(-2) = - l

Tdm lai he da cho cd bo'n nghiem:

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6.25: Gidi hd phiTdng trinh:

xy{x-y) = 2'

Giii

Day Ik he phiTdng tfinh dang cap (bac ba).
• V d i X = 0: h6 trd thanh: =

[0 = 2
• Vdi x^O: Dat: y = kx, ta c6:

(kh6ng th6a)

He 4^ x'-k'x' =1 
kx^ {x-kx) = 2

x' (l-k') = 7 
x'k{\-k) = 2

. T i r d d t a c d :

2 ( 1 = 7^(1-it) <^2^'-7Jt'+7^-2 = 0

^{k-l)(2e-5k + 2) = 0^ 
k = l 
k = 2 
k='-.

2

• V d i k = l : T h 6 ' v a o ( l ) , tadufdc:0.= 7 (v6 nghiem).
• V d i k = 2 : T h 6 ' v k o ( l ) , t a d U ' d c :

- 7A: ' = 7 < ^A : = - 1 (>' = 2 . ( - l ) = - 2 ) .
• V d i A: = - : T h 6 ' v a o ( l ) , tadifdc:

=S^x = 2 y = -.2 = \

T d m l a i h ^ da cho cd hai nghiem: (x = -\,y = -2);(x = 2,y = l ) .

168

§ 4. HE PHJONG TRiNH CO DANG DAC BIET

x — y — xy = 3 
0 6 6.26: Giai he phifdng trinh:

x^ +y^ +xy = l. 
x^ +y^ +xy = l.

Gi^i

Dat:

Hd ^
u = X

v = -y

M + V + MV = 3
+ - MV = 1

(M + v) + MV = 3
(M + v)^

—

3uv

—

1

S + P = 3 (1) 
5 ' - 3 P = l (2)
(vdi: S = u + V ; P = u.v)

T a c d : (1)<^P = 3-S.

The vao (2): 5 ' - 3(3 - 5) = 1 <^ 5 ' + 35 - 1 0 = 0
^ ^\S^2 (P = l )

V V d i : S = 2 ; P = 1 thi u, V la nghiem ciia phifdng trinh

X^ - 2 X + l= 0 < i^ X = l < ^ M = l

v = l

x = l 
y = -\ 
• V d i : S = - 5 ; P = 8 thi u , V la nghiem ciia phiTdng trinh:

X ^ + 5 A : + 8 = 0 phiTdng trinh nay v6 nghiem
T d m l a i he da cho cd mot nghiem: (x = 1 ; y = - 1).
m 6.27: Giai cac he phiTdng trinh:

3 6

1)

2x — y x + y 1 !_

2x-y x + y I

= -l

(I). 2)

= 0

1

2x-y x-2y 2

_2 L_ = . J

-2x-y x-2y

(11).

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1) D i l u k i e n :

Dat:

u =

x~2y^0 
x + y^O

1
Gidi
(*)
V = 
2x-y 
^ thi:
x + y

4^

3M- 6V = - 1
M - V = 0

2x-y = 3 
x + y = 3 
Vay he da cho c6 mot nghiem: (x = 2; y = 1).

2x-y^0 
x-2y^0

1
2) D i l u k i e n :

Dat:

V = •

2x-y 
I 
x-2y

thi: (//) <^

2M + 3V = 
-2
2M - V = —

4^ 2A: - > ' = 12
J: - 2 > ' = 9

X =

y = 
V i y h6 da cho c6 m6t nghidm: (x = 5 ; y = -2).

6.28: Gidi hd phtfdng tiinh:

;c + y + - = 5

{x + y)-= 6. 
y

Dieu kien: y^Q. 
u = x + y 
Dat:

V — thi he <!=>

M + V = 5
M.V = 6
Do dd: u , V la nghiem ciia phiTdng trinh: 
170

- 5X + 6 = 0 <J=>
X = 2

M = 3
V = 2 V

V d i :

M = 3
V = 2

x + y = 3

^ = 2 x = 2y

« = 2
v = 3

+ y = 2

^ = 3

A; + >' = 2
X = 3^

M = 2
v = 3.

jc = 2 
y = l .

_ 3
2

T d m lai he da cho c6 hai nghiem: [x = 2;y = l ) ; 3 1
2 ' ^ 2

) l 6.29: Giai he phufdng trinh:

x - y =

[4y

-4^){\ xy)

x' +y^ = 54

GiSi

•^-y = (V>'-V^)(i.+ ^ ) (1)

x ' + / = 5 4 (2)

Dieu kien: A ; > 0 ; y > 0

• Na'u: x>y ^ 4x> d o d d : x - > ' > O v k 
( 7 y - V ^ ) ( l + ; c y ) < 0

n6n (1) sai. Vay x > y khong thda.
• N6ii: x<y=^yfx <^Jy d o d d :

x - y < 0 va -4x){\ xy)>Q 
n6n (1) sai. Vay X < y khdng thda.
• V d i : x = y.

Tilf (2) ta cd

x' +x' =5A^2x' =5^^x' =21^x' =3' ^ x = 3 {y = 'i)

I V d i X = 3 ; y = 3 thda phiWng tnnh (1).

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6.30: Giai he phiTcfng trinh: {x + yf -4{x + y) = l2

{x-yf~2{x~y) = 3.

G i i i

{x + yf-4{x + y) = \2 (1)

{x-yf-2{x-y) = 3 (2)

Datu = x + y ; v = x - y

T i r ( l ) t a c 6 : - 4M = 12 - 4M- 1 2 = 0

u = 6

u = -2 • Vay

TO(2)tac(3: - 2 v = 3 <^ - 2 v - 3 = 0

x + y = 6 
x + y = -2

v = - l

v = 3 • Vay:

(1)

(2)

TO(l)va(2) tacd:

x + y = 6,x-y = - l 
x + y = 6,x-y = 3 
x + y = -2,x-y = - l

:^+>' = - 2 ^ - l = 3

He phiTcJng trinh cd bon nghiem la:

(5 7](9 3]( 3 I] (I 5

'x-y = -l\ 
x-y = 3

j

- 1 .

-1

''~2'^~2

9 3

x = —;y = —

_ _ 3 , _J_ 
2 ' - ^ " 2

1 5

x = —;y = —.

2 2

U' 2 A 2 ' 2 J' i 2 ' 2 j ' i 2 ' 2 j

-6.31: Tim ta't cd cdc gia tri x, y thoa he: ^ ^ + / < 1

GJii

'x'+y'<l (1)

x'+y > 1 (2)

172

( 1 ) ^ x' ( A- 1 ) < 0

y'{y-\)<0

x'{x-l) + y\y-l)<0 (*) . 
x' ;c < 1

/ < 1

1

(3)

(1)

vk

(2)

= » A : ' ( A: - 1 ) + / ( } ' - 1 ) > 0 ( * * ) :

TO (*) va (**) ta c6: x' {x-l) + y^ {y-l) = 0 '

Ket hdp vdi (3), ta diTdc: ,

x' {x-l) = 0

/ ( y - l ) = 0

Thu" lai, nhan thay

x^OV x-\ 0 
y=Oyy-\=0

x = 0, *

hoac

x = l , 
x = 0, *

hoac thdahe:
.>' = 1 • 1 y = 0

x = 0\/x = l

y = Oyy = l, • ?

X* +y^ <1

x'+y'>l

Bi 6.32: Giai he phiTdng trinh: {x-yf+3{x-y) = 4 (1)

2x + 3>' = 12 (2)
G i i i

Dat: u = x - y . Ta cd:

(1)<^M'+3M = 4 < ^M' + 3M- 4 = 0

M = l

u = —A>

x — y = l 
x-y = -4

Vdi: x - y = 1 ket hdp vdi (2) ta diTdc he:

x — y = l

2x + 3y=^\2^'

Vdi: x - y = -4. k^'t hdp vd,i (2) ta difdc he:

x-y = -4 
2x + 3y = \2

x = 3 
y = 2. 
x = 0 
y = 4.

Vay he da cho cd hai nghiem: [x = 3;y = 2);(x = 0;y = 4).

JfB^ 6.33: Tim tat ca cac cap so' (x, y) thoa phiTdng trinh:

5x-2^f^ [2 +y) + y^ +1 = 0^

I

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Taco; 5x-2yfx {2 +y) + + \ 0

<^ (^4x-4^ + l^ + (^y^ - lyyfx +x^ = 0

< ^ ( 2 V I- l) ' + ( > ' - ^ / I ) ' = 0

(l^-lf =0

^ 2

[y-^f=0

2yfx-\ 0

y-y[^ =

0

27^

= 1

Vay cap s6' (x, y) phai tim la

'• =

2-J_ 1^

4'2

6.34:

Giai he phi/dng trinh: - 5 x

+ y =

0 - 5

A: + = 0

A:-V)^ + 1 = 0

x^-5x

+ y=:Q (1)

(2)

• He phiTdng trinh xac dinh vdi A: +

1 > 0 ;c > - 1 (*)

Khidd: = ^ + 1 <^>' = ;c'+2A; + 1 (3)

Thay vao (1) ta diTdc: 2A

;^-3x + l = 0

1 thoa (*).

Thay vao (3) c6 y^ =4,y.=-.

4

Vay nghiem cua he phu-dng trinh la: [x = l;y = 4);

91

174

PC'

6.35:

Giai he phtfdng trinh:

3 _

X

y — 2

2 1

X

2

-y

- 2

= 1.

Gi§i

Dilu kidn: x^O

y-2^0' . Dat:

1 M = —

X

V =

2

-y

thihe ^

M + 3v = 2

2M^V = 1

5

7

5

< ^

• 3

7

3

7 
1

Vay he da cho c6 mot nghiem

7 1

6.36:

Giai he phtfdng trinh: +y^+xy^l

/ + / + ; c V = 2 1 .

x^+y^+xy=^l

x' - f / + ; c ' > ' ' = 2 1

x^ +y' = l-xy

{l-xyf-xV =21

Dieu kien: xva y cung dau. Khi do: (1)

4^

x'^ + y^ = l-xy

[x'+/f-2x'y'+x'/=2l

'x^+y^=l- xy

49_14

;cy =

21

x

xy = 2

'+y'=5

x'+/=5

xy=4

vay x\y^ la nghiamcua phuTdng trinh : t^-5t + 4 = 0.

t = l

t = 4

(1)

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Do dd hoac

V i X va y cOng dau nen ta c6:

x'=4 ;c = ± l

2 ^ hoac

y = ±2 •

x = ±2

x = l 
y = 2 hoac

x = -\ x = 2 
y = -2' hoac
y = -2' ^ = 1 •

x = -2 
y = - i 
Vay nghiem cua he phu^dng trinh da cho Yk:

x = l x = -l 'x = 2 x = -2 
y = 2' y = -2' y = i'' y = - i

f>€ 6.37: Giai vk bien luan theo tham so a he phiTdng trinh:

X -xy + ay = 0 (1)

y -xy~4ax = 0 (2)

Na'u a = 0,.he trd thanh:
x^-xy = 0

y^ -xy = 0 
• Neu a^O:

x{x-y) = 0 
y{y-x) = 0

T a c d : ( l ) ^ ; c ^ = ; ^ ( ; c - a ) ^ 3 ; =

x=0 V x=y 
y=0 V x=y

(vdi x^a) 
^ x = y

x-a 
Tha'y v a o ( 2 ) ta cd:

x-a x-a •-4ax = 0

•a)

-x^ [x-a)-4ax[x-af = 0 
^ ax^-4ax(^x^ -2ax + a^) = 0

^ax[-3x^ + Sax-4a^) = 0

<r^x = 0 V 3;c^-8ajc + 4a^ = 0 (via^O) 
^x = 0 V x = 2a V A: = — ( t h o a d i l u k i e n

Vay ta cd: * Neu:a = 0 thi x = y (vdi xeR)lk nghiem. 
176

* N e u : a ^ O t h i nghiem gdm:

[x^y^0);{x = 2a,y = 4a); la Aa\

6.38: T i m m de he phu'dng trinh sau v6 nghiem:

X + Imy = 1 (1) 
2mx-6my = 4m + 3 (2)

Gi§i

* K h i m = 0: thi (2) v6 nghiem nen he v6 nghiem.
* K h i m ^ 0 :

T i i r ( l ) t a c d : 3 x + 6my = 3 (!')
C6ng ve theo ve (1') va (2) thi cd:

{2m + 3)x = 4m + 6 (3)

• Neu 2OT + 3 ^ 0 (tiJcla m ^ - - ) t h i (3)<^ x = ^ ^ ^ ^ ^ t ^ = 2
2 2 m + 3
( l ) ^ y = l-x

2m (vdi m^O). Luc dd he cd nghiem 
Neu m = -^ thi (3) ^Qx = 0 (v6 dinh)

1 — X 1

Con y = = •—(1 - x) n^n he cd v6 s6'nghiem. 
2m 3

T d m l a i he phiTcIng trinh v6 nghiem khi m = 0.
6.39: Giai phu'dng trinh: ;c' + 1 = 2lJ2x-\.

Phufdng trinh duTdc viet thanh: x^ = 2^2x-l - 1 . 
Dat t = U2x-l^t' = 2x-\.

Do dd phiTdng trinh trd thanh he phu'dng trinh sau:

x' = 2 f - l (1) .^

= 2x-\)

Trilf ve theo \€ (1) cho (2) thi:

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<=^t = x

V

x"- + tx + t^ + 2^0

^ t = x 'vix^ + tx + t^ +2 = 0 c6 A = t^-4[t^+2) <0:vdnghiem) 
The : t = x vao (1) ta difcJc:

x' -2x + \ 0^{x- -l)(x'+x- - 1 ) = 0

f

(1)

6.40: G i a i he phifdng trinh: 2x' 
' = l + x^

[ 1 + /

(2)

(3) •

Gidi

V d i

X = 0

thi

(1)

=^ z = 0 va v d i

X = 0

thi

(2)

cho y =

0.

Do dd X = y = z = 0 la mot nghiem cua he

Bay gid ta x ^ t x>0,y>0,z>0:

(Lm y rang (1), (2), (3) cho ta ^ > 0,^ > 0,z > 0 )

Ta biet rang l + a^>2a<=^ < 1 va dau " = " chi xay ra khi

1

+ a
a ^ = l .

Do dd t a c d : (\)=^x = z.- 2z 
l + z' 
i2)=^y = X.

(3)^z = 
y.-2x 
\ x

2y 
2 -<x

•<y

1 + /

Suy ra: y<x<z<y nen phai cd:

y = X = z \uc &y y^ = x^ = z^ = \ y = X = z = 1. 
K e t luan: N g h i e m cua he gom: {x = y = z = 0);{x = y = z = l) •

178

CHl/OfNG VII: DO THI CUA HAM SO

7.1:

V e do thi cua ham so": y =

yjx^

-4x + 4 .

Ta cd: y = V - ^ ^ -4x + 4 = ^(x - 2 ) ' = |x - 2

x — 2 neu x>2 
2 — x neu x <2 
• M X D : R.

• Bang gia t r i :

• V e :

0

2

4

2

0

2

Nhan xet: D o thi cua ham so la hai tia A B va A C v d i A ( 2 , 0),
B(0, 2), C(4, 2).

De 7.2: Cho ham so y = f(x) = 2-yjx'' -2x + \.

1) V e do thi cua ham so' tren.

2) T i m tat ca cac gia tri cua x sao cho

f(x) <

1.

G i i i

1) T a c d : y = f(x) = 2-yjx^-2x + \ 
= 2-yj(x-\) = 2 - | x - l

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2-x + l neu x>l

2 + X — 1 neu x <l 
T X D : R

Bang gia tri:

3 — X neu x>l

x + \u x<\

V e :

X 0 1 2

y 1 2 1

• Nhan xet: Do thi ham so f(x) la hai tia A B , AC vdi A ( l ; 2 ) ,
B(0;1),C(2;1).

2) / ( . x ) < l < ^ 2 - x - l <1<=^ x - 1 > 1

x-l>l

j c - K - 1

x>2 
x<0

Vay X > 2 hoat jc < 0 thi f(x)

De

7.3: Cho ham so: y =

yf]^ + ^Jx^

-4x + 4 .

1) T i m tap xac dinh cua ham so'.

2) Rut gon y (loai bo dau V~ va dau 11).
3) Ve do thi ham so.

4) T i m gia tri nho nhaft cua y ya cac gia tri tifdng iJng cua x.
5) TCr do thi hay chi ra true do'i xiJng cua do thi da ve d cau 3 va

dung phep toan de ch^ng minh dieu nay.

1) Ham so" xac dinh <^

G i i i

x^>0

x^ - 4A: + 4 > 0

x^>Q

{x~2f>Q ^ XER

180

Vay : T X D c i l a y Ik R.

2) y = 4 ^ + ^{x-2f = | ; C | + | A: - 2

x + x-2 neu x>2

= • x-x + 2 neu 0 < ;c < 2 =

- x - x + 2 neu j c < 0
3) T X D : R .

• Bang gii tri: ,j

2 x - 2 n6u x>2

2 n€u 0<x<2 
-2x + 2 neu x<0

Do thi ham so' la hai tia A C , BD va doan t h i n g A B v d i A(0; 2),
B ( 2 ; 2 ) , C ( - 1 ; 4 ) , D ( 3 ; 4 ) .

4) Nhin vao do thi ta tha'y gia tri nho nha't cua y la 2, dat duTdc
<l=!>0<;c<2.

5) • TO do thi nhan ra true do'i xitng ciia d6 thi la x = 1.
• Goi M (.«o!3'o) 1^ diem thuoc do thi

Goi M'{x',;y',) la diem doi x i f n g c i i a M qua durdng t h i n g 
= 1.

K h i d6: yo = / o v« ^ x', =2-x, ^ =2-x',

=^ y0 = >o =

1

^0

1 +1

^0

2 h

|2 - xo'

1

+1(2

- ) - 2

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M' thu6c do thi ham so.

Vay do thi ham so c6 true doi xiJng la x = 1.

De 7.4: Ve do thi cua ham so: >' = x- \ \

Ta c6: y =

Giii

- 1 - l | neu x > 1
1 - J: - l | neu x < 1

x-2 neu x>2 
2-x neu l<x<2

X neu 0 < ;c < 1

-X neu X <0

x-2 neu x > 1
neu A: < 1

TXD: R.
Bang gia tri:

Ve:

X - 1 0 1 2 3

y 1 0 1 0 1

NhSn xet; Do thi cua ham so la hai tia OC, BD va hai doan OA,
AB vdi A ( l ; 1), B(2; 0), C ( - 1 ; 1), D(3; 1).

182

Dt' 7.5:

1) Ve do thi cua cac ham so' sau day tren ciing mot he true toa do:
y^x^-l (1) y = -x^-2x + 3 (2)

2) Chitng minh giao diem cua hai do thi n6i tr6n luon thuoc do thi
ciia ham so':

1
y =

k + i {l-k)x^-2kx + 3k-l ydik^±l.

Gi^i

1) Ve (P,):y = - 1 .
• TXD: R
• Bang gia tri:

X -2 - 1 0 1 2

y = x'-\ 3 0 - 1 0 3

• Do thi la mot parabol c6 dinh (0; -1) nhan true tung lam
tnic do'i xiJng cat true hoanh tai cac diem (1; 0), (-1; 0).
• Ve {P^):y = -x'-2x + 3.

• TXD: R.
• Bang gia tri:

- 3 - 2 - 1 0 1

y = -x^-2x + 3 0 3 4 3 0

• Do thi la mot parabol c6 dinh ( - 1; 4) nhan dtfdng thing
x = - 1 lam true do'i xiJng.

B6 thi cat true hoanh tai cac diem (1; 0), ( - 3; 0).

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2) Hoanh do giao diem cua (fj) va [P^) la nghiem ci5a phU"dng 
trinh: -1 = -x" -Q.x + 3 2;c^ + 2;c - 4 = 0

<^ = l , x = - 2
n6n cac giao diem la A ( l ; 0), B ( - 2; 3).

* X ^ t dd thi (C) cua h^m %6: y = " ^ ^ [ ( l - ^ ) ^ ' - 2/:x + 3ik - 1
Taco A G ( C ) < ^ 0 = l - i t - 2 y t + 3 ) t - l . Dieu nay lu6n diing .

5 € (C) ^ 3 = - ^ [ 4 ( 1 - i t ) + 4 / : + 3 i t - 1

<!=> 3it + 3 = 4 - 4/: + 4)t + 3/: - 1

3)t + 3 = 3A: + 3. Dieu nay luon dung.

Vay cac giao diem A va B luon thuoc do thi ( Q s j

7.6: Trong cung he true vuong gdc cho parabol ( P ) : j = — va
4
du'dng thdng (D) qua diem / cd he s6'gde m.

1) Ve (P) va vie't phu-dng trinh cua (D).
2) Tim m sao cho (D) ti6p xiic vdi (P).

3) Tim m sao cho (D) va (P)cd hai diem chung phan biet.

1) • T X D : R .

m

• Bang gia tri:

X

- 4 - 2 0 2 4

x''

' = 4 4 1 0 1 4

Ve:

184

-4 -3 -2 -1 q

• Nhan x^t: Dd thi ham s6' y = ^ la mot parabol cd dinh

0(0; 0) la diem ctfc tieu, nam phia tren tnie hoanh ; tnic
tung la true do'i xiJng.

• Dqdng thing (D) cd he so' gdc bang m

nen phu'dng trinh d^dng thang (D) cd dang: y = mx + b.

3 3
Theo gia thie't: I e (D) ^ - I = -m + b =^ b ^ — m - 1 .

^ ^ 2 2
3

Vay phu'dng trinh du'cJng thang (D) \k y = mx — —m — l . • 
2) Phu'dng trinh hoanh d6 giao diem cua (D) va (P):

x^ 3

— = mx m-l 4^ x^-4inx + 6m + 4 = 0 (*) 
4 2

Ta ed: A ' = .4/7ji - 6m - 4
(D) tiep xiic (P) ^A' = 0^

m = 2 
m — — 
3) (D) va (P) ed hai diem ehung phan biet

<^ (*) cd hai nghiem phan biet

• <^ A ' > 0 0

m > 2
1
m < - - .

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De 7.7: Trong ciing he true vuong gdc, cho parabol (P):

y = - — x^ va du'dng thang (D).: y = mx - 2m - 1.

4

1) Ve(P).

2) Tim m sao cho (D) tiep xilc vdi (P).

3) ChiJng to (D) luon luon qua diem co dinh A e (P).

GiSi

1) • TXD:R

• Bang gia tri:

X

- 4

- 2

0

2

4 ' - 4

- 4

- 1

0 - 1

- 4

Ve:

' • Nhan x6v. Do thi ham so' j = - ^ x ^ la mot parabol cd

dinh 0(0; 0) la diem ciTc dai, nam phia du'di true hoanh ;

triic tung la triic do'i xiJng.

2) Phyong trinh hoanh do giao diem cua (D) (P)

x^

=^mx — 2m — \<^x^— Amx — 8m — 4 = 0

4 A' = 4m' +8m + 4 = 4(m + l)'

(D) ti6p xiic (P) <^ A' = 0

4=^

(m + if = 0

<^m + l = 0<^/n = - l . ?

3) Goi ) la diem co' dinh thuoc (D)

^0

—

'^^0 — 2m — 1 diing vdi mpi m.

m(xo - 2 ) - l - = Odung vdi moi m.

x,-2 = 0

x,=2

[-l-y^=0 [y,=-l

Vay (D) lu6n qua di^m c6'dinh A(2; -1)

-2^

Matkhac: A e (P) - 1 = — ^ - 1 = - 1 (dung). Vay A e (P)

Tdm lai: (D) luon luon qua diem co'dinh A e (P).

Oi 7.8: Trong mat phing toa do cho parabol (P) -.y = va diTdng

thing (D): y = x + 2.

1) Khaosatvaveddthi(P)cuahams6': >' = x \

2) Ve (D).

3) Tim toa do giao di^m A va B cua (P) va (D) bl[ng 66 thi va

phep toan.

4) Tur A va B ve AH ± Ox ; BK lOx. Tinh dien tich ciia ti?

giac AHKB.

I

I) Khaosdtva ve (P): y = x\

• TXD: R

• Vi a = 1 > 0 n6n ham s6' nghich bien trong khoSng x < 0

va dong bien trong khoang x > 0.

• Bang gia tri:

X

-2

- 1

0

1

2 y = x' 4

1

0

1

4 • Diem cifc ti^u la 0(0; 0) (do a = 1 > 0).

E)6 thi la mot parabol c6 dinh O (diem cufc tieu) n^m phia

tren true hoanh va c6 true do'i xitng la Oy.

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2) (D): y = X + 2 la diTdng thing qua hai diem C(x = 0 ; y = 2) va

E ( x = - 2 ; y = 0).

3) * Bang do thi: Can ciJ vao do thi cua (P) va (D) hai giao diem

A va B

CO

toa do la: A ( - 1; 1), B(2; 4).

* Bang phep toan: Toa do giao diem cua (D) va (P) la nghiem

cua he phu'dng trinh:

'y = x' (P)

y = x + 2 (D)

Ta suy ra phqcfng trinh hoanh do giao diem la:

=x + 2^ x^ -x-2 = 0.

Phu'dng trinh nay c6 a - ^ + c = 0(l - ( - 1 ) - 2 = O), nen c6

hai nghiem la: ;c =—1 ; x — l

Thay x = - 1 vao (D) hoac (P) ta dqdc: >- = - 1 + 2 = 1

Thay x = 2, ta du'dc y = 4.

Vay nghiem cua he phu'dng trinh tren la:

( x = - 1 ;y = l).VayA(-l; 1).

(X

= 2 ; y = 4) . Vay B( 2 ; 4 ).

4) Theohinhve: AHKB la hinh thang vuong c6 hai day la AH,

BK,dirdngcaolaHK.

Dt{AHKB) = ^{AH + BK)HK

• Dya vao do thi ta c6:

188

BK = \

1

= 4.

HK^HO + OK

(vl O nam giffa H, K)

--Vay: D<(A//0) =-^-(l + 4).3 = y (dvdt).

+ \x, = 1 + 2 = 3

p6 7.9: Cho (P) la do thi cua ham s6' y = ax^ va diem A ( - 2 ; - 1)

trong cdng he true.

1) Tim a sao cho A € (P); ve (P) (vdi a viTa tim difdc).

2) Gpi

B

e (P) c6 hoanh do la 4; vie't phu'dng trinh du-dng thing

AB.

3) Viet phu'dng trinh dufdng thing tiep xiic vdi (P) wk song song :

vdi AB.

1 1) Tacd: A(-2;l)G(P)<^-l = a-(-2) <^« = - 7

Vay iPy,y = ~ x \

• TXD: R.

• Bang gia tri:

X

- 4 - 2 0

2

4

1

2

- 4 - 1 0

- 1

- 4

• Ve:

y /

-4 -3 -2 -1

1 2 3

' / -4

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• Nhan x6t: Do thi ham so y = ~^ la mot parabol c6 dini, 
0 ( 0 ; 0) Ik diem eye dai, nam phia difdi true hoanh, true 
tung la true do'i xiJng.

2) B{x,;y,)e(P)^y,=-^xl^y,=-4.y'^yB(4;-4). 
Phu-dng trinh dudng t h i n g A B eo dang y = ax + b (vi A B kh6ng

song song v d i Oy).
Ta eo: AeAB

BEAB

-l = -2a + b 
-4 = 4a + b

6a = - 3
b = -\ 2a. 
2

b = -l+2

__\_

2 
b^-2.

Vay: Phu-dng trinh dudng t h i n g A B la : j = ~ x - 2 .

3) Phtfcfng trinh du'dng t h i n g ean t i m c6 dang y = a'x-\-b' (d) 
{d)IIAB:^a' = ~-b'^~2^{d):y = -]^x + b'. 
Phufdng trinh hoanh do giao diem eiia (d) va A B :

x^ 1

—- = x + b' ^ x^ -2x + 4b' = {) 
4 2

A ' = l - 4 ^ '

(d) tiep xuc (P) A ' = 0 1 - 4 ^ ' = 0 <^ 4Z?' = 1 <^ Z?' = i 
4
PhiTdng trinh du'dng t h i n g ean t i m la : = - i A; + .

-2 4

—"1

7.10: Cho parabol (P) : y = jx^ va du'dng t h i n g (D) qua hai 
d i e m A, B tren (P) c6 hoanh do Ian lUdt la: - 2 va 4.

1) K h i o sit sir bien thien va ve do thi (P) eua ham so'tren.

190

2)
3)

V i e t phUdng trinh du'dng thang (D).

T i m diem M tren eung A B cua (?) (tufdng \ing hoanh do

X g f - 2 ; 4 ] ) sao eho tam giae M A B eo dien tieh Idn nhaft.

Gi§i

1) T X D : R

1
• SU bien thien: ham so eo dang y = ax v d i a = — > 0 
nen ham so nghich bien khi x < 0, dong bien khi x > 0, bang 0
khi x = 0.

• Bang gia t r i :

x - 4 - 2 0 2 4
x'

4 4 1 0 1 4
Ve:

• Nhan xet: Do thi ham so >' = — la mot parabol c6 dinh
4

0 ( 0 ; 0) la diem eUe tieu, nam phia treri true hoanh, true tung
la true doi xiJng.

2) PhiTdng trinh dudng t h i n g AB CO dang y = ax + b (vi durdng ^
thiniT A B khong song song vdi Oy).

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Be{P)=^y, = ^ = 4. Vay B(4; 4)

Ta cd: AeAB

BeAB

l^-2a + b

4 = 4a + b

1 a = —

2

-2.- + b = l

2

6a = 3

-2a + b = l.

1 a = — 2

b = 2.

Phifcfng trinh difdng thang AB y = + 2.

3) PhiTdng trinh diTdng thang (d) song song AB cd dang

y^^x + b.

Phifdng trinh hoanh dp giao diem cua (d) va (P) la:

Lx^ =-x + b^ -2x-4b = 0

4

2-A' = l + 4b.

(*)

(d) tia'p xuc(?) ^A' = 0 ^ l + 4b = 0^b = - - .

4 Khi: b = - - nghi^m kdp cua (*) la x = 1.

4

Khi dd y = - . l ' M

4 4

la diem can tim. That vay moi

di^m M' khac M thupc cung AB ciia (P) deu cd khoang each tu" M'

den AB nho hPn khoang each tOf M den AB (vi M' nim giffa hai

dirdng thang song song d va AB)

Diem M can tim la M

De 7.11: Cho ham so y = ax^ cd dd thi (P).

1) Tim a biet rang (P) qua diem A(l; - 1). Ve (P) vdi a vffa tim

dffPc.

2) Tren (P) lay diem B cd hoanh dp - 2, tim phffdng trinh cua

192

dUdng thang AB va tim tpa dp giao diem D cua dudng thang

AB va true tung.

3) Vie't phu'dng trinh dufdng thang (d) qua O va song song vdi

AB, xac dinh tpa dp giao diem C cua (d) va (P). (C khac O).

4) Chu'ng to OCDA la hinh vuong.

1)

* (P): y = ax^

A(l;-1) e(P)^>'^ ^ax\\ a{\f <^a = - l .

Vay phiTdng trinh cua (P) Va y = -x^.

* Ve(P):

* TXD: R

X

-2

- 1

0

1

2 y = -x^ - 4

- 1

0 - 1

- 4

(P) la parabol cd dinh 0(0; 0) (diem cUc dai vi a = - 1 < 0), nam

phia dirdi true hoanh ; true doi xiJng la true tung Oy.

'2) Ta cd: B G (P) ^y,= -x\ - ( - 2 ) ' = - 4 . VSy B( -2; -4).

* Phu'dng trinh du-dng thing AB ed dang y = ax + b .

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TOd6,ta CO he:
b = -\ a 
- 3 a = - 4 + 1

a + Z? = - l \b = ~\-a

-la + b = -A [ - 2 a + ( - l - a ) = - 4 
a = l

Z7 = - l - l = - 2

Vay phiTcfng trinh cua dicing thang A B la: y = x - 2.

Giao d i ^ m D ciia dtfdng thang A B va true tung c6 hoanh do
= O v a = ^ 0 - 2 = 0 - 2 = - 2 . Vay: D ( 0 ; - 2 ) .

3) X d t du-dng thang (d) qua O c6 dang : y = a'x

• {d)ll[AB) ^a = a' (hai he so'gdc bang nhau).

V d i a = 1 (he goc cua (AB) : y = x - 2) nen (d) c6 he so gu,

a' = 1 . Vay (d) : y = x.

• Toa do giao diem C cua (P) va (d) la nghiem cua h ^ phu-dng

trinh:

y = x 
[y = -x\

Suy ra: -x^ = x<i=4>x^+jr = 0<^A: = 0;jc = - 1 .

X = 0 la hoanh do giao diem O.

X = - 1 la ho^nh do giao diem C.
Suy ra: tung do cua C la : = jc^ = - 1 .

V a y : C ( - l ; - l ) .
4) O C D A la hinh vuong.

= >'c = - 1 (A va C cung tung do) n6n dtfdng thang A(
song song vdi true Ox, suy ra: AC _L OD tai I .

• I la trung diem cua doan A C vi cd:

IA=x, = • i = 1

IC= . - 1 = 1 
I cung la trung diem OD vi cd:

10 = - 1 = 1

ID = OD- 01 (I nam giffa O va D).

1 |

y, - 2 — 1 = 2 - 1 = 1.

Vay: Tu" giac O A D C cd hai dudng cheo cat nhau tai trung diem
chung I , vuong gdc vdi nhau va bang nhau (AC = OD = 2) nen
la hinh vuong.

7.12: Trong cung mat phang toa do cho hai du'dng thang:

{D,):y = x + \ (D^): x + 2^ + 4 = 0 .

1 ) T i m toa do giao diem A cua ( D , ) va ( D ^ ) bang do thi va 
kiem tra lai bang phep toan.

2) T i m a trong ham so' y = ax^ cd do thi (P) qua A ; khao sat va
ve do thi (P).

3) Tim. phUdng trinh cua du'dng thang tiep xuc vdi (P) tai A .

Giii

1) • Bang do thi:

Xet hai difdng thang: ( D , ) : = X + 1 va ( D 2) : X + 2y + 4 = 0
hay: y = — — x — 2 .

• Bang gia tri:

X 0 2 .

y = x + \ 1 3

1

- 2 - 3
Ve:

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. (Z),) la audng thang qua (0; 1) va (2; 3) 
{D^) la du-dng thang qua (0; - 2) va (2; - 3)

Nhin vao do thi ta thay {D^ ) cat (D^ ) tai A ( - 2; - 1).
• Bang phep toan:

PhiTdng trinh hoanh do giao diem cua (DO va ( D 2 ) :
X + 1 = X -2 <^ x = - 2 (y = - l ) 
Vay (Z),) cat ( D j tai A ( - 2 ; - 1 ) .

2) (Py. y = ax'qua A^-l = a.{-2f ^a = ~^ 
Way: {P):y = -^x\

• T X D : R

• S\i bien thien: ham so c6 dangy = ax^ c6

a = - - < 0

nen : Ham so dong bien khi x < 0, nghich bien khi x > 0, bang 0
khi X = 0.

• Bang gia tri:

X - 4 - 2 0 2 4

- 4 - 1 0 - 1 - 4
• Ve:

• Nhan xet: Do thi ham so' y = la mot parabol c6 
4

dinh 0(0; 0) la diem cu'c dai, nam phia difdi true hoanh,
true tung la true do'i xiJng.

3) Du'dng thang (D) khong song song Oy nen phtfcfng trinh eo
dang y = ax + b.

Ae{D)^-\ a.[-2) + b^b = 2a-\. 
Vay: [D): y = ax+ 2a-l.

Phu-dng trinh hoanh do giao diem cua (D) va (P)
--x'^ =ax + 2a-\^ x^+4ax + Sa-4 = 0

A ' = 4 a ^ - 8 a + 4 = 4 ( a - l ) ^

(D) tiep xiie (P) A ' = 0 4^ (fl - 1 ) ' = 0

< ^ a - l = 0 < ^ a = l .

a = l taco b = 2.1 1 = 1 
-Vay: phu-dng trinh du-dng thang (D) la : y = x + 1.

i 7.13: Trong eung he true toa do goi (P) la do thi eiia ham so'
'•y = ax^ va (D) la do thi ciia ham so' y = - x + m.

1) T i m a biet rang (P) qua A(2, -1) va ve (P) vdi a t i m du-de.
2) T i m m sao cho (D) tie'p xiic vdi (P) (d eau 1) va t i m toa do

tiep diem.

3) Goi B la giao diem cua (D) (d eau 2) vdi true tung; C la
diem do'i xiJng cua A qua true tung. Chitng to C nam tren
(P) va tam giac ABC vuong can .

Gi^i

A(2; -l)e{P)^ y^^axl^-\ a.2^ ^a = -^ 
Vay {P):y = ~ x '

• T X D : R.
• Bang gia t r i :

197

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X

- 4

- 2

0

2

4

- 4

- 1

0 - 1 - 4

Ve (P):

Nhan xet: Do thi ham so y^--x^ la mot Parabol c6

4 dinh 0(0; 0) la diem ciTc dai, nam phia diTdi true hoanh,

true tung la true do'i xiJng.

2) PhUdng trinh hoanh do giao diem ciia (D) va (P):

= - x + m<^jr - 4 x + 4«i = 0

4 A' = 4 - 4 m .

(D) tiep xiie (P) A' = 0 ^ 4 - 4m = 0 ^ 4m = 4 ^ m = 1.

Khi do nghiem kep eua phu"cfng trinh la x = 2.

Vdi x = 2thi y = --(2f = - 1 .

4 Toa do tiep diem la A(2; - 1).

3) C doi xiJng eua A(2; - 1) qua true tung nen C(-2; - 1).

• Gia siiCe{P)<=>y,=axl^-\ -^{'^f ^ - 1 = - 1

(diing). Vay Ce(P).

Ta eo AC cat true tung tai M(0;-1)

Vi BM = MC = MA = 2 AA5C vuong tai B.

198 Ma EC = BA (Do A, C d6'i xitng nhau qua true tung va B thuoc

true tung).

Do d6 tarn giac ABC vuong can tai B.

i

7.14: Cho do thi {C):y = x'^-2mx~4 va diTdng thang (D):

y = 2x.

1) ChiJng minh (D) luon cat (C) tai hai diem phan biet A va B.

2) Tim M de khoang each AB la nho nhat.

Gi^i

1) Hoanh do giao diem giffa (C) va (D) la nghiem eila phtfdng [

trinh:

x'-2mx-4 = 2x^x'-2(m + l)x-4 = 0 (*)

Taeo A' = (m + l ) ' + 4 > 0 ;Vm.

Vay (D) luon c^t (C) tai hai diem phan biet A va B.

2) Dat A{x,;y,),B{x,;y,) thic6:

= 2x, ; y, = 2x, nen:

AB^ = {x^ x,f + (2x, 2x, f =5 (x^ x, f

-= 5

ma (*) cho ta: x^+x^=2{m + \),x^x^ = - 4

nen AB' ^5 4(m + l f + l 6 ] = 2o[(m + l f + 4

Do vay doan AB ngin nhat k h i m + l = 0 < ^ m = - l .

DC 7.15: Trong he toa do vuong goc Oxy cho ba diem A(2; 5),

B ( - l ; - l ) v a C ( 4 ; 9 ) .

1) Chu-ng minh ba diem A, B, C th^ng hang.

2) Chifng minh du-dng thing AB va cac difdng thing

(^i ),(^2)

c6 phtfdng trinh la y = 3, = -^{x -1) la ddng quy.

Gi§i

1) Xet du-dng thang AB cd phu-dng trinh dang y = ax + b

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(1)
(2)
Dirdng thing qua A nen: 2a + b = 5

Di/cing t h i n g qua B nen: — a + b = — 1

Giai hd (1) va (2) ta c6 : a = 2, b = 1. ,„
Do 66 d\Xdng thing A B c6 phu-dng trinh la : y = 2x + 1

Ta c6 : 9 = 2.4 + 1 = nen diem C thuoc diTdng t h i n g A j
Vay ba diem A, B, C t h i n g hang.

2) To a do giao cua [d^) va du'dng thang A B thoa :

y = 3 
y = 3x + l

'.^ = 1

y = 3.

Ta l a i cd 3 = - 1 ( 1 - 7) nen giao di^m ay thuoc ( j , ) .
Vay cac dtfdng thing A B , {d,),{d,) ddng quy.

7.16: Cho do thi (C) cua ham so' y = ax^ +bx + c.

1) Dinh a, b, c de (C) d i qua ba diem sau: A(0 ; - 4 ) ;B(1 ; - - 6 ) ;
C ( - 3 ; 1 4 ) .

2) T i m phufdng trinh du-cJng t h i n g di qua M ( 0 ; - 8) va tiep xiic (C).
3) T i m toa do tiep diem trong cau 2.

1) Do thi (C) di qua A , B, C nen ta c6:

c = - 4

a + b + c = -6 <^ 
9a-3b + c = 14

a + b = ~2

9a-3b = \S<^a = l,b = -3,Q^-4.

c = - 4

Vay a = l , b = - 3,c = - 4 .

2) Du-dng t h i n g (d) qua M(0, -8) c6 phu-dng trinh dang: y = mx - 8.
Phu-dng trinh hoanh do giao diem giffa (d.) va (C) la:

x^ -3x-4 = mx-S^ x^ -{3 + m)x + 4 = 0 (*)

De (d) tia'p xuc (C) thi ph^i c6 A = 0 <^ {3 + mf - \6 = 0 
^ ( m + 7 ) ( m - l ) = 0

4 = >/ n = l , m = — 7.

200

Vay cd hai dtfdng thing (d) can tim vdi phu'dng trinh la:

{d,):y = x-S 
[d,):y = -lx-%.

3) * V d i (J;) thi (*) tra thanh: ;c^-4jc + 4 = 0<i=>jc = 2
nen tiep diem cd toa do la : (2, — 6).

* V d i [d^) thi (*) trd thanh: x^ -^4x + 4 = Q ^ x = -2 
nen tiep diem cd toa do la : ( — 2; 6).

De 7.17: Cho hai Parabol [P^):y = x^

{P,):y = x'-2x-\.

Vie't phu'dng trinh du-dng t h i n g (d) tiep xiic v d i ca {P^) va (P^) •

Giii

• Du'dng t h i n g (d) cd phu'dng trinh dang : y = ax + b.
• Phu'dng trinh hoanh do giao diem giiJa (d) va (Pj) la:

x^ =^ax-\-b x^ -ax-b = 0.

Dieu kien d^ (d) tiep xuc vdi (P^) la Ai = 0

^a'^4b = 0 (1)

• Phu'dng trinh hoanh do giao diem giiJa (d) va [P^) la:

A: ' - 2x - 1 = ax + ^ ^ - (2 + a);c - (1 +Z J) = 0 .

Dieu kien de (d) tiep xuc vdi (P^) la A 2 = 0
^ ( 2 + a ) ' + 4 ( l + Z7) = 0

+ 4 a + 4Z? + 8 = 0 (2)
Tuf (1), ta cd: 4b = - a ' . The 4b vao (2) ta du'dc:

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CHlTdNG V I I I :

MOT SO BAI TOAN

V E

TO

H0P

P H E P E ) E M - D O N G

T i / . . .

D e 8 . 1 : H a i dia d i e m A , B each nhau 60km. Ngifdi d i xe dap khcii
hanh tu" A den B r d i tiT B trci A ngay v d i van toe nhif luc dau,
nhifng sau k h i d i tu* B du'dc 1 g i d thi nghi 20 phut r d i d i t i e p ve A v d i
van toe tang t h e m 4km/h. T h d i gian d i va ve bang nhau. T i n h van toe
ban dau.

Gi^i

G o i X km/h la van toe ban dau ciia ngtfdi d i xe dap.
(dieu k i e n : x > 0).

T h d i gian d i tu: A den B : — g i d .

X

T a c 6 : 2 0 p h u t = i gicf.

T h d i gian d i tu" B ve A : 1

+ i +

3 x^A

J

gid , hay :
;4 ^ 6 0 - ; c '

3 ;c + 4 , gid.

Theo dau bai ta c6 phu-dng trinh:

60 4 , 60-x x = 2Q

X = - 3 6 ( l o a i )
Vay van toe liie ban dau: 20km/h.

D e 8.2: M o t ngu-cfi d i oto tir A tdi B v d i van toe 30km/h. Sau do mot
thdi gian mot ngu-di d i moto v d i van t6c 40km/h va neu gii? nguyen
_van toe nhtf the^ thi se b^t kip olo d t a i B . The^ nhtfng k h i d i dtfde nij-a
202!

quang du"dng A B thi moto tang van toe len 45km/h va sau do 1 gid
da bat kip oto. T i n h quang dtfdng A B .

Gi§i

Dat quang du'dng A B = x (x > 0)

T h d i gian ma oto ehay h e l quang du'dng la: =
30
T h d i gian ma moto ehay het quang du'dng v d i van toe
4 0 k m / h l a : L =

40;

De ea hai gap nhau hie vtfa t d i B thi thdi gian moto ehay sau
oto la: L = — - — = .

0 1 2 Ar\ 30 40 120

TiJ* lue k h d i hanh luc gap nhau moto can thdi gian la:

40 80

NiJa quang du'dng dau, moto ehay v d i van toe 40km/h
va quang du'dng moto da d i du'de la: — + 4 5 .

Til Mc k h d i hanh t d i luc gap nhau oto can t h d i gian la:
- + 45 „

= ^ + 1.

30 60 2

X

V i moto k h d i hanh sau oto thdi gian la: = nen ta c6:

_x_ 3 _

60 2 ~ 80 • + 1

120

X X , ^ X , ^ X

+ ^ - + 1 5 - - + 10 + — ,
120 6 8 12

A: x X

i 12 6
= 5
< ^ x = 120.

Vay quang du'dng A B dai 120km.

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Dd 8.3: Mot r^hom hoc sinh diTdc giao nhiem vu trdng 80 cay thong
con. NhiTng khi thuTc hien nhdm ay diTdc tang cifdng them 4 ban, do
d6 moi ban da trdng it hdn 1 cay so vdi du" dinh. Hoi hic dau nhom

CO bao nhieu hoc sinh ? (Biet rang so' cay moi ban trdng nhuf nhau).

Gi^i

Goi X la s6'hoc sinh c6 lilc dau cua nh6m (x nguyen du'dng).
80

Luc dau moi ban du'dc giao trdng — cay.
X

Nh^ng khi thi/c hien c6 them 4 ban nffa so'hoc sinh tham gia
80

trdng la (x + 4) ban, do do moi ban trdng du'dc cay.
x + 4

Theo dau bai cd phifdng trinh:

- - — = l h a y x ^ + 4 x - 3 2 0 = . 0 .
X x + 4

Giai ra diWc x = 16, thda man dieu kien bai toan.
Vay sd'hoc sinh tham gia trdng cay liic dau la 16 em.

Dd 8.4: Cho tam giac ABC vudng tai A c6 chu vi 12m va tong binh
phu-dng cua ba canh la 50. Tinh do dai ba canh cua tam giac ABC.

Giii

Dat: AB = x (m), AC = y (m), BC = z (m).
Giai su": x <y <z

TJieo dau bai ta c6 he phrfdng trinh:

X + y + x = l2 (1)

x^+y^+1^=50 (2)

Theo dinh ly Pitago, trong AABC : x^ =^z^. 
Thay + y^ = vao phifdng trinh (2) :

204

2z^ = 50 <^ = 25 <^ x= 5
Z = - 5 (loai) 
D o d 6 : x ' + / = 2 5 (3)

Thay z = 5 v a o ( l / . x + y = 7 (4)

Tiir(3) va (4) ta c6 he:
x + y = l 
x'+y'=25

x + y = l

{x + yf -2xy = 25.

5 = 7

S- -2P = 25 P = 12.
Vay X , y la nghiem cua phu'dng trinh bac hai sau:

fx = 4

Vay:

-1X + 12 = 0^ 
x = 3 \x = 4

V
y = 4 [y = 3

X = 3

•Vi: X <y<z nen chon x = 3 
y = 4 
Vay canh AB = 3m, canh AC = 4m va canh huyen BC = 5m.

De 8.5: Mot ngu'di mang binh 8 lit di mita 6 lit sffa. Chu quan lai chi

CO mot binh 12 lit day sffa va mot binh khdng Slit. Lam the' nao de
dong dffdc 6 lit sffa vao binh 8 lit ?

Binh 12 lit Binh 5 lit Binh 8 lit

Lan 1 0 4 8

Lan 2 4 S 3

• Lan 3 9 3 0

Lan 4 1 3 8

1 5 6

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-D e 8.6: Chia mot hinh vuong cd canh
bang 10^/2 cm l a thanh 100 hinh vuong
nho bang nhau (xem hinh v e ) . X e p 201
d i e m vao ben trong hinh vuong (cac d i e m
deu d ben trong cac hinh vuong nho).

D e 8.6: Chia mot hinh vuong cd canh
bang 10^/2 cm l a thanh 100 hinh vuong
nho bang nhau (xem hinh v e ) . X e p 201
d i e m vao ben trong hinh vuong (cac d i e m
deu d ben trong cac hinh vuong nho).
D e 8.6: Chia mot hinh vuong cd canh
bang 10^/2 cm l a thanh 100 hinh vuong
nho bang nhau (xem hinh v e ) . X e p 201
d i e m vao ben trong hinh vuong (cac d i e m
deu d ben trong cac hinh vuong nho).
D e 8.6: Chia mot hinh vuong cd canh
bang 10^/2 cm l a thanh 100 hinh vuong
nho bang nhau (xem hinh v e ) . X e p 201
d i e m vao ben trong hinh vuong (cac d i e m
deu d ben trong cac hinh vuong nho).
D e 8.6: Chia mot hinh vuong cd canh
bang 10^/2 cm l a thanh 100 hinh vuong
nho bang nhau (xem hinh v e ) . X e p 201
d i e m vao ben trong hinh vuong (cac d i e m
deu d ben trong cac hinh vuong nho).
D e 8.6: Chia mot hinh vuong cd canh
bang 10^/2 cm l a thanh 100 hinh vuong
nho bang nhau (xem hinh v e ) . X e p 201
d i e m vao ben trong hinh vuong (cac d i e m
deu d ben trong cac hinh vuong nho).
D e 8.6: Chia mot hinh vuong cd canh
bang 10^/2 cm l a thanh 100 hinh vuong
nho bang nhau (xem hinh v e ) . X e p 201
d i e m vao ben trong hinh vuong (cac d i e m
deu d ben trong cac hinh vuong nho).

u-iiung iiiinu l a i i g iiic K C uuyu i i i y i uuung

tron CO ban kinh 1cm chita it nha't ba d i e m trong 201 d i e m n d i tren.

Giai

Phan pho'i 201 d i e m vao trong 100 hinh vuong
nho thi c6 it nhat mot hinh vuong nho chiJa it
nhat ba d i e m (Nguyen ly Dirichlet).

Ta CO canh hinh vuong Idn la 1 0 > ^ c m thi canh
hinh vu6ng nho la ^/2 cm, suy ra ban kinh du'dng

tron ngoai tie'p hinh vuong do la 1cm. Du'dng tron nay chiJa hinh
vuong nho nen chiJa ca ba d i e m ben trong hinh vuong nho ay.
V a y ke dtfdc du'dng tron ban kinh 1cm chtta it nhat ba d i e m trong

201 d i e m da cho.

D e 8.7: Chia mot hinh vuong thanh 5 hang va 5 cot (cd 25 6 vuong
nho). Trong m o i 6 nho ta ghi mot con so' Irich tur tap hdp so

• 0 , 1 , - 1 } . Chu'ng minh rang trong 5 hang, 5 cot va hai du'dng cheo
cua hinh vuong thi cd it nha't hai du'dng cd tong cac so' ghi tren 6 ctia
nd la bang nhau.

Gi§i

Ta cd: So'hang la 5, so'cot la 5 va du'dng cheo la 2 nen chiing gon'
ta'tca la 12 du'dng. .

-T r e n m o i hang hoac m o i cot (hoac du'dng cheo) deu cd chiJa 5 6
ma m o i 6 cd gia t r i la 0 hay ± 1 nen tong S cac 6 dd phai thoa

- 5 < 5 < 5 nghla la S cd the nhan 11 gia t r i tir -5 de'n 5.

NhuT vay: chi cd 11 gia t r i gan cho 12 du'dng nen cd it nha't 2
du'dng nhan cung mot gia t r i . (Nguyen ly Dirichlet).

206

' 8.8: Cac so' nguyen tuT 1 de'n 9 du-dc sap xe'p vao mot hinh vuong

3 x 3 6 sao cho tong m o i hang, m o i cot va long cac du'dng cheo deu
bang b o i cua 9. Chu'ng minh rang so' ct tam hinh vuong phai la b o i
cua 3.

Gi§i

Theo de bai ta cd:
1) A + 5 + C = 9^i,
2) D + E + F = 9/:,,

3) G + // + / = 9/:3,

4) A + D + G = 9A:4,

5) B + E + H ^9k,,

I; 6) C + F + I^9k„ 
p) A + E + I = 9k,,

8) C + £: + G = 9^,,

Tuf cac phu-dng trinh 2, 7, 8 ta du'dc:
E = 9k^_-{D + F)

A B C

D E F

G H I

E = 9k, -{A + I] 
E^9k,-{C + G)

=^3E = 9k^ + 9/7 + 9k^-{A + D + G + C + F + l)\k ket hdp v d i
cac phu'dng trinh 4, 6, la cd:

3E = 9k^+9k, +9k,-9k,-9k, ^E = 3{k,+k, +k,-k,-k,). 
V a y E la b o i so cua 3.

i

{

8.9: Trong hinh vuong 4 x 4 , viet d m o i 6 mot so sao cho tong
bon so' m o i hang bang tong bo'n so' theo m o i cot, bang tong bo'n so
theo m o i c u'dng cheo va deu bang a. Tinh tong ciia bon so'd bon
clinh cvia hinh vuong theo a.

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Ta gia su* cdc so' diTdc dien vao hinh vuong la:

«11 «12 «13 «M 
«22 «23 «24

«3I «32 «33 «34

«41 «42 «43 «44

Tong cac s6' d 4 dinh la 5 = a,, + a,^ + + a^,.

Theo tinh cha't cac so'ciia hinh vuong nay, ta c6 tong cua 12 so'con

lai la 4a - S.

Mat khac, siJ dung hai hang dau va cuo'i, ta c6 :

a,2 + + '^42 + ^ 4 3

= 2a - 5.

Suf dung hai cot dau va cu6'i, ta c6: a^i + "31 + <^24 + ^34 = 2a - 5.

Su: dung hai du'dng ch^o ta du'cJc: a^^ + a^^ + a^^ + a^^ —2a —S .

TO cac phtfdng trinh tien suy ra:

4 a - 5 = 2 a - 5 + 2 a - 5 + 2a-5<i=>4a-5 = 6 a - 3 5 < ^ 5 = a

Vay tong cila bo'n so'd bo'n 6 dinh cila hinh vuong la a.

DC' 8.10: Trong moi 6 ciia bang kich thu-dc 25x256, dien vao cac

so' +1 hoac - 1. Ta ky hieu tich tat ca cac so'd dong thu" i la a,., tich

tat ca cac so'd cot thi? j la bj.

ChiJng minh rang : + + +... + a^^ + b^^ ^ 0.

Goi a la sd'cac so' a. bang 1

b la so' cac so' a. bang - 1.

c la so' cac so' bj bang 1.

d la i; *' "-I'; so' bj bang — 1.

2081

Ta c6 a + b = 25 va c + d = 25.

Tich cdc so'trong bang (—l)'' va b^ng (—1)''

Do d6:

(-1)" = (-1)^

bwk d cOng tinh ch§:n \6 ^b + d chKn.

Do dd a^+b^+a^ + b^+... + a^+b^=a-b + c-d

= {a + b)-2b + {c + d)-2d = 25-2b + 25-2d

= 2[25 - (Z? + d)\ 0 (vi 25 le va b + d chSn). '

Bi

8.11: Cho da giac loi c6 10 canh.

1) Tinh s6'du-dng ch6o.

' 2) ChiJng minh rang cd it nhat hai diTdng ch6o tao vdi nhau mot

L^....,.,,,,....l<^c_nhohdn^6°. ^ '

G i i i

1) Qt ndl hai dinh ciia da giac lai thi c6 mot canh hoac m6t du'dng

ch(5o, do dd tong s6' canh va drfdng ch6o ciia da gi^c 10 canh 1^:

1 ^

= 45.

2 Vay s6'dufdng ch^o Ik: 45 - 10 = 35..

2) * Ne'u trong 35 duTdng chdo c6 hai dtfdng chdo song song vdi

nhau thi hai dufdng chdo dd tao nhau gdc 0° (nh6 hdn 6°).

* X6t 35 du'dng chdo trong dd hai dufdng bat trong chiing la

khong song song. TO mot diem O bat k^ trong mat phdng k6 cic

dtfdng thing song song Ian lu'dt vdi 35 dufdng chdo cda da gidc

thi tao n6n 70 gdc cd tong b^ng 360°.

^ . 360°

, 0

Tacd: = 5°,...

70 Do dd cd it nha't mot gdc trong s6' 70 gdc ndi tren

la

nhd hon 6°. ;

T T T r r m f i

8.12: Cd the c^t da giac idi 17 canh thanh 14 tarn gi^c difdc

khong ?

</div>
(106)<div class='page_container' data-page=106>

Gidi

Tdng cdc g6c trong cila mot da giac Idi 17 canh bkng

180° (17-2) = 15.180° , c6n t^ng cac gdc trong cua 14 tam giac

bing 14.180°.

Khi chia m6t da giac thanh cac tam giac thi tdng cac gdc trong

ci5a cic tam giac khong the nho hcfn tong cac gdc trong cua da

gidc. Vay, khdng the thtfc hien dtfdc phep c^t nhu" de bai.

8.13: Trong m6t phdng cd 288 ghe'diTdc xep thanh cac day, moi

day diu cd s6' ghe' nhu" nhau. Neu ta bdt di 2 day va moi day con lai

them 2 ghe' thi

viTa

da cho 288 ngufdi hop (moi ngu-di mot ghe). Hoi

trong phdng cd may day ghe' va moi day cd bao nhiau ghe ?

Goi so" day ghe trong phdng la x (x nguyen du'dng).

288 Moi day cd — ghe, sau khi bdt di 2 day t u con lai x - 2 day

(x>2).

288 Luc dd moi day c6n lai cd ghe'.

x — 2

Theo bki ra ta cd phtTdng trinh: _ ^ = 2.

x-2

X

^288x-288(x - 2) = 2x(x - 2) <^ -2;c-288 = 0.

GiairadufOc: x^=lS,x^=-l6 (loai).

Vay trong phdng cd 18 day va moi day cd 16 ghe'.

Hi 8.14: Trong mdt buoi hoi thao cd 9 nha khoa hoc tham dU". ba

ngtfdi bat k>^ trong ho thi cd it nhat hai ngu^di ndi cilng mot thu" ti^'ng.

Hon nffa moi ngffdi trong cuoc hdi thao ndi dtfdc khdng qua 3 thit

tie'ng. Chrfng minh cd it nhat 3 ngffdi ndi cilng mot thff tie'ng.

•_210

Giii

Ta chffng minh bang phan chffng.

Gia su" r^ng khdng cd 3 ngffdi nko ndi cing thff tie'ng.

Goi

X

la mot ngrfdi bat ky trong cuoc hoi thao. Vi x

chl

ndi

dffOc

td'i da 3 thff tie'ng ma thoi va do dieu gia tren moi ngtfdi chi

ndi cho mot ngtfdi khac cung

thff

tie'ng nen ngoai x va 3 ngffdi ma

X cd the ndi vdi ho cung thff tie'ng thi 5 ngtfdi cdn lai khdng the

ndi b^ng thff tie'ng nao vdi ho ca.

Goi mot trong 5 ngu-di ay la y. Vi y chi ndi dtfOc td'i da vdi 3

ngffdi nSn trong 4 ngffdi con lai cd it nha't mot ngffdi la z khong

ndi chuyen du'Oc vdi y.

Luc dd ta cd trong nhdm (x, y, z) thi: x khong ndi du'Oc vdi y.

y khdng ndi diTOc vdi z.

z khdng ndi dffOc vdi x.

Dieu nky mau thuan vdi gia thie't cu* 3 ngu^di thi cd it nhat 2 ngffdi

ndi

Cling

mot thu" tie'ng.

Vay cd it nhat 3 ngu-di ndi cung thff tie'ng.

8.15: Trong mot giai bdng da gdm 11 ddi bdng tham dff thi da'u

theo the thtfc vdng trdn IffOt di va- IffOt ve (nghia la hai ddi bat ky

ludn gap nhau hai tran) va tinh didm bdi: tran thang du'Oc cong 7

diem, tran thua bi truf 6 diem va tran hda la 0 diem.

1) Tinh s6' tran thang, tran thua va tran hda cua doi dufOc 13

diem khi ket thiic giai.

2) Ket luan doi dffOc 13 diem la trung binh'cong cua diem cua

cac doi la dung hay sai ?

Giii

1) Vdi the thiJc thi dau vdng trdn ItfOt di va vi thi doi phai thi da'u

20 tran.

Goi X la sd' tran thing, y la so' tran thua thi ta cd:

</div>
(107)<div class='page_container' data-page=107>

7 x - 6 j ' = 13

x + y<20

(2)

(1)

Tiir(l)tac6: 6y ^Ix-13 = 6x+ {x-\3) =^ y = x + x-13

x — \3

Dat: / = — - — thi / e Z va tacd: x = 6t + l3,y^7t + l3.

(2)chota: 13/ + 2 6 < 2 0 o

t<-13

Mat khac X > 0 nan: 6/ + 1 3 > 0 = ^ r > - 13

V d i - — < / < - — t h i chi CO t = - 1 .

6 - 1 3

Suy ra

X =

7, y = 6.

Vay d6i dMdc'U diem se thang 7 tran, thua 6 tran, hoa 7 tran.

2) Moi doi khi ket thuc giai da thi dau 20 tran nen toan giai gdm c6:

20x11 ,

= 110 tran.

2 (Khi ta lay 20x11 = 220thi ta da tinh lap lai hai Ian so' tran dau

giffa hai doi n^n phai chia 2 de cd so'tran da'u that suf).

Trong mot tran dS'u ne'u hai doi h6a nhau thi khong c6 di^m nao.

Trong mot tran da'u c6 thang thua thi tong so' diem ciia hai dpi la

bang 1.

Do dd vdi 110 tran

thi

tong so' diem ciia cac d6i se khong Idn

hdn 110; va nhiT

the'

xay ra ^ < 13 tifc la ket luan dpi c6 13 diem

bing trung binh cpng diem cua tat ca cac dpi la sai.

8.16: Xet tap hpp A gdm 2005 du-dng thing trong mat phing

thoa cic dac diem sau day:

1) Hai dtfdng thing ba't ky trong A thi khong song song vdi nhau.

2) Ba dufdng thang ba't trong A thi khong ddng guy. Bat lAI la

212 s6' tam giac tao bdi cac du'dng thang trong A mk khong bi c^t

bcti ba't ky mot du'cfng thang khac ciia A. ChiJng minh r^ng

A > 2003.

ai

• X6l

A3

gdm ba dufdng thing thi c6 duy nhat m6t tam giac can

timtiJc

la|A3|

= l.

• X^t A^gdm bo'n du'dng thing. Xdt drfdng thing ghi la thoa

tinh chat:

Tat ca cac giao diem cua cac du'dng thing trong A^

d

cung

trong mot niJa mat phing ma bd la . Gpi I la cac giao diem

d

gan nha't thi diem I se t^o vdi mot canh

d

tren a, mot tam

giac thoa de bai. Ngoai ra ba du'dng con lai da tao mot tam giac

can tim niJa nen it nha't cd hai tam giac can tim, vi vay 1 > 2

• tong quit vdi

A2005 = A2004

U {^} (d

a

ngoai

A^^ ).

Hdn nffa vdi diTdng thing (d) thoa tinh chat cac giao di<^m ciia cac

dufdng thing trong

A^QOS

ci cung mot nufa mat phing cd bd la (d) (luc

nao trong Ajoojta ctlng chpn difdc mot du'dng thing (d) nhu* the ).

Gpi J la giao diem d gan (d) nhat thi cd them mot tam giac niTa vdi

J la dinh va canh la d tren (d), do dd ta cd:

•^005 ^

Tijf cacke'tqua:

A3

= 1, A^

2004

>

+ 1

+ 1 = 2,|A,

>

A, +1 = 3

- ^005 > ^004 + 1 suy ra rang

</div>
(108)<div class='page_container' data-page=108>

Vay:

^ 0 0 51 > (2004 - 2) + 1 = 2003 
\A\= Ajoos > 2 0 0 3 .

8.17: Ta ghi bon con so' 0 va nSm con s6' 1 l^n trdn mot difdng
tron theo mot thu* tu* tily y. Sau 66 ciJ giiTa hai con s6' bkng nhau ta
ghi so' 0, gii?a hai con s6' khac nhau ta ghi so' 1; k6' 66 la xoa cac s6'
ghi vao luc dau; va lap lai viec lam ban nay. ChiJug minh rkng n6'u
ciJ tiep tuc lam mai thi khong the nao nhan du'Oc tren dtfdng tron
toan chin so' 0.

Giii

Gia su" khi la tdi Ian n ta cd du'dc 9 so' 0 tren du-dng tron. Suy ra
cl Ian thu" n - 1 cac so' tren du'dng tron phai bkng nhau (va di nhi6n
khac 0, vi n^'u da cd 9 so' 0 rdi d Ian n - 1 la mau thuan ) va deu bkng
1, do dd d Ian thif n - 2 tr^n du-dng tron phai cd 9 sd'dlu khac nhau ddi
mot lien tiep. Mud'n dieu nay xay ra thi cac so' 0 va cac so' 1 phai bang
nhau. Dieu nay la trai vdi d^ bai vi tong cac con so'ghi trSn du'dng tron
la 9 (day la mot s o l e ) .

Vay khong the xay ra tren du'dng tron gdm toan 9 so'O.

8.18: Tren du'dng tron ta vi6't 30 so', sao cho moi so' trong chung
bing gia tri tuyet dd'i cua hieu hai s6'ke'ti6'p theo chieu ngtfdc chieu
kim ddng hd. Bie't rang tong tat ca cac so' hlng 20. Hay tim ta't ck 
cac so' nay ?

Giii

Theo gia thie't ta» suy ra cac so' da cho la
khong am, goi cAc sd'theo iM t\S ngrfdc chieu
quay kim ddng hd bat dau tuf so' Idn nha't 1^

â,ậ..,ậ (oj >a,.;i = 2,3,...,30)

Theo gia thie't = « 2 - « 3 mk* < « 2 
^2 ~ ^3 ^'h 
214!

odd :

Ng'u ^2 - « 3 <a2=^a^<a^ ma a^ > do dd =
V d i : = , th^o gia thid't suy ra

= 0,0^ = a i , a j =ai,ag = Q\...\a^ =a^,a^g = « P« 3 O = 0 , tong cac
sd'b^ng 20ai = 2 0 = > a i = 1 .

Do do Oj = = flg = . . . = = 0 ; cAc s6'c6n lai b^ng 1.

• Néu

- I

< a, ^ â <â ma a, > dj do đ =ậ >t
V d i : a, = 0 , theo gia thid't suy ra

=0,a,=a^,a^= 0,a^ = a„...,a^ = a„a^g = O . O J Q = a , . 
Tong c^c s6' b^ng 20a, = 20 =^ a, = 1.

Do dd = ô 5 = ^8 = ããã = ^29 = 0' '^^^ h\n% 1.

8.19: Ba hoc sinh Idp X: Dan, Mao, Thin di chdi, tha'y mdt ngu-di
lai xe 6 t6 v i pham luat le giao thdng, khdng ai nhd s6' xe Ik bao
nhi^u, nhufng mSi ngufdi ddu nhd mot dac diem cua s6' xe. Dan nhd
r^ng hai chff s6' dau gid'ng nhau, Mao nhd Ik hai chff s6' cud'i cdng
gio'ng nhau. Thin thi qua quye't r\ng so xe cd bd'n chif sd" la mdt s6'
chinh phtfdng. Chiing ta hay thiJ tim so' xe.

Giii

Goi cdc chi! s6' thi? nha't va thif hai \k x, cdc chi? s6' thit ba vk thiJ tiT

J a y. Sd'xe se la:

W 1000A: + 100A:+ 10>' + ) ' - 1 1 0 0 ; c + !!>' = 11.(lOOx + y) (*)
So' nay chia h6't cho 11 va cung chia h^'t cho 11^ v i theo bki ra nd Ik

#

6t s6' chinh phiTdng. NhuT tha' lOOx + y chia ha't cho 11.

Dijfa vko dau hieu chia h6't cho 11 thi sd* x + y cung chia h6't cho 11.
Bidu dd cd nghia Ik x + y = 11 ,vi moi chff so' x vk y 6iu nhd hdn 10.

</div>
(109)<div class='page_container' data-page=109>

y = 6 thi

= 5, y = 9 thi x = 2.

Vay ta chi tim s6' xe trong 4 s6' sau: 7744, 6655, 5566, 2299.

Trong b6n s6' nay cd s6' 7744 = 88Ma so' chinh phuTdng.

Vay s6'xela:7744.

8.20: Trong m6t mang hidi lien lac giQ"a viing A va B c6 n tram

d vilng A va k tram d vi^ng B. Moi tram d vung A c6 ihi liin lac

diTdc vdi it nha't k - p tram 5 vilng B. ChiJng minh r^ng n6'u: n.p < k

thi cd it nha't mot tram d vung B cd the lien lac duTdc vdi moi tram d

vilng A.

Goi s6' mang hidi lien lac giffa n tram d A vk k tram d vdng B \k s.

Ta chitng minh bkng phan chiJng.

Gia suf khong cd tram nao d viing B lien lac vdi ca n tram d A. Ta

tha'y ngay 1^:

s<k{n-l) (1)

Mat khdc theo gia thie't thi moi tram d vilng A cd the lien lac diTdc

vdi It nha't k - p tram d vi^ng B.

Dodd: s>n{k-p) (2)

Ta lai tha'y do : np < k n6n : nk - np > nk - k, hay:

n{k-p)>k{n-\) (3)

Tir (1) va (2) ta suy ra : n{k- p)<k.(n-1) trai vdi (3).

Vay phai cd it nha't 1 tram d v^ng B li6n lac difdc vdi ca n tram d

vilng A.

216

ll/dNG IX: TRICH DE THI TUYEN SINH LOfP 10

DEI

ll/dNG

PTTH CHUYEN LE

H6NG

PHONG 2004 - 2005

J»h^n chpn : Hoc sinh chon mot trong hai cau sau day :

L la : (4d) Cho phiTdng tfinh: - 3 (m +1) x + 2m' -18 = 0 (c6 an

la

X).

h) Tim

m de phifdng trinh cd hai nghiem deu am.

b) Goi ;ci, la hai nghiem ciia phifdng trinh. Tim m de cd:

so

X, -X,

< 5 .

cau lb: (4d) Rut gon cac bieu thiJc sau:

x' +

a) A = p 7=

—

7

X

+ yJx

+l

X

-y/x

+l

+X+1

2

+ yfx yfx -2]iXyfx + X - \fx -I

h) B=

p

II Phan bat buSc:

CSu 2: (4d) Giai phifdng trinh:

a) ^3x^

+X

-4 =2-2x.

(X >

0).

(x>0)

b)

3 - V 9 + 2x)

2x'

^u3:

(4d)

- = x + 9.

^ a) Cho A: > 1, y > 1. Chitng minh: x^ly^ + y-Jx^ < xy.

b) Cho X > 0, y > 0 va X + y = 1. Tim gia tri nho nha't cua bieu thtfc:

. A =

1 -

1

</div>
(110)<div class='page_container' data-page=110>

C a u 4: T i m cac so'nguyen x, y thoa he: y

-X — -X - 1 > 0

y-2 + x + 1 - 1 < 0

Gi^i

C a u l a :

a) T a c d : A = (m + 9) > 0 ; V m nen phifdng trinh luon cd hai nghiem
la: = m — 3 ; :iC2 = 2m + 6

Phifdng trinh cd hai nghiem deu am
A > 0

m - 3 < 0 ^
2m + 6 < 0

m<0 < ^ - 9 ^ m < - 3 .

m < - 3

b) T a c d :

<5<;=>m + 9 < 5 < ^ - 5 < m + 9 < 5 < ^ - 1 4 < m < - 4 .

C a u l b : 
a) Ta cd:

A =

^i^^fx-ij(x + ^fx+lj

VI(VJC+I)(A:->/I + I)

x + yfx+l x-sfx+1 
-^X-ylx-X-y[x+X + l = i^yfx-lJ .

+ X + 1

b) T a c d : B = ' 2 +

^/7 ^ / I - 2

;C(N/]^

+ I

) - (

V I + I)^

2 + V]^)(V^ -

1 )

- (V^^ -

2 )(V^ +

1)

^ V r + i ) ' ( V I - i )

• ( V I + i ) ( . - i

)l

7^

218j

2^ ]fiiZll]_^^(^"^^-2

i 2 :

2 - 2 x > 0

3; , 2^ ; c - 4 = 4 - 8 x + 4 x '

h) D i e u k i e n :

K h i dd :

•\

9 + 2;c > 0

^ 9 + 2

J C

- 3 ^ 0

x^O

^ x = l.

2;c' (3 + V9 + 2X 
\

•\

3 - ^/9 +

2 A:

3 - ^ 9 + 2 ^ ) ' (3 +

N

/9

T

2^)

- = jc + 9

2X + 1 8 : : 6 V 9 T 2 ^ ^ ^ + 9 ( ^ ^ 0 )

- 6 ^ 9 + 2x = 0 <^ ^ = - 2 '^''^^''^

^ , . . i ± ( p l ) = ^ (*)

y , / r ( 7

: i

) < y - 2 2

l + _ xy ^^^^

Cong (*) va (*•*) theo ve ta cd: x ^ / y ^ + y^lx-^ <

xy-Dau " = " xay ra k h i va chi k h i x = y = 2.

b) T a c d : xy < \x + y

1 1

— —

4 xy

> 4

</div>
(111)<div class='page_container' data-page=111>

_ (^ + 1)(>'+1)A:>> _ (x + \){y + l) _ ^ + ^ + y + i

(xyf xy xy 
- - ^ + 2 ^ 1

+

A > i

+ 2.4 = 9.

xy xy

Dau " = " xay ra khi x = y = — . Vay min A = 9.

CSu 4 : Tim cac so' nguyen x, y thoa he

( 1 ) ^

x'-x 1 > 0 (1)

X —X

y-2 + x + l - 1 < 0 (2)

<y-l=^y-\>0^y>l (3)

y-2\<l [ l < 3 ; < 3
( 2 ) ^ y-2 + x + l <1=^

;c + l < l [ - 2 < A : < 0
Do do ta suy ra xE { - 2 , - 1 , 0 } \h y e {1,2,3

Thijf lai ta difdc tap nghiem can tim la: {(-1;3);(0;2)

(4)

B E 2

L O P 10 C H U Y E N T O A N

TRl/OfNG T H P T C H U Y E N L E H O N G P H O N G 2004 - 2005

C a u l : (4d)Giaihe: 2x — y x + y 
1 1
2x-y x + y

• = - 1
(/)
= 0

220

(A2: (3d) Cho X > 0 thoa:x^ + \ 1. Tinh :x' + \

L ' * X X

3x

rek 3: (3d) Giai phifcfng trinh: ,

V3;c + 10

= V 3 x + l - l (1)

C3u4: (4d)

a) Tim gia tri nho nha't cua P = 5x''+ 9y^ - 12;c>' + 24x - 4Sy + 82 .

, , \x + y + z = 3

b) Timc^cso nguySnx, y, zthoahe , , , „
x'+y'+z =3.

Gi§i

CSul : Dieu kien: j c - 2 > ' ^ 0
x + y^O

Dat:

1
u = 2x-y

thi: (/) ^

V = •

3 M - 6 V = - 1 1
M - v = 0 3
x + y

2x-y = 3 
x + y = 3

x = 2 
y = \'

| 2 : ^ ' + ^ = 7

$ nen A : ' + - ^ =

1
x +

-X 
X

- 2 = 7

= 9=^;c + - = 3 {do x > 0 )

X

1
;c +

-X

X - x - + x'—-x — -\

X X X X

= 3 1 x' + 1 + 1

= 3 x' + 1 - 2 - 7 + 1 = 3[49_ 81 = 123.

</div>
(112)<div class='page_container' data-page=112>

CSu 3 : Dieu kien: 3x + 1 > 0 <^ ;c >

Dat: t = yl3x + l ^ t>0

=3x + l

• e

+ 9 = 3x + 10.
Tacd: (1)<^ f ^ = f - l 4 » ( / - l ) f < + l - V f ' + 9

- ^

= 0

t = \ (2)

f + l - # T 9 = 0 (3)

(2)<^ V3;c + 1 = 1 ^ 3 ; c + l = l4^;c = 0

0) ^ yjt^ + 9 t + \ + 9 ^ + 2t + \ 2t =

<^ / = 4 <^ ^ / 3 ^ T T = 4 <S=> 3;c + 1 = 16 <^ X = 5. 
Vay ( l ) < ^ x = 0V;c = 5.

C a u 4 :

a) Tacd: P = (2x-3>' + 8 ) ' + ( ; c - 4 ) ' + 2 > 2 .
Dau bkng trong bat ddng thitc tren xay ra k h i :

2;c-3>' + 8=:0 y =

x = A

16
3

x = A.

Vay min P = 2.

I) Tacd: x' +y' +e ={x + y + z)^-7>(x + y){y + z)(z + x)

^ 3 = 21-?>(x^y){y + z)(z + x)^\^9-{x + y)(y + z)(z + 
^(x + y){y + z){z + x)^^^{2,-z){2>-x)(?,-y) = %

Suy ra 3 - z, 3 - X, 3 - z la cic ifdc so cua 8. 
Ma cic Mdc so cua 8 la ± 1 , ± 2 , ± 4 , ± 8 .

Nhir vay 3 - x , 3 - y , 3 - z nhan mot trong gi^ tri da neu.
LSp bang:

222

- 1 +1 -2 2 -4 4 -8 8
3 - x

^ 3 - y
[ 3 - z

Thur tren bang ta diTdc:

x = \ A: = 4 A: = 4 ;c = - 5

y = 4 ; > = - 5 ; > = 4
z = l z = - 5 z = 4 z = 4.

B E 3

T R I J 6 N G T H P T C H U Y E N T R A N D A I N G H I A 2004 - 2005

C a u l : (4d) Cho phifdng trinh: / - ( 3 m + 14);c^+(4m + 1 2 ) ( 2 - m ) = 0
(cd an so" la x).

a) Dinh m de phifdng tiinh cd bo'n nghiem phan biet.

b) Dinh m sao cho tich so' cua bo'n nghiem tren dat gia tri Idn nhat.
'Su 2: (4d) Giai phifcfng trinh: .

a) x" + 2;c + l| - 1 = 2-x\

\2x-%

b) V'2A- + 4 - 2 V 2 - X = .

V 9 x ' + 1 6

|^§u 3: (3d) Cho x, y 1^ nai s6' thifc khac 0. Chtfng minh: 
2 2

^ + ^ + 4 > 3 ^y x^ (1)
CSu 4: (3d) Tim c^c so' nguyen x, y thoa phtfdng trinh:

Gi§i

C a u l : / - ( 3 m + 14);c\+(4m + 1 2 ) ( 2 - m ) = 0 (*)

a) E-inh m de phifcfng trinh (*) cd bon nghiem phan biet:

</div>
(113)<div class='page_container' data-page=113>

Bat t = x^.

(*) <^ - ( 3 m + 14)t + (4m + 1 2 ) ( 2 - m ) = 0 (* *) 
/ = 4m + 12

r = 2 - m

(*) c(5 bo'n nghiem phan biet :
4m + 1 2 > 0

2 - m > 0 - 3 < m < 2
4m + 1 2 ^ 2 - m

b) Dinh m sao cho tich cua bon nghiem tren dat gid tri Idn nhS't.
T a c d bo'n nghiem cua (*) la ±^,±^[1^, v d i 1^,1^ 1^ nghiemciia(

x^X2X^x^ =t^.t2 = ( 4 m + 12)(2 —m)

= - 4 m ' - 4 m + 24 = - ( 2 m + if +25 < 25Vm 
=^ Gia tri Idn nha't ciia jCjjCj ^1:3^:4 la 25.

difcfc khi m = — ^ (thda dieu kien d cSu a).

Cau2:

a) x^ + 2x + l - 1 = 2-x^ ^

2-x^>0

x" +|2x + l | - l = 2 - x '
x ' + 2;c + l - l = x ' - 2 .
2X + 1| = 3 - 2 J : '

x'<2 
2x + \

x'<2 (VN)

3-2x^ > 0

x'<2

2x + l = 3 - 2 x '
2X + 1 = 2 X ' - 3

224

x'<-" 2

2; c' + 2 x - 2 = 0
2; c' - 2 x - 4 = 0

x'<-2 
x =

x = -\ 
x = 2

X = - 1
X =

^2x + 4-2^2-x = 12;c-8

V97

+ I 6

6x-4 1 2 X - 8

yl2x + A + 2s!2-x ^ 9 x ' + 1 6

( - 2 < ;c < 2)

_ 2
^ ~ 3

'42x + 4 + 2 V 2 ^ ] = V9^' + 1 6
(1)

(2)

(2) =^ 4 (2x + 4) + 1 6 (2 - jc) +16^8 - 2J:' = 9x' + 1 6 
^ 1 6 V 8 - 2 J C ' - 8 X = 9 X ^ - 3 2

8 ( 2 ^ 8 - 2 ; c ' - ;c) - 9A:' - 32 
8 ( 3 2 - - 9 x ' ^ ,

2 V 8 - 2 ; c ' + j r

4x/2

9 x ' - 3 2 = 0

2 ^ 8 - 2 x ' + J c = - 8

JC = ±

-2 ^ 8- 2 x ^ =-%-x (vdnghiemvl : - 2 < x < 2 )

X = ± — . M lai ta difcfc = .
3 • 3

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C a u 3 :

Dat t = - + ^=>\t\= — X

+

I ma X

+

I

y X y X y X y X

> 2 (do bat danc
thiJc C6si) ^ I H > 2 ^ ? < - 2 hay 2 < /

K h i d o : = ^ + ^ + 2

y X

Bat dang thtJc (1) <^ + 2 > 3< <!=^ - 3< + 2 > 0

4^ ( / - ! ) ( / - 2 ) > 0 (2)
(2) la hien nhien ditng do t<-2 hay 2 < r'. <

x''

Vay : ^ + ^ + 4 > 3 x_^y_ \y

X

C a u 4 : +xy + y^ = ;c'/ =»(2x + 2}')' = (2xy^\)^ - 1

=> [2xy + 1 + 2x + 2>') [2xy + 1 - 2A; - 2>') = 1

^ 2xy + 1 + 2x + 2y = 2x}; + 1 - 2x - 2^

•=^x + y^Q.

Thay vao phifdng trinh ban dau ta c6:

X = 0, y = 0 hoac x = 1, y = — 1 hoac x = — 1, y = 1.

B E 4

L(5P 10 C H U Y E N T O A N

T R U C K N G T H P T C H U Y E N T R A N D A I N G H I A 2 0 0 4 - 2 0 0 5 
Cau 1: Cho phiTdng trinh: x^ + px + 1 = 0 cd hai nghiem phan biet 01,^2

va phifcfng trinh x^ +qx-\-\ Q nghiem b^,b^.

ChiJng minh: (oj -b^){a^ -b^)[a^ + ^2)(«2 + ^2) = 9^ - •
Cau 2: Cho cac so' a, b, c, x, y, z thoa: x = by + cz, y = ax + cz,

z = ax + by, x + y + z .
2261

Chu-ng minh: — h — ^ + — ^ = 2.

\+a l+b 1 + c

t

a) Tim x,y thoa 5x^ +5y^ + 8x>' + 2 x - 2 > ' + 2 = 0. 
b) Cho cac so' dUdng x, y, z thoa jc^ + + = 1.

ChiJng minh:

vrr7

^/r7

y > 2 .

Cau 4: Chvfng minh rkng khong the' c6 cac so' nguy^n x, y th6a phiWng
trinh x ^ - / = 1 9 9 3 .

Gi§i

C a u l :

Theo dinhly V i e t t a c o :

^ 1 + 0 2 = - ^ ; a^a^=\ b^-\-b.^=-q ; b^b^=\

(«1 + ^ ) ( « 2 +^2)

= (fljOj - ( r t , + rtj+ bl \a^a^ + (fli + 02)^2 + ^2 
= (1 + pfo, + ^,^)(l - pb, +bl) = [pb, - qb,){-qb, - pb,)

= {p-(i)K(-p-^)K^^^ -

p^:au2:

-Cong ve vdi ve edc d i n g thiJc ta difdc: x + y + z = 2 (ax + by + cz).

Do x + y-\-z^Qx\Qn ax + by +

cz^Q-Cong hai ve cua cac d^ng thiJc Ian liWt cho ax, by, cz ta diWc:

+ \)x = ax + by + cz \{b-{-\)y = ax + by + cz ;(c + l)z = + fay + cz

X

y_ . z - + •

1

+ a,' \ b '

1

+ c ax + by + cz ' ax + by + cz ' ca + by + cz

x + y + z

ax + by + cz • = 2.

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(115)<div class='page_container' data-page=115>

a) 5x''+5y^+8xy + 2x~2y + 2 = 0

^x + y = 0,x + l = 0^y_^^Q^^^_^

x V T T ^ - ^ 2 ^ 1 _ ^ 2 - 2J:' . C h i l n g m i n h tiTdng tiT:

C3u 4 :

y =l993^{x-y)[x'+xy + /) = m^

D o 1993 la so' n g u y e n to', n e n ta c6 cac he phiTdng trinh:
f j r - 3 ; = 1993

x-y = l

X +xy + y^ = 1 9 9 3 A: + + = 1

MUC LUC

CHlTOfNG I : D A N G T H l f C 
C H U O N G I I : B A T D A N G T H l f C

§ 1 : Phep bien doi tifdng diTdng. T i n h chat cua ba't dang thiJc
§ 2 : Bat dang thiJc Cosi (Cauchy)

§ 3 : Bat dang thuTc trong tarn giac 
§ 4 : Phifdng phap l a m troi

CHl/dNG I I I : SO H Q C

CHUCfNG IV : G I A T R I LCfN N H A T VA G I A T R I NHO 
N H A T C U A H A M SO

CHUCfNG V : P H U O N G T R I N H

§ 1 : Phifdng trinh bac hai - PhiTdng t r i n h bac ba
- D i n h ly V i - e t

§ 2 : Phifdng trinh qui ve bac hai

§ 3 : Phu'dng trinh chufa gia t r i tuyet doi. PhiTdng trinh 
chiJa can thufc

§ 4 : PhiTdng trinh v d i n g h i e m so nguyen

CHl/CfNG V I : H E P H U C I N G T R I N H

§ 1 : H e hai phifdng trinh bac nhat hai an. H e ba phuTdng
trinh bac nha't ba an

</div>
(116)<div class='page_container' data-page=116>

§ 3 : He doi xiJng. He dang cap 161
§ 4 : He phUdng trinh c6 dang dac biet 169
CHUCfNG V I I : D O T H I CUA H A M SO 179
CHl/ONG V I I I : M O T S O B A I T O A N V E T O HQfP P H E P

D E M . D O N G T l f 202
C H U O N G I X : T R I C H D E T H I T U Y E N S I N H LCfP 10 217

230

R E N L U Y E N TOAN NANG CAO DAI SO 9

* •

TS Nguyin Cam (Chu bien) • ThS Nguyin Van Phifdc

N H A X U A T B A N

DAI H O C QUOC GIA T P HO C H I MINH

Khu pho 6, Phi/cfng Linh Trung, Quan Thu Dure, TP.HCM
DT: 7 242 181 - 7 242 160 +(1421, 1422, 1423, 1425, 1426)

Fax: 7 242 194 - Email:
* * *

Chiu trdch nhiem xudt ban

P G S - T S N G U Y E N Q U A N G D I E N

Bi&n tdp

N G U Y E N V I E T H O N G
Su'a ban in 
T R A N V A N T H A N G

Trinh bay bia

M I N H D I E N

TK .01. T(V)

D H Q G . H C M - 0 5 011/ 304 T.TK.453 - 05(T)

I n 3000 cuon, kho 16 x 24cm. Giay ph6p xuat ban

011/ 304/XB-QLXB do Cue Xuat bdn cap ng^y 17/ 03/ 2004.
Giay trich ngang so: 564 / K H X B ng^y 28/ 09/ 2005.

</div>

Rèn luyện Toán nâng cao Đại số 9 - Nguyễn Cam

TS.

N G U Y E N

CAM (Chu bien)

^ ThS.

N G U Y E N V A N

PHUCfC

<b>REN LUYEN TOAN NANG CAO </b>

<b>DAI</b>

<i> SO</i>

<b> 9 </b>

* Phan loai cac dang toan nang cao dinh cho hoc sinh khi gi6i.

* Tuyen chon cac de thi tuyen sinh Idp 10 vao cac triTdng chuyen.

<b>THLT VIENTIfvHBINH THUAN </b>

<i><b>LdlNOIDAU </b></i>

<i><b>Cung vdi viec bien soan bo sach Ren luy^n todn can ban LOP 9, </b></i>

<i><b>chiing toi gidi tliieu cuon Ren luyen todn ndng cao DAI SO 9 nham giup </b></i>

cac em hoc sinh nang cao trinh do de tham gia cac ki thi tuyen vao Idp 10

<b>thuQC cac trirdng chuyen, Idp chuyen. </b>

Cuon sach bao gom 9 chifcfng :

ChUdng 1 : Dang thiJc

ChiTdng 2 : Bat dang thiJc

Chufdng 3 : So hoc

ChiTdng 4 : Gia tri Idn nhat va gia tri nho nhat cua ham so

ChiTdng 5 : Phifdng trinh

ChUdng 6 : He phifdng trinh

ChiTdng 7 : Do thi ham so

ChiTcfng 8 : Mot so bai toan ve td hdp ; dpng tiJ

ChiTdng 9 : Mot so de thi vao Idp 10

<b>Vdi muc dich h5 trd dac life viec tuT hoc ciia minh, cuon sach diTc^c </b>

trinh bay theo bo cue nhiT sau : md dau mSi chufdng la phan torn t^t cac

kien thilc can ye'u ma hoc sinh phai nam vffng; tiep theo la mot he tho'ng

b^i tap CO hiTdng din each giai diTdc sap xep theo thiJ tif kho dan.

De viec hoc cd ket qua to't, sau khi doc va tim hieu that ky cac bai

tap CO hiTdng din hoc sinh nen co gang tif minh giai cac bai tap ay mot

cdch doc lap cho den khi thuan thuc. Vdi each lam ay, hy vong cac em se

tif minh kham pha ra nhieu each giai khac mot each thu vi.

Raft mong nhan diJdc cac y kien phe binh cua quy doc gia de cuon

<b>sich ngay cang diTdc hoan thien va phuc vu ban doc tot hdn. </b>

<i><b>TM.nhdm tdc gid </b></i>

<i><b>TS. Nguyen Cam </b></i>

<b>f </b>

<b>CHl/OfNGI:</b>

<b> DANG THlfC </b>

I . HANG DANG THlfC

1)

<i> (a + bf =a'+2ab + b\ </i>

2)

<i> {a-bf=a' </i>

<i>-2ab + b\ </i>

3)

<i> {a + bf =a' -h^a'b + ^ab^ +b\ </i>

4)

<i> {a-bf=a' </i>

<i>-3a'b +3ah^-b\ </i>

5)

<i> a' +b' ={a + b){a'-ab + b'). </i>

6)

<i> a'~b'={a--b)(a^+ab + b^). </i>

7)

<i><sub> a"-b"={a--b\a"-' +a"-'b + a"-\b' +... + ab"-' +b"-'). </sub></i>

8)

<i><sub> {a + b + cf = +b^ + 2(ab + bc + ca). </sub></i>

9) (a + Z) + c)' =

<i>a'+b'+c'+ 3(a + b)(b + c){c + a). </i>

<b>n.</b>

CAN THlfC

1)

<i> ^A^ =\A\=< </i>

<i>AneuA>0 </i>

1)

<i> ^A^ =\A\=< </i>

<i>-AneuA<</i>

0

2)

<i> A+B±lylAB=(^±4B^{A>0,B>Q) </i>

3) ^ + 5

<b>' ± 2 5 V I = ( V I</b>

± 5 ) ' ( ^ > 0 )

1