luyện giải đê trước kì thi 3 miền bắc tring nam môn toán

Nh^ gi^o uu tij - TTi^c siToAn hpc
Gl^ng vi«n chfnh - Thac sITo^n hpc

Ph6 gi^o su - mn si:

NGUYEN VAN THONG
NGUYEN VAN MINH
NGUYEN VAN HIEU

(Chu bifen)

••a

LIIYeNGIIUIliniUIICKVTHIIilUHOCSMliN

TOAN
HOC
ii@®

Ddnh cho hpc sinh 12 luy^n thi BH-CB ^
Biin so$n theo n0i dung djnh hudng ^
i mdi cua BO Gi^o Due & B^o Tao

THLT MIEN TiWH BINH THUAN
MZ

DVL

J My^6' /I5

I

NHAXUJTBiiNTONGHQPTHiiNHPHOHOCHlMINH


Left noi ddu
Chiing toi la cac anh em ruot thjl, muo'n vie't quyen sach nay cho the he sau on
tap de chuan bi thi v^o dai hoc, vi bU'dc v^o triTdng dai hoc, ngtfcfi hoc sinh b^l
dau mot ddi song mdi vk c6 tiTdng lai tiTdi sang khi chpn dtfdc mot triTcJng dai hoc
tot, ch^c ch^n r^ng triT^ng nay se chon diem cao. NhiT vay trong qua trinh on
luyen cac em can mot tai lieu tUcfng thich. Dieu nay se diTdc thoa man neu cAc em
chiu kho on tap theo cac chuyen de mh chung toi bien soan, ch^c ch^n thi se dai
diTpc diem cao.

I 'K^mdt m MJOl C«0
GI(3l HAN

Mong muon ciia chiing toi l ^ m the nko de c^c em tif hpc to't mon Toan, va
kha nang thi dau v^o cdc tri/dng Dai hpc Idn phai nh5 vko sir kien tri cua cic
em "c6 cong mai s^t c6 ng^y nen kim" cd nhan day that diing vay.
Quyen sich nay cung 1^ ky nipm 3 nim cic em hpc Toan vdi thay: Bich
Lien, PhiTdng Thao, Anh ThiT, Mai HiTdng, Thiiy Hien, Th^o Uyen. Cic em da
giiip thay chinh sufa ban thao, mpt cich nhipt tinh trdch nhipm.
Chiic c^c em se th^nh cong my man trong ky thi s^p idi
Chu bien: Nguyen VSn Thong
(To trirdng To Todn trirdng chuyen Le Quy Don - Da NJng)
Nhd sdch Khang Vi$t xin tran trgng giai thi^ tai Quy dgc gia va xin
Idng nghe tnoi y kien dong gop, de cuan sdch ngdy cdng hay han, bo ich hart.
Thu xin giH ve:
.„,.^-..
_

Cty T N H H M p t Thanh Vien - Djch
Van Hoa Khang Vi?t.
71, D i n h Tien Hoang, P. Dakao. Qu^n 1, TP. H C M
Tel: (08) 39115694 - 39111969 - 39111968 - 39105797 - Fax: (08) 39110880
Hoac Email:



z^r TWM I \ F M

TRONG DI^M

vA TfNH UfeN TUC C6A

1. C a c d j n h If ve g i d i h a n

M O T H A M S6

••1 ( '

Gia sur lim f ( x ) = L, va lim g(x) = L j , khi do
X-^X()

lim

[f(x)

X-+X,)

+ g ( x ) ] = lim f ( x ) + lim g(x) = L | + L 2

x->X()

lim

"->"()

lim

"-+"()

x-^xo

[ f ( x ) . g ( x ) ] = l i m f ( x ) . lim g(x) = L|.L2

x^X()

x-^xo

f(x)

lim

X-^X()

Neu

g(x)

lim

f(x)

lim

g(x)

X^X()

,
_ ^1

X-*X()

='--'v,V'

''->X()

L2

vdi

3s > 0 sao cho f ( x ) < g(x) V x € (xo - e; x,) + s) va ton tai

lim

f ( x ) , lim g(x) thi lim f(x) < lim g ( x )

x->X()

a.

"-^^o

[ f ( x ) - g ( x ) ] = lim f ( x ) - lim g(x) = L , - L 2

x->X()

4 trong bang xep

Cic chuyen de chiing toi deu soan tilf can ban den nang cao, dp kho dii de
cho cac hpc sinh khd va gioi tU' luyen, neu thong hieu taft ca, ch^c ch^n ring cac
em se giai du'pc de thi Dai hpc mpt cdch de dang, nh\i chiing toi da tilfng luyen
rat nhieu em do thu khoa cic tru'dng Dai hpc danh tieng. Dieu quan trong hdn
nifa vdi each vie't ciia mOt gido vien day chuyen Todn lau nim, nen each gpi
md, d i n d^t mang dam n6t t\i duy nang cao, hpc sinh se diTcJc phat trien kha
nSng Toan hpc khi hpc quyen sdch n^y.

inx-r

I . T 6 M TAT L I THUYET

Quyen sdch nay gom 20 chu de trong diem theo cau true de thi cua Bo giao
due hang nam, cac vi du duTa ra ttfcfng doi kh6 va c6 hiTdtng dan giai. Tie'p theo
la 28 bp de thi cho cac khoi A, B, D cua tri/cfng chuyen Le Quy Don - Da N^ng
dilng de thi thijr tiTng nSm qua 66 danh gid diTdc hpc sinh,
nit ra kinh nghiem
giang day. TriTdng Le Quy Don - Da NSng hang nam deu c6 ti 16 dau vao cac
trirdng dai hpc la 100%.
Rieng nam 2010 c6 so hpc sinh dat diem cao diTcfc xep thu"

hang cua Bp giao due va Dao tao

mill

''-»X()

x^XQ

1 ,

li •

x->X{)

NguyinIfgidinank?p:
Neu

3 E > 0: f ( x ) < h(x) < g(x) V x e (x„ - E ; X„ + e) \}

va lim f ( x ) = lim g(x) = L thi lim h(x) = L .
x^X()

x-»X()

x-»X()

b. Cdc dang gidi han dQc bi$t
lim
X->X()

lim

X->X()

= 1 ; lim (1 + x ) x = e
X

X^XO

= e ; lim
x-»X()

I,,,,,.,.,.;,.,:

e^-l
X

= 1 ; lim

ln(l

+ X)

= 1

x->X()

2. G i d i h ^ n d a n g v d d j n h : —
0

P(x):dathu'c, P(x„) = 0 >
P(x)
a. Dang 1. I = lim — ^ vdi <.
x ^ x „ Q(x)
[Q(x): da thiTc, Q(Xo) = 0
PhiTdng ph^p:
7 — - . i'
1= Hm
lim i ^ i ^ ^ i o l W ^
P ^ ^ P ^
x^xoQ(x)
x->Xo(x-Xo)Qi(x)
X^X()Q,(X)
Qi(Xo)

vdiQ,(x«)^0

,


.

J I IWtg,

„

Ne'u Pi(x„) = Qi(x,)) = 0 thi phan tich tie'p •

Qu^

|,W

C6ngiyTNi:ii

P,(x) = ( x - X o ) P 2 ( x )
Q,(x) = ( x - X o ) Q 2 ( x )

0
trinh khuT dang v6 dinh ^ la qua trinh k h u r cac nhan tuT chung ( x - X o ) ' ' s e

difng l a i k h i nhan du'dc g i d i han xac dinh ttfc la Qk 5^ 0.
Khid6I=

x->xo

' f ( X ( , ) = g(x„)

vdi \

g(x)

Ng'u p < q thi ton t a i g i d i han

•

N e u p > q t h i khong ton tai g i d i han

; . W / / A / . U ; . ^

1 0>t

4. Gi6i han dang v6 dinh « - oo
Phifdng phap: B i e n d d i diTa ve dang g i d i han —

l i n . - ^ = l i m A W ^ M ^
>'^''()Q(x) x ^ x „ Q , , ( x )
Qk(xo)

b. Dang 2. l i m

•

:

'

T i m g i d i han sau

= 0

O O .

f ( x ) , g ( x ) chtfa can thtfc dong bac

lim

(x + V x j - x
.
= lim
x^+«>^x + ^/^+^/x

"

Vx + V x - V x = l i m
J

x-++<»L

PhiTdng phap: SuT dung cdc hKng d^ng thtfc de nhan l i e n hdp 6 tuT va mau
nh^m true cac nhan tijr (x - x,,) ra k h o i can thtfc.

5. Gidi han dang v6 djnh oo . 0

,

.

Phifdng phap: Du'a ve dang v6 djnh —
00

A-B

~.

^ ' J

C h i n g han, t i m g i d i han
lim

A + B

X-»+a>

x

- - lim

Vx^ + l - x = l i m

X->+oo .

mil tioub fi?. i m

Vx^ +1 +:

1
2

l + -^+l
X

A-B

6. Gidi h^n dang vd djnh ham lUging giac
2"M/A+2n+^^

c.

5.1 =
'

PhtTdng phap: Suf dung cac ket qua g i d i han cd ban sau dSy:

A + B

hm
vdi
x^xog(x)
[ f ( x ) chtfa can khong dong bac

•

,.
sinx
, ,.
X
,
hm
= 1; h m
=1
x->0 X
x->() sinx

•

lim

x->0

sin ax
X

= lim

x-»0

sin ax
I

ax

V

,. sinax
= a. h m
x->() ax

,.
sinax
a => h m
=a
x^O

X

„

sin ax
^^"0

g(x)

g(x)

x^XQ

h
m
mil

.g(Xo) = 0

x->()sinbx

Bi6n doi

1=

lim x->xo

sinax

'?yu(v)-ci-rjyv(x)-c
- lim

g(x)

•^uW-c

5!/v(x)-c

g(x)

g(x)

X-+XO

lim
x-»o

D e n day cac g i d i han diTdc tinh theo dang 2 .
•

3. Gi6i hain v6 djnh —
00

PhiTdng ph^p: X 6 t I =

tan ax
x

=
— iUm
iiii

ax

- ^

a , .
=

— -—( a , b e R * )
x-+()sinbx
b

= ^^lim

smax

.—:—-— — — —• 11111

x-»()bx sinbx
bx

b

a
sinax
,.
tanax
hm
.
= a => h m
=a
x->0
X
x^ocosax
ax

tanax
,.

ax
ax
a
Hm
= hm — •
=x->otanbx
x->{)bx tanbx
b
tanax

tanax
,.
hm
x->()tanbx

a
b

bx
lim ^

v d i P(x), Q(x) la cac da thi?c hoSc cic

ham

daj s6. G p i bac P(x) = p; bac Q(x) = q v^ m = min(p, q ) , k h i d6 chia ca
mau cho x " ta c6 k e t luan sau:

va

x-»xo

Sinax

g(x)

hm

sinax

x-*() tanbx

,
sinax
,.
ax
. a x
hmcosbx.
= hm—.cosbx. . .
x->()
sinbx
x->obx
sin Dx
bx

^
b

r

,UI


Luyfn gidi di truOc kp ihi DH 3 miin Bdc, Trung, Nam Todn hoc - Nguyin Van ThOng
Hifdng din giai

7. Gidi han dang v6 djnh 1°
Ta

Phifdng phap:
f
a. Sijfdung: lim(l + x ) " = e ;

lim

x->()

= e ,,,,
t'l

• •

•

.

I-.,,

lim

''1

"

X-»X()

^JZ^

x-»3

=1

>

X
I-

v l ^ - 1

x->0

US;' fili'd i&ili ffH'T

(u-ir

X \

X^

Hifdng d§n giai

(l + x 2 ) - l

Ta c6: lim

•+ X

—

x-.(.^2

l + (u-l)"-'

Bie'n doi lim

'^^^^ =
(X-3)X

Bai 3. Tim hm

b. Xct lim uCx)"*"' CO dang T
X^X()

lim

x^3

1

1+ -

x^+«>

CO

8. Gidi han d?ng v6 djnh cua ham mu va Idgarit: ^

3/^1777+^/i77^^1

- --

- == lim
iim

,

,

= —

''-*"^(i+x2)+^/iT^+i

Bai 4. Tim hm

^^^
' *

x-»()

X

Hi/dng d i n giai
Phifdng phap: S^rdung l i m ^
x-»()

^ = 1; lim
X

^

x->()

=1

Ta C O

X

II. B A I T A P M I N H H Q A

,. (2N/rT7-2) + ( 2 - ^ / 8 ^ )
hm

x-»()

X

,.
lim

X

x^O

Bai 1. Tim gidi han lim f ( x ) , vdi r(x) =
x^-1

X->1

,„;.,.

Htfdng din sial
Ta

CO

Vs-x^ - 2

lim
x^l

= Hm
x-^o

\/x^ + 7 - 2 ^

x^-l

x->()

hm

= lim X

Va lim
x->l

X[4 + 2 W ^

= hm

Matkhac:

X ^ l

>/l + x + l )

Vx" + 7 - 2
X

-1

= lim
X->1

" - l / v ^

-1

- f x ^ + x + l)
\

'

^ / ^ ^ + 1^4 + 2

.100

v2

x^+7

x^ +7 + 4

12

^

^

^^100

2

^

j ^ ^ N

100

+ ...+ 1 +

lim -

= 100
.100

(3)

19^

loo^ll^

+ ^ 1 0 + ^HX.

•a
Bai 6. Tim gidi han lim

Thay (2), (3) vao (1). la difdc L = - - - — = - —
8 12
24
. . ™ ,. x'^ - 4 x ^ +4x - 3
B a i 2 , Tim l i m —
"-^^
x--3x

,^100

1+

X-»+QO

1
3

12

Hi/dng din giai

Ta c6
x^l

""12

.
(x + 1)'"" + (x + 2)'"*' + ••• + (x + 9)'"" + ( X +100)
Bai 5. Tim lim
X-»+oo
x'"" + 10x'" + 100'"

^"'(x^-lWx^+vf+2^/?77
x^+7+4

= lim

+^(8-x)2J

^

l(X)

(2)

x^-l

= lim
"->!

+ N/(8-X)^

(1)

x^-l

1-x-^

fijsi

2x

x-»l

m
1-x""

, (m, n e N * , m
1-x"
Hi^dng dan giai

lim

X->1

f_n.

m

Al-x'"

l-xj

i_^]

:;

n) , ,

3


Luy(n gUU d6 trade

lim

thi DH 3 miin Bdc. Trung. Nam Todn hoc - Nguyen Van ThOng

m - ( l + x + x^+... + x'""')

n - ( l + x + x^+... + x""')

l-x""

1-x"

X-+1

lim

2 sin

( l - x ) + ( l - x ^ ) + ... + ( l - x " ' " ' )

( l - x ) + ( l - x ^ ) + ... + ( l - x " - ' )

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

(1 - x ) ( l + x + x^ +... + x"~')

x-»l

lim

2sm

l + ( l + x) + ... + ( l + x + ... + x " ' ' ^ )

l + ( l + x) + ... + ( l + x + ... + x""^)

l + x + x2+... + x " ' - '

1 + x + x^ +... + x " '

t

m(m-l)

l + 2 + ... + ( m - l )

n(n-l)

=

m-n

2

2

2

2 " '

sma

1-

2 sm—

,

•

cos

= Hm

smcx

l-iVcosbx

l-'^cosax

(cx)2

x-*()

A p dung (1):

2"

C=

(cx)2

u

. e
B a i 2. T i m g i d i han sau: I = Hm

BaiS.Tim

2

m

2c^

n

m ,

a
a
a
cos—.cos—...cos2"

lim
n->-H»

e-^"
I

.1-

^

a

a

a

A = cos —.cos — . . . c o s
2
2
2^^
1
2 sin

^
2"

2^^

. a
a
/sm — cos
.cos
2"
2" j

;a^0

•= Hm 3cos^ X .

x->()

x->()

. e"
B ^ i 3. T i n h g i d i han I = l i m

.cos

,n-l

3x2

i•

' Sin X

^

I

;

X

- V c o s x + l n ( l + x^)

' •!

x-»()

a

a

ijji'j

= 3-1=2.

a
= COS

x"

- 1 Icos^x + c o s ^ x - l

= Hm

lAvtdng dSn giai
A

cos x - 1

H i M n g d S n g i a i '"'^

1 + "Q^cosax +... + \/cos"' ' a x

aM

n

,

l i m ^ i M L • = ... = 0
x->o sm(tanx)
tanx

1-cos ax

l + !ycosbx+... + N/cos"-'bx

2

jj,,

—cosx

x^O

1-cosbx

bM

.,

Hifdng d i n giai

r

x-X)

,,.

cos — cosx
2
B a i 1. T i m g i d i han sau: C = Hm — ;
x-»o sin(tanx)

(1)

(cx)"

C

I ' , - j j •••

n

a

r

= -^lim

sma

2"

III. B A I T A P T i ; L U Y E N C O D A P S O .

(l-jycosbx)-(l-".ycosax)
x^O

i'.'Jir]

sm-

Hifdng d i n giai

lim

sma

2"

I

1-cos ax

2".sin-^
2"
a

'

„
%/cosax-Vcosbx
^,
,^
.
B a i 7. T i n h L = l i m -^^
(a, b, c la cac so thifc khac khong) m , i
"-»('
sin cx
<• m cac so tir nhien khac nhau Idn hdn 1.
''
^

Bdde: lim

sma

2sin—cos—
2
2

2 " sin ^
2"

m

m
n-1

a
a
...cos-^.cos—
2^
2

2"

l + 2 + ... + ( n - l )

m-1

2n-l

rcos

a

HrfdngdSngiai

a

, n - lr . . . C O S - T - C O S —

a

a

1= lim

2

x-+()

r...COS-T-COS—

'

e"

-1

x^

^^"'^

2(l + V ^ ) Y x 2 ' l
, 2 ,

+

l n ( l + x^)

-x

4

,


Luyfn giil di trudc thi DH 3 miin Bdc, Trung. Nam Todn hpc - NgiQ^n Van Tlidng

C6ng ty TNHH MTVDWH Khang V

toan xdl giao diem cua do thi y = f(x) vdi drfdng thing y = m (vuong gdc vd
true tung va c^t true tung tai diem cd tung dp m).
4. SI/ tiep xuc cua hai dudng cong.
•

C A C B A I T O A N K H A O S A T H A M s6
I.

a. Dinh nghla: Hai ham so' f(x) va g(x) tiep xiic nhau tai diem M(xo; y,,) neu M

la mot diem chung cua chung va hai di/dng cong cd chung tie'p tuye'n vdi
nhau tai M .

T 6 M TAT LI THUYET:

1 . Cac bu6c khao sat si/ bien thien va ve do thi cua ham so
Bifdc 1. Tim tap xac djnh cua ham so
Btfdc 2. Xet sU' bien thien cua ham so
-

Tim gidi han tai v6 cifc va gidi han v6 cifc (neu c6) ciia ham so. Tim cic
dircJng tiem can ciia do thi (neu c6). l>

-

"

Lap bang bien thien cua ham so, bao gom: Tinh dao h^m cua ham so, xet
dau dao ham, xet chieu bie'n thien va tim ciTc tri cua ham so (neu c6), dien

.

cac ket qua vao bang.

..^.^

BMcJ. Ve d6 thi cua ham so.

b. Dinh It: Hai diTdng cong y = f(x), y = g(x) tiep xuc vdi nhau khi va chi khi he
ff(x) = g(x)

phiTdng trinh an x sau day cd nghiem: \
[f'(x) = g'(x)
(He phiTdng trinh nay la h? phiTdng trinh \ic dinh hoanh dp tiep diem cua
haidirdng).
'
- •
II. C A C B A I T O A N M I N H

HQA.

^

^^

j

Bai

1.

Cho ham so y = f(x) = - x ' + 2x^ + | x

"^^

-

Ve cac du'dng tiem can cua do thi (neu c6).

a. Khao sat h^m so.

-

Xac dinh mot so diem dac biet cua do thi, chang han tim diem uon, giao
diem cua do thi vdti cac true toa dp (trong triTdng hdp do thi khong c^t cac

b. Tiep tuyen cua do thi (C) cua ham so tai goc tpa dp, cat (C) tai diem M .
Tinh tpa dp diem M .

true toa dp hoSc vice tim tpa dp giao diem phtJfc tap thi bo qua phan nay),...

c. Bien luan theo k so giao diem cua (C) va du'dng thing (d): y = kx.

-

Nhan xet ve do thj: Chi ra true va tam doi xuTng cua do thi (neu c6, khong
yeu cau chufng minh).

2.

Diem uon cua do thj: la diem U(X(); yo) ciia do thj sao cho lie'p tuye'n tai dd
di xuyen qua do thi, tiJc la ton tai mot khoang (a; b) chuTa diem Xo sao cho
tren mot trong hai khoang (a; x,,) hoSc (X(i; b) tiep tuye'n cua do thi tai diem
U nam phia tren do thi, con tren khoang kia tiep tuyen n^m phia du'di do thi.
Ta thU'dng suT dung ket qua sau day de tim diem uon cua do thi:

n

a. Khao sat ham so

3. Giao diem cua hai do thi.

a. Phu"dng trinh (xac djnh) hoanh dp giao diem ciia hai do thj y = l(x) va
y = g(x) (ciing ve tren mot mat phing tpa dp) la f(x) = g(x) (1)
b. So nghiem cua (1) cung chinh la so giao diem cua hai do thi y = f(x) va
y = g(x). Dac biet: PhiTdng trinh (1) cd nghiem (v6 nghiem) khi va chi khi hai
do thj citt nhau (khong cat nhau).
i
c. Ta thudng gap u-irdng hpp phiTdng trinh (1) cd dang f(x) = m, trong dd m la
tham so va ham so f(x) khong chtfa tham so m. TriTdng hdp nay ta cd bai

Hi^ngdlngiai
1

vi£a>t,

l.Tapxacdinh:D = R

J/

2. Sir bien thien:
a. Gidi han:
lim y = lim
X-»+00

Neu ham so y = r(x) cd dao ham cap hai tren mot khoang chuTa diem Xo,
f"(xo) = 0 va r'(x) doi dau khi x di qua diem Xo thi U(X(); f(X())) la mot diem
uon cua do thi ham so'y = f(x).

Ia

;

-X

^

o 2
+2x

5

- -00

+ - X

X->-H»

. ...2
lim y = lim (-x 3 +2x
+ -5x ). =
X->-QC

X->-<»

+oo

3

b. Bang bien thien
y' = - 3 x ' + 4x + —, y' = 0 o X = — hoac x
3 '

3
•
X

y'

-00

-1/3
—

0

+

+00

0
100/27

to

^
-00

-8/27

Ham so dong bien tren khoang

3 • ^'

5/3

y
^

5

f

\_5
Ob

i;-!


Luy?n gidi dS IruOc kp thi DH 3 miin Bdc, Trung, Nam Todn hgc - NguySn Vdn Th6n^

g
H a m so" nghich bien tren cac khoang

—;+oo

TrUdng h<tp 3. A ' > 0 < : > k <

, . .
5 .
100
H a m so dat cifc d a i tai x = - va y^^ = —
1

mms^datcirctieutai x - - -

y^=—i.''>

3. D6^thi (Hoc sinh W ve)
+ D i e m uo'n

,

-/.v

+ Ne'ug(0) = 0 < = > k = -

= ^ _^^y.

^ g,^,^ ^0^3^,}^ jj^H

+

NhSn x6t: D 6 thi nhan diem I

k ;t - thi (1) c6 dung 3 nghiem o
3

(C) va (d) c6 diing 3

,1)/, .

,

.,

B a i 2. Cho h a m so y = f(x) = - x ^ + 3x^ + 3(m^ - l ) x - 3m^ - 1 (1), m la tham so.

la d i e m uon.

a. Khao sat s i f b i e n thien va ve do thi ham s 6 ' ( l ) k h i m = 1.

J

b. T i m m de h a m so' (1) c6 cifc dai, ciTc tieu va cac d i e m cUc t r i cua do thi ham
46^

3'27

l a m tam d o i xtfng.

mi

b. PhiTdng trinh tiep tuyen (A) cua (C) tai O (0; 0) la:

y = f ( 0 ) ( x - 0 ) + 0 = ^x

thi(l)c6dung2nghiem.

d i e m chung.

^''^>''< l-^^''''"

'2

N e u g(0) 9t 0 o

^^J^^Q J|

hi

x = 0=>y= —

+

,,

i « ,r/-if>'t)

^
.
n
2
/ 2 46
y " = - 6 x + 4, y " = 0 o x = - . D i e m l
-;—
3 ,
.3 2V

+

a»arb m y i i . o u i l i l

-

so (1) each deu goc toa do O.
HrfdngdSngiai
a. Khao sat h a m so (Hoc sinh tif giai).
„ ,
b. Ta c6 y ' = - 3 x ^ + 6x + 3(m^ - 1 )
;
'
rx

.,4,)(„f,
. >

lv*i-.--K;vM.!:... •

= l + m

.

.Won

y'=0«
[x = 1 - m

SB" 0*

D i e u k i e n de h a m so c6 ctfc dai, cifc tieu la y ' = 0 c6 2 nghipm phan biet
PhiTdng trinh hoanh do giao d i e m ciia (A) v^ (C) IS:
-x^+2x^+-x--x
3

3
•

x = 0=>y = 0

•

x = 2=>y =

o

- x ^ ( x - 2 ) = 0c^

om?
"x = 0

G o i A ( l + m ; - 2 + 2 m ' ) ; B ( l - m ; - 2 - 2m') la hai d i e m ciTc tri.
Ta c6 O each deu A va B o O A = OB

x = 2

o

—
3

r,

m = 0 (loai)

10^
o

'3

c. Phu-dng t n n h hoanh do giao diem cua (C) va (d) Ik:
= k x ( l ) o x x^-2x + k - - = 0 o
3J
x2-2x + k - - - 0
3

So' giao d i e m cua (C) va (d) l i so nghiem cua phiTdng trinh (1).

k >

^

K e t luan: m = —; m = - — .

ihih

In i\)

ff

If 1;'•nciial'wi,.:;'^^

B a i 3. Cho (C J : y = f(x) = - x ^ + 2mx2 - 2 m + 1
b. Chtfng minh r^ng (C^) luon di qua hai d i e m co dinh A, B v d i m o i m.

c. T i m m de cdc tiep t u y e n v d i (C J tai A, B vuong g6c v d i nhau.

o

/. A' < 0 o

m = -l/2

a. Khao sdt s\i b i e n thien v^ ve do thi tfng v d i m = 1.

^

D a t g ( x ) = x ^ - 2 x + k - - la tam thiJc bac hai CO A' = - - k
3
•
3
Tmyn^

8 m ' = 2 m <=> m = l / 2

^

x = 0
-x''+2x^+|x

(1 + m)^ + (- 2 + 2mV = (1 - m)^ + (- 2 - 2m')^

Hifdng d i n giai

-

a. Kh3o sdt hkm so i?ng v d i m = 1. (Hoc sinh tif giSi)
K h i do (1) CO dung 1 nghiem <=> (C) va (d) c6 dung mot d i e m chung.
Tmyrtfi hap 2. A' = 0 <:> k =

b. G o i A(x,); yo) la d i e m c o d i n h cua (C), k h i d6:
yo = - x ' * + 2 m x o - 2 m + 1 ; V m ,

-

K h i do (1) C O dung 2 nghiem <^ (C) vk (d) c6 diing 2 d i e m chung.

H

0

2m(x2 - l ) - ( y „ + x ^ - l ) = 0 ; V m

'
^; ;..;); ^ /o,^

'


o

X„

xf,-l =0
yo+x^-l =0

=1

:

X() = - 1

•
;•

•

, E }\ V '= (O'k tr-.v; +

( 1 ) Ub

yo=l-Xo

x„= 1 z ^ y o = 0 = > A ( l ; 0 )
X„

a. K h a o sdt v a v e do thi (C) c u a h a m so ( I ) .

2m(xf, - 1)= y„ + XJ5 - 1 ; V m

.gfirMJD

c. T i e p tuyen tai A ( l ; 0) c6 he so g6c k, = r ( l ) = - 4 + 4 m

Vrr'Jid

T i e p tuyen tai B ( - l ; 0) c6 he so goc k j = f ' C - l ) = 4 - 4 m
T i e p tuye'n tai A , B vuong goc v d i nhau <=> k|.k2 = - 1

(-4 + 4m)(4 - 4m) = - 1

T i m d i e m M e (C) d e long k h o a n g e a c h liT M d e n hai du'dng t i e m c a n c u a

a. K h a o sdt v a v e do thi (C) c u a h ^ m so

= - l =^y„ = 0 = : > B ( - l ; 0 )

o

c

HU
f

3
m =—
4
o

T i m c a c d i e m thuoc (C) c6 loa do la nhiTng so n g u y e n .
(C) nho nhat.

^

,

b

1. T a p x ^ c dinh: D = R\ 1}

m,;;;!:
. i f ) S. iiiS

. ;,,,; oiJd:-! ?

_ ;

, ,.
,. 2 x - l
+ o o ; Iim y = l i m

i

,. .

hm y = lim y = 2

5

=> y = 2 la t i e m can ngang cua do thi ham so.
b. Bang bien thien: y ' =

, < 0 ; Vx e D
(x-1)^

+00

-00

H i f d n g d§n g i a i

+00 •

a. K h d o sat ham so (Hoc sinh tif giai).

do a t

b. Ta c6: y ' = 4 x ' - 4m^x = 4x(x^ - m^)

—00

H a m so nghich bie'n tren cac khoang ( - o o ; 1); ( 1 ; +oo)

rx=o

3. D o thi (Hoc sinh M ve hinh)

=m

D i e u k i e n de do t h i ham so (1) c6 ba d i e m cifc t r i l i y ' = 0 c6 ba nghi^m
^ .

f

= ('

G p i A ( 0 ; 1); B(m; 1 - m"); C(-m; - m'*) la cic d i e m cifc t r j .
T a c 6 : A B = Vm^+m**, A C = Vm^+m**, B C = N / W

i

y =0

o 4m^ = 2m^ + 2m'' o m* - m^ = 0 <=> m = l
m =- l

'
1

=> X =

-

A' i ....

2
Nh^n xet: Do thi nhan giao diem 1(1; 2) cua hai di/cJng t i ^ m can l a m tam doi xufng.
b. G o i M ( x „ ; y o ) 6 ( C ) , X o + l .

'^O - 1

X„ - 1

B a i 5. Cho h a m so y = f ( x ) =

^
x-1

(1)

,

Do X(), yo e Z n e n ta c6 cac triTdng hdp sau:
•

X o - 1 = 1 c^x„ = 2 = > y „ = 3 = > M , ( 2 ; 3 )

•

Xo-

1=-1 0x0 =0

=>yo=

c. T a c o :

Ke'tluanims l ; m = - l

v

Tac6y„= ^ ^ " " ^ ^ 2 + — ! —

D i e u k i e n de A A B C vuong can la BC^ = AB^ + AC^
m = 0 (loai)

^J'^ ' '

x =0 ^ y = l

•T->

= -m
m ?i 0.

= -oo

o>'

= > X = 1 la t i e m can duTng cua do thi ham so.

'?

vuong can.

phan bi?t o

,

j '

,.
2x-l
lim y - l i m

b. T i m m de do thi h^m so (1) c6 ba diem cifc t r i 1^ ba dinh ciia m o t tam giac

X

jjiiiiodl . v

a. G i d i han, t i e m can:

IT

a. K h a o sat ham so (1) khi m = 1.

X

<

2. Sy b i e n thien c u a h a m so

m =—
4
B a i 4. Cho h a m so y = f ( x ) = x " - 2 m V + 1 (1) v d i m la tham so

y' = 0 o

•

: . g ,.,^.,.,\

igj a!?;;

•,

<

^

1 =>M2(0; 1)
;;'

Z

' <: '

T i e m c a n duTng (d,): x - 1 = 0
T i e m c a n ngang (dz): y - 2 = 0
15


U^ngua ae miOc Ic m uii t mien oat;, irung, num ivunrm- lyguyen vun i tivngGoi M (X(,; y„) e (C). Khi d6:
d(M;d,)= |xo-l|
'2x0-1

.

1

-2
Xo-1
-1
Khi d6, tong khoang c^ch \.\S diem M(X(); y o ) den hai di/cfng tiem can Ik

d = d(M;d,) + d(M,d2) ^
d(M;d2)= y o - 2

Xo

. . .

Xo-1

+

x„-l

•dn,i„ = 2khi X Q - I

III. C A C BAI T O A N T I ; L U Y f N c 6 H U 6 N G D A N
Bai 1. Khao s^t sir bie'n thien vk ve do thi (C) cua hkm so y = f(x) = x^ - 3x^ + 1.
1\i do, bien luan theo m so nghiem cua phiTdng trinh sau:

Xn

-1

-1 o ( x o - i r = 1 0

Xo=0
Xo-2

Hi/dng d i n giai

a. Khao sdt hkm so (Hoc sinh tiT giai).
b. Gpi M (xo; yo) 6 (C); y„ = f(xo)
PhiTdng trinh tie'p tuyen2x^cua (C) tai M(xo; yo) la (d): y = f (xu)(x - XQ) + f(xo)
y=
-x + (Xo+1)'
(Xo+1)'
Toa dp giao diem A cua (d) vk Ox Ik: A( -Xo ; 0)
(
2x2
Tpa dp giao diem B ciia (d) va Oy la: B 0;
^

1-1^2
4

Xn.

2x2

2 "(xo+1)^

Ket luan: M,

S^QAB

=^0A.0B

4x^-(Xo+1)2=0:
;M2(1;1)

o^Ti,

HiMngdSngiai

in,

?

,

• m < - 2 hokc m > —: Phi/Png trinh c6 mot nghiem.
2
• m = - 2 hokc m = —: PhiTPng trinh c6 hai nghiem.

• xo = 0 = > y „ = 1 =>M3(0; 1)
•
Xo = 2 z:> y o = 3 =^ M4(2; 3)
Bai 6. Cho h a m s o y = f(x) =
Aihly.-x2x+ 1
a. Khdo sdt sif bie'n thien va ve do thi (C) ciJa ham so' da cho.
b. Tim toa do diem M e (C), bie't tie'p tuyen cua (C) tai M c^t hai true toa dp
Ox, Oy tai A, B va AOAB c6 di?n tich bkng - .
_
4

De thay AOAB vuong tai O

.v!'

-x^-x2+m+2=0

,1

=2

1
Xn

,

1 S m -J hod t.
2
• - 2 < m < —:
f
, ,
3 PhiTPng trinh c6 ba nghipm.
• u) ;.u ~ f ..'
B a i 2 . C h o h k m s o y = f(x) = x ' - m x + m - 1 (1)
a. Chtfng minh r^ng tie'p tuyen cua do thi hkm so (1) tai diem Xo = 0 c6 he so
goc nho nhat.
b. Vdi gia tri nko cua m do thi cija hkm so (1) tiep xuc vdi Ox. Khao skt vk ve
do thi hkm so (1) gik tri tim diTpc cua m.
c. Xkc dinh m do thi hkm s6'(l) c^t true hoknh tai 3 diem phan bi^t.
HUdng dSn giai

'

a. f ( X ) = 3x^ - m > 0 - m = f (0)
=> hp so gdc tie'p tuyen tai diem x = 0 Ik nh6 nhat.
b. m = 3; m = —
,41

'^^ '
^^' *
li -

4

c. x' - mx + m - 1 = 0 o (x - l)(x^ + x + l - m ) = 0=>

r m > 04

Bai 3.

sat si/ bie'n thien vk ve do thi (C): y = f(x) = x' + x - 1.
b. Gpi Xo Ik nghipm cua phiTPng trinh f(x) = 0. Chtfng minh r^ng:
2Xo - X(, - 1 < 0
M (trr
a. Khio

H i ^ n g d i n giai

2xo + X() + 1 = 0

Xo=--

2 x 2 - X o - l = o'

Xo=l
if

2

f(x) lien tuc tren

( n
V

2

=

17 .

r
ViEN T I W H B I N H

THUAN

'i'"' '

. J i i f P ' M '

k -


Luyfn giii di trade kp thi DH 3 miin Bdc, Trung, Nam Todn hoc - Nguyln Van ThOng

b Phu'dng irinh hoanh do giao diem
f(x) CO n g h i e m X(,

•(xo-

2

(2x,l-6xo)x--x^3xi,+- =^ - 3 x

1) x„ + — < 0 => 2 x f ) - X ( , - 1 < 0 (dpcm)

B a i 4 . C h o ( C J : y = f(x) = mx' + ( m - l ) x ^ + 1 - 2m

+ -

'

.-:,,„•

x/r

o(x-xo)'(x^+2x„x +3x^6) =0

p

Bai 7. Cho ham so y = r(x) =

a. Tim m dc ham so' co mot diem cifc tri.
b. Viet phiTcfng trinh tiep tuycn cua f C , ^ di qua g6'c toa do.

>f{ £ > tu »

. 2)

a. Khao sat sir bien thicn va ve do thi (C) cua ham so.
b Tim m de di/c^ng thang (d): y = 2x + m c^t (C) tai 2 diem phan bi^t A va B.
t. m + A« j

tim tap hdp trung diem I cua AS.

Hrfdng dSn giai

a. m < 0 hoac m > 1
b. (d,): y = 0; (d:): y = ^ x ; (d,):

b

"'^

) .S

/.m - ''X •-- (xYt
B a i 5 . C h o ( C ) : y = f(x) = (x + 1)^ ( x - 1)^

tiep xuc vdi (C).

a. •

a < 10

: Phi/cfng trinh v6 nghiem

•

a = 10

: Phu'dng trinh c6 1 nghiem. _

•

10 < a < 11

: Phu'dng trinh c6 4 nghiem. j

•

a = 11

: PhiTcfng trinh c6 3 nghiem.

•

a > 11

: Phifdng trinh c6 2 nghiem.

m + 4'\

;
v

4

•m =

7^

-2

^

r

^'g;

*

ft)

Bai6.Cho(C):y = f(x)= - x ' * - 3 x ^ + - .
2
2
a. Viet phiTdng trinh tiep tuyen cua (C) tai diem x = Xo.

(x - x,,)^ (xf, + 2x„x + 3xf, - 6) = 0
Hifdng d§n giai

. ik>M..4

vdim<-4ho$cm>4. ,
2

y

.

:,,[•••.

— -

* Dat t = sin x; x e [0; 71J ^ t e [0; IJ

•^rmd^ fa; ;

n « (,.,

= 2m - 1 (1)

"

'

Dieu kien de phiTdng trinh c6 dung 2 nghiem x e [0; n] la phtfdng trinh (1) co
dung 1 nghiem t e [0; 1).

19^'-

b. Chufng minh rang hoanh do giao diem cua (C) va (d) la nghiem ciia phiTdng

a. (d): y = (2 x;', - 6x„)x ~ ~ + 3x1 + x

-fi:^r.:

2x — 1
. Tilf do, tim m de phiTdng
x +2
, . 2sinx-l
.
, ,
...
'* 'sil »tn -.i •
trinh
= 2m - 1 c6 dung 2 nghiem x
x;
sin X + 2
HiAJngdSngiai
al =.(x)t mil
Khao sat va ve do thj (Hoc sinh tiT lam)

PhiTdng trinh trd thanh:
4

-f +

Bai 8. Khao sat va ve do thi ham so y = l(x) =

phiTcfng Irinh

0

rn + 4

nhffngdiemM

.^j,,,

b. (P) tiep xuc vdi (C) khi va chi khi hoanh do tiep diem la nghiem cua he

2x

,

Tap hdp trung diem I cua AB la phan diTdng thang 2x + y - 4 = 0, gom

,,

Htfdng d§n giai

x^-2x^ + l = m x 2 - 3

'(.m.utll

[ 2 x ^ + ( m + 4)x + m + 4 = 0

..x-

b. Tim m de parabol (P): y = mx^ - 3(m^0)

trinh:

= 2x + m o <^

x +1

phi/dng trinh x"* - 2x^ + 11 - a = 0.

-4x=2mx

—=-

'

Dieu kien de (d) c^t (C) tai 2 diem phan biet la m < - 4 hoSc m > 4.

' a . Khao sat va ve do thi (C) cua ham so'. TiT do, bien luan theo a so' nghiem cua

4x

8- ^ (£)^j '""

Khao sat ham so g(t) = = ^

iren [0; 1) => - < g(t)

<3z:>-
KHAO SAT MpT SO HAM SO DA THllfC NANG C A O
Bai 1. a) Bie't r^ng d6 thi ciaa hhm so'y = (3a' - l ) x ' - (b' + l ) x ' + 3c'x + 4d co
hai diem cifc tri la: (1; - 7 ) , (2; - 8 ) . Hay xac dinh tong M = a" + b' + c' + d'
b) Chu-ng minh rang do thi ham so y = x ' + 2 m V + 1 luon citt difdng thang
y = X + 1 tai diing hai diem phan biet vdi moi gia trj m.

I ¥unf^, t^urri

i utin

n\n, - ii^ujfcri

run

i riurig

COngtyTNHHMl

^ L v vii KHung vict

Hifdng d i n giai
H^m so nghjch bien tren moi khoang (-oo; - 2 ) , (0; +oo) va dong bien tren

a) De ddn giSn, ta dat A = 3a^ - 1, B = -(b^ + 1), C = 3c^ D = 4d, h^m so da
cho chinh m y = Ax^ + Bx^ + Cx + D
Ta

CO

y' = 3Ax^ + 2Bx + C

Ham so dat ciTc dai tai x = 0 va dat ciTc tieu lai x = - 2 .

y' = 0 « • 3Ax^ + 2Bx + C = 0
•

PhU"dng irinh nay c6 hai nghicm phan bi^t la x = 1, x = 2 nen ta c6 h? sau
f3A + 2B + C - 0

_

Ta cung c6

ry(l) = _7

„x

I

i .

.•

0

-2

—00

'

+C0

0
+00

[ A + B + C = -7

y(2) = - 8 ' ^ l8A + 4B + 2C + D = -8

rniT
f>

/

-00

0

'

diTcIc cdc gia tri tiTcfng tfng la a = ±1, b = 2, c = ±2, d = -3.

ta tinh
-

^

•

Ta c6 y" = -6x - 6, y" = 0
o X = - 1 nen diem uon cua
do thj la (-1;2)

V a y M = a^ + b^ + c^ + d^= 1^ + 2.2^ + 3^= 18 x ^ i

Do thi cii true lung tai diem

b) Phi/dng trinh hoanh dp giao diem cua hai do thj Ik
x" + 2 m V + 1 = X + 1 o x(x' + 2m^x - 1) = 0

.-j.,,

De thay phifdng trinh nay co nghiem x = 0, ta can chifng minh phifdng trinh
c6n lai c6, nghipm duy nha't khac 0.

(0; 4) va cat true hoanh tai
diem(l;0),(-2;()).
Do thi cua ham so:

t

b) Ham so da cho nghich bien

That vay, xet ham so f(x) = x ' + 2mx^ - 1.

tren khoang (0; +co) khi va

Ta c6 f (x) = 3x^ + 2m^ > 0 nen ham so nay dong bien tren tap so thi/c.

Va

'•

^

TCf cdc dieu kien n^y, ta tim diTdc A = 2, B = - 9 , C = 12, D = -12

Hdnnffa

Bang bie'n thien

^(^)i=^= Y o ^ m i f i i , . . ! r , i « a

_12A + 4B + C = 0 ,
g

khoang (-2; 0).

chikhi

lim f ( x ) = lim (x"^ + 2m^x - 1 ) =-oo

^

, X

y' = -3x^ - 6x + m < 0. Vx > 0
»

lim f ( x ) = lim ( x ' ' + 2 m ^ x - l ) = +oo

Hinh 7. Do thi ham soy = - J C ' - 3 X ^ - ^ 4

3x^ + 6x > m, Vx > 0

Ta CO bang bien thien cua ham so g(x) = 3x^ + 6x tren (0; +oo).
Nen phi/dng trinh f(x) = 0 luon c6 nghi$m.

+00

0

Suy ra phi/dng trinh f(x) = 0 c6 dung mot nghiem nay cung khdc 0 do
f(0) = - l . T a c 6 d p c m .

+00

i_

Bai 2. Cho ham so y = -x^ - 3x^ + mx + 4 vdi m 1^ tham so thifc

Tir do, ta difdc dieu kiOn cua m la m < 0.

a) Khdo sat sir bien thien va ve do thi do thi hhm so khi m = 0.

Bai 3. Cho ha m so y = f(x) = Sx'* - 9 x ^ + 1

b) Tim tat ca cac gid tri cua tham so m de h^m so da cho nghich bien tren
(0; +00).

a) Khao sat sir bien thien va vc do thi ham so tren.

Ht/dng dSn giai

^

''

- •

HU
Tap xdc dinh D = R .

a) Tap xdc dinh D = R .
X-»+oo

X-»-00

,^

Chieu bia'n thien: Ta c6 y' = -3x^ - 6x,
y' = 0 o -3x^ - 6x = 0

1

8cos'x - 9 c W x + m = 0 vdi X e [0; 71].

Gidi han cua hjim so la l i m y = - o o , l i m y = +oo
•

- i

•

b) Dira vao do thi tren, hay bicn luan iheo m so nghicm cua phiri^ng trinh

a) Vdi m = 0, ta c6 h^m so y = - x ^ - 3x^ + 4.
•

r ; .

X

= 0, x = - 2 .

Gidi han cua ham so

lim (Hx-*-9x^ + 1) = lim ( 8 x ^ - 9 x ^ + 1) = + o o .
x->-a!

X->-KC

* Chieu bien ihicn:
y

Ta c6 y' = 32x' - 18x. y' = 0

3
x = 0, x = ± -

. ii 0

m c:^ i

Li^n

COng ty TNHH

giii di tniOc Aj> thi DH 3 miin Bdc, Trung, Nam Todn hQc - NguySn Van Thdng

—

4

Va nghich bi6'n tren (

I

Ncu 0 < 1 - m < 1 o

3

H a m so dong b i c n tren
3^

-co;

;+tx3

A',,

Neu

0; —
4j

4;

H a m so dat ciTc dai tai (0; 1) va dat ciTc tieu tai
X

-00

y'
y

+00

-

0

+

r3.

49 ^

(

3.

.

4 ' 32 J

49^

. 4 ' " 32

Ta CO y " = 9 6 x ' -

^

49

49

32

32

18, y " = 0 o

x = ± — ,

y = - —

^^/3
4 '

32

4

1

lim y =
X->-OD

do

1

;0

,

n

*

y = 1 - m
(d)

1/
3 /
4 /

1

1

D a y chinh la phifdng trinh hoanh

81

Neu

1- m = 1o

, 5}^ , ,

l i m ( x - ^ + 3 x ^ - 1 ) = - . , va j i m y = l i m ^ ( x ^ + 3 x 2 - 1 ) = + « ) .
x->+oo

Sy bien thien ciia ham so:
CO

2

y' = 3x + 6x, y' = 0 o

x = 0, x = - 2 .

-1

1

/o

\

\

I

It
>.

i
2

x

*

Do thi h a m so:

h) Ta x e l c a c trU'ctng hdp sa

K h i m = 0 thi y = X - 1

-

m = 0 ihi phiTdng trinh co diing mot n g h i e m .

''^

V d i m ^ 0 thi y' = 3 m x ' + 6mx - m + 1.
Him

ihi phiTdng trinh vo nghiem.

'

nen h a m so k h o n g CO cifc trj.

-2-

Htnh 8. Do thi ham soy = 8x^-9x^

'

p h a n biet.

X

1

X =-1

( - 1 ; 0) v a c a t true h o a n h tai ba d i e m

"V

-

1-m.

Bang b i e n thien:

Do thi h a m so c a t true tung tai d i e m

~ 32

Di/a vao do thj, ta thay r^ng

m < 0 V ni > —

,..

/

n e n d i e m u o n c u a do thi la ( - 1 ; 1)

do giao d i e m cua hai do thi

o

>

X->-00

Ta CO y " = 6x + 6, y " = 0 o

+

-2

49
32

*

"K

nha't mot gia tri cua x. phiTdng

l - m > l v l - m < -

.i.'-yr

4

'y

cua t (t e [ - l ; 1 ] ) cung cho ta duy

Ncu

^x,^

(-2; 0).

t e [ - l ; 1] va ro rang m o i gia trj

y = 8x''-9x'+l,y=

h a m so y = x-* + 3 x ^ - 1 .

32

[(); n ] , ta c6

f(t) = St' - 91^ + 1 = 1 - m

'

H a m so nay dong bi en tren cac khoang (0; +00), ( - c o ; - 2 ) va nghjch bien tren

.2N/2

trinh da cho ti/dng diTdng v d i

ihi phi/dng trinh c6 dung h a i n g h i e m .

13^

D o t h i ham so

b) Dat I = cosx, X e

CO

G i d i han cua h a m so:

^

n e n hai d i e m uon cua

•3

•

32

T a p x^c dinh D = M .

+CO

Do Ihi cua h ^ m so c^t true tung tai d i e m (0; 1) va c^t true hoanh tai bon
d i e m phan biet la ( - 1 ; 0), ( 1 ; 0)J

<=> m = | i

Ihi phifdng U-inh c6 d u n g b o n n g h i e m .

Hif(}ng d§n g i a i

Ta
thi ham so nay la

32

b) Xac djnh taft ca cac gia t r i m dcf ham scTy = l ( x ) khong c6 cifc tri.
a) V d i m = 1, ta

+

0

1< m < ^

a) Khao sat sir bi en thien va ve do thi cua ham so khi m = 1
+00

^ 1
^

J

4

0

Khang Vi?t

B a i 4. Cho h a m s 6 > = f(x) = m x ' + 3 m x ' - (m - l ) x - 1 v d i m la tham so'. '

,X

0

4

< 1- m < 0 o

32

DWH

0 < m < 1 Ihi phiTdng Irinh c6 dung hai n g h i e m .

32

Neu I - m =

MTV

+ 1^

so n a y k h o n g c6 cifc trj khi v4

c h l k h i phiTdng Irinh y' = 0 k h o n g c6
n g h i e m h o S c c6 n g h i e m k e p , tiirc la
A ' < 0 « . 9m^ + 3 m ( m - 1 ) = 12m^ - 3m < 0 <=> 0 < m <

^.


Ceng ty TNHH MTV DWH Khang Vi(t

Luy(n giii di trade kj> thi DH 3 miSn Bdc, Trung, Nam ToOn hoc - NguySn Van ThOng

Phildng tiinh nay c6 ba nghi^m phan bi^t Ik 1 - 2>/3 , 1 , 1 + 2%/3 thoa man de bai.

VSy dieu kien can nm la 0 < m < - .
^

't> 0 > m

>

Vay gia tri can tim cua m la m = 11.

• •

B&i5.Chohamsoy = - 2 x ' + 6 x ^ - 5 c 6 d 6 l h i ( C )
a) Khaosatv^ vedo Ihihamso (C)

Bai 7. Cho hkm so y = f(x) = x ' - 2x^ c6 do thj (C).

'•

•

,-.Vl

b) Vie't phiTdng trinh ticp tuye'n cua (C) di qua diem A ( - l ; - 1 3 ) .
ji:

•

a) Khao sat va ve do thj (C).

'i--,i,.-r

,v.,

'"^

vdi nhau.

b) Ta CO y' = - 6 x ' + 12x. Goi M(x„; y„) \k tiep diem cua (C) vdi tiep tuy^n can tim.
^

Phifdng trinh tiep tuycn cua (C) tai M

t-f v j !

'

Tim dieu kipn cija a, b de tiep tuye'n ciia (C) tai cAc diem A va B song song

a) Hoc sinh tiT khao sat.

_

' ' <•

b) Tren do thi (C) lay hai diem phan bipt l i A v i B c6 hoinh dp Ian li/dt \h a, b.

Hi^ng d§n giai

Khi do y,i = -2xi'', + 6xf, - 5 .

t,j

f""

,,, „,,

Htf^ngdSngiai
mm so y = f(x) x" - 2x^ c6 tap xac djnh D = R .

y - y,, = f (X())(x - x„) hay

S\ibien thien cua ham so':

y = (-6xf) + 12x,) )(x - x„) - 2xf) + 6xo - 5

i

Gidi han cua ham so: lim y = +oo ; Um y = +oo
N

d -uyr

«

i

x .

T a c 6 y ' = 4 x ' - 4 x = 4 x ( x ^ - l ) , y ' = O o x = 0,x = + l .

Tiep tuyen nay di qua diem A ( - l ; -13) nen

"

^

Ham so dong bie'n tren ( - 1 ; 0), ( 1 ; +oo) va nghjch bien tren (-oo; - 1 ) , (0; 1)

-13 = (-6x,2, + 12xo)(-l - x„) -2xf, + 6x?, - 5

Diem cifc dai ciia do thi Ik (0; 0), diem cifc tieu cua do thi ham so' la ( - 1 ; -1),
I'Vi

o - 1 3 = -2xo+6xo-5 + 6xo+6xo-12x„-12xo
o xf, -

3X()

+2=0o

X(,

(1;-1).
Bang bie'n thien

= 1 V x„ = -2

+00

0

0

V d i M ( l ; - l ) thi phiTdng trinh tiep tuyen can t i m i a :

+00

+00

y + l = 6 ( x - l ) c ^ y = 6x-7
-

'*

-1

-00 '

Ta CO cdc tung do tiTcJng iJng la y ( l ) = - 1 , y(-2) = 35.
-

iftj'

Vdi M ( - 2 ; 35) thi phu'dng trinh tiep tuyen can tim Ik

'

y - 35 = - 4 8 ( x + 2) o y = - 4 8 x - 61

Ta c6 y" = 12x^ - 4 = 4(3x^ - 1), y" = 0 «

x=±

Bai 6. Cho ham so y = x ' - 3x^ - 9x + m vdi tham so m.
a) Khao sat sif bic'n thien va ve do thj hkm so da cho khi m = 0
b) Tim tat ca cac gid tri m de do thi ham so c^t true hoanh tai 3 diem phan biet

c6 hoanh dp lap thanh cap so cpng.
Hifdng dSn giai
a) Hoc sinh tiT khao sit.

fS
I

Theo djnh li Vi-et thi tong ba nghi^m cua phU'dng trinh (*) l i 3. Do d6, theo
tinh chat cua tong cua cap so cpng thi 1 chinh la nghi$m cua phu'dng trinh (*)
hay l ' - 3 . 1 ' - 9.1 + m = O o m = l l .
Vdi m = 11, phiTdng trinh (•) trd thanh
(X

- l)(x^ - 2x - 11) = 0

5^

3'

^

(J3

5]

[3-

9J

* Do thi cua ham so'

b) Ta

b) Do thi h i m so da cho c^t true hoanh tai ba diem phan biet c6 hoknh do lap
thinh cap so'cpng khi va chi khi phiTdng trinh sau c6 ba nghi^m phan biet lap
thanh cap so cpng: x"^ - 3x^ - 9x + m = 0 (*)

x ' - 3x^ - 9x + 11 = 0 o

Cac diem uon cua do thi la

CO

f

(X) =

4x^ - 4x.

Gpi a, b Ian li/dt la hoanh dp cua A vk B.
Hp so gdc cua tiep tuye'n cua

-3

(C) tai A va B Ian li/dt la
kA

= f (a) = 4a^ - 4a, ke = f (b) = 4b' - 4b.

Tiep tuye'n tai A va B Ian lUdt c6 phiTdng trinh la
y = f (a)(x - a) + f(a) = f (a)x + f(a) - af (a)

y = f(b)(x - b) + f(b) = r(b)x + f(b) - bf(b)
Hai tieo tuve'n n^v song song hokc trung nhau khi \k chi khi


Luy^n gidi di truOc
ICA

= ke «

thi DH 3 mijn Bdc. Trung. Nam Todn hoc - NguySn Van Thdne

Ceng ty TNHH MTVDWH

4 a ' - 4a = 4 b ' - 4b c:>(a - b ) ( a ' + ab + b ' - 1) = 0
(2m^-l)x-m

Do A va B phan bict n c n a * b, suy ra a ' + ab + b ' = 1
M 3 l k h i c , hai tiep tuyen cija (C) t a i A va B irung nhau k h i va chi k h i
a- + ab + b^ = l,a ^ b

^

f ( a ) - a f ' ( a ) = l"(b) - bl"'(b)

a^ + ab + b^ = l,a ;t b

\ •
[-3a^ + 2a^ = -Bb"* + 2b^

V a y dieu k i e n can va du dc hai tiep tuyen ciia (C) t a i A va B song song v d i
nhau la a"^ + ab + b^ = 1, a 9t+1, a b.

De thay he nay luon c6 nghiem x = m

x - l = ±(m-l)

1 n6n ta c6 d i l u phai chiJng minh.
2
. . .
.
,
^
nen t i e m can xien cua do t h i
x+2

ham so da cho chinh la y = x + m. du'dng thang nay tiep xuc v d i y = x ' - 3 x '
_ 8x khi va c h i k h i he sau co nghiem:

KHAO SAT MQT SO HAM SO PHAN THaC NANG CAO

+ m= x-^-3x^-8x

X

B a i 1. BicTt rang del thi ham so' y = ^

(x + 2)(x + m ) + 2
=x+m +
x+2

u\l CO y-" ^ ' • ^

](x-m)^=()

x - l = ±(m-l)

(m-ir ^

; ^j,. ^)|,;f>{

( h%

_^
[(2m-l)x-m^ =x^-x

^ . .j.

G i a i he nay, ta diTdc hai nghiem la (a; b) = ( - 1 ; 1), (a; b) = ( 1 ; - 1 ) .

la r 2 ; 2

va d i qua goc toa do. X a c djnh tung dp cua d i e m c6 hoanh dp la

m = x-^ - 3 x ^ - 9 x

Vdi

X=

Vay

CO

o

x^ - 2 x - 3 = 0

l = 3x^-6x-8

, ac ^ 0. ad - be ^ 0 c6 t a m d o i xtfng

thupc do t h i .

m^x"* - 3 x ^ - 9 x
X =

-1V

X=

- 1 , ta CO m = 5 va v d i X = 3 Ihi m = - 2 7 .

3

^

hai gia t r i m can tim la m = 5, m = - 2 7

Do do thj CO tam doi xufng la I

;

j , - i)Oi ;

Tim la't ca cac gia t n p, q sao cho khoang each giiya hai d i e m cifc t r i la VlO .
nen 2c + d = 0, - = ^ 2

d = -2c, c = 2a ^
b) ChiJng minh rang v d i m o i m t h i do thj cua ham so y =

+ ( m + l)x + m + l

X

X +

.
a) Ta

CO

y' =

(2x + p)(x^ +1) - 2 x ( x ^ + px + q) _ - p x ^ - 2 ( q - I)x + p

luon tiep xiic v d i diTcfng phan giiic cua goc phan tu" thi? nha'l.

^
tiep

x +2
xuc v d i dirdng cong (C): y = x ' - 3 x ' - 8x.

i'

X- 1

CO hai nghiem phan biet p x ' + 2(q - l ) x - p = 0 (*).
0 o (q - 1 )^ + p^ > 0.

V i p ^ nen do thj luon cd hai d i e m ciTc t r i .
G p i X , , X2 Ian liTdl la hai nghiem cua phiTdng trinh (•), day cung chinh la cac
cifc tri cua ham so da cho.

Hifdng dSn gial

m ?t 1.

Do do, dieu k i e n dc do thj ham so nay c6 hai d i e m ciTc t r i \k phi/dng trinh sau
D i e u nay liTPng difdng v d i A' > 0, p

+ ( m + 2)x + 2m + 2
0) l i m m de t i c m can xien cua do thi ham so y =
1*

—^^^"^

(x^ +1)^

(x^ + l ) '

2

a) ChiJng minh rang v d i m o i m ^ 1 t h i do t h i cua h a m so y = ^?^^^—^——?^

a) Ta can churng minh r^ng y =

'

Hifdng dan giai

B a i 2.

'

1 '

luon CO hai d i e m ciTc trj va khoang each giila chung khong d o i .
c) Khao sat va ve do thj h a m so (1) k h i p = 1, q = 2

-

(1)

'

Do thi di qua goc toa do n c n b = 0. Do do y( 1) = ^ ^ ^ =
^
= -—

^
c+d
2a + (-4a)
2
V a y tung dp can t i m la

'

Bai 3. a) Cho ham so y = ^ '^^^ ^ ^ trong do p ^ 0. p^ + q^ = 1.
Hiidng d§n giai

K\

Khang Vi?,

luon tiep xuc v d i y = x v d i m o i

T

' , «

•

,

.V

1 a tinh diTdc gia tri tifdng iJng la

2X| + p

'
2x,

2x2 + P
~
2x.

Tir do suy ra khoang each giffa hai diem cifc trj chinh la

D i e u k i e n de hai diTdng nay tiep xuc nhau la he phiTPng trinh sau c6 nghicm
d^=(x,-X2)^ +

2X|+p
2x,

2x2+p
2X2

=(x,-X2r

+

2x,

2x2


Luy(n gidi dS trudc thi DH 3 miin Bdc. Trung, Nam Todn hoc - Nguyin Van Thdng

= (xi - x j r

Nen ham so da cho nghich bien tren tiifng khoang xdc dinh.
Ham so da cho khong c6 cifc tn.
, -v^ » v>
V

4xfx^ J

,

Theo dinh li Vi-et cho phuTdng trinh (*) thi x, + X j = -^^3_i2 ^ XjXj = - 1
P
Dod6 10 = ((x,

+X2)^-4x1X2)

1+

4(q-ir
4xfx2^

a c dir&ng tiem can: Ta c6
'

X

Urn y = hm

^

4(x-3)J
^

+4

lim = lim

- -00,

Vai

:

= 1o

= 1 _ q^, ihay vao ding thuTc iren. ta diTdc

10 = ( ( q - l ) ^ + l - q ^ ) 1 + 10(l-q^) = ( 2 - 2 q ) ( l - q ^ + 4 ) o q - % 4 q ^ - 5 q = 0

(x + i y

ill :m ,1 > A <>r •

Bii-J,nh

1;

+00

-

+00

1

y

1

4

4

—00

x +1

, y'= 0 o

= +00

4(x-3)

1

-00

ta tha'y thoa. Vay cac gia tri can tim la (p; q) = (1; 0), (p; q) = ( - 1 ; 0).

y'=l--

^

4(x-3)j

y'

Giai phifdng trinh nay, ta thu diTdc nghiem thoa man de bai la q = 0. Thuf lai

1

X

la I 3 ; - .
V 4j
* Bang bien thien
X

b) Ta CO y = x + m +

'

lim y = lim

. 1 > X > !• O S2:jx m

+

M . fr. < ; T

D6 thi CO X = 3 la ti?m can dilng va y = j la ti?m can ngang, tarn doi xuTng

1 +

Theo gia ihiet

< - i'- a . i . j

pB t ar 6. r > u v

* Do thi ham so c^t true tung va true hoinh tai (0; 0). (ve h\nh)
(X + 1 ) - = 1

o

b) Goi

X = 0 , X = 2.

M(X(,;

y„) la mot diem thuoc do thi. Khi do. tiep tuyen cua do thi tai M \k

-3

Do do, hai diem cifc trj cua do thi chinh la (0; m + 1), (-2; m - 3) nen khoang

y=

K..

-3x

-(X-X(,) +

4(Xo-3)

4(xo-3)

. 4xf)-9X()

_,

= y

4(Xo-3)'

4(xo-3)'

each giffa hai diem nay la d = V(0 +2)^ + (m + 1 - m + 3)^ - 2 y / 5 khong doi.

fo.4xf,-9x„

Ta CO diem phai chiJng minh. | p + »:} +^x>yC.--(l ^ ^xVo + xS)
c) Hoc sinh tirkhao sat

Giao diem cua tiep tuyen vdi true tung c6 tpa dp la

Bai 4. Cho hamso y =

diem cua tiep tuyen vdi true ho^nh c6 tpa dp l i

'4(Xo-3)^j

f4xS-9x„.
4(x-3)

c6 do thi (C).

;0
ft

a) Khao sdt va vc do thi cua ham so da cho.

„ ,. , , v .

Do d6, tir dieu ki^n de b^i, ta c6

b) Tim tpa do diem M thupc (C) sao cho tic'p tuycn cua (C) tai M c^t hai true
tpa do Ox, Oy Ian liftJt tai hai diem A, B v& dien tich tam giac OAB \h -.
8
c) Tim nhurng diem u-en (C) co khoting each den true hoknh gap 3 Ian den true tung.
Hif(}ng dSn giai
a) Tap xac dinh D = ( - 0 0 ; 3) u ( 3 ; + 0 0 ) .

* Sirbienthiencuahamso:
Ta CO y' = • -3
<0
4(x-3)^

, giao

. , .

,

,,

,

,^

1 (4xf)-9x„)^_3

2

'"^
.2_a..

_3)2

j

lit.

Gidi phirpng trinh nky, ta thu diTpc 3 nghicm 1^ x„ = ^ , x,, = ^ ( l ± ^ )
2
4

I

Ttr d6 ta xac djnh diTpc cic diem thoa man de b ^ i .
B i i 5. Cho h^m so y =

c6 do thi (C).

A) Khao sdt vk ve do thj ham so.

,

29


o o n g ly iisnti

b) V d i m o i d i e m M ba't k i thuoc do thi (C), t i m gia tri nho nha't cua tong khoang
each tCr M den hai true toa do.

,

>,

j , n ;, ^

c) T i m nhOrng d i e m tren (C) C O toa do la cap so nguycn.
HMng

t,tij

dSn giai
'

'

3

+ 2x + 3

A =

A = -x +

Taco

d>2

Xa

Ta C O l ' ( x ) - —
f(x)>lXO)=

!
on

.

2x + 3

„ ,

4(x^ + 3 x + l )

U(j

Hoac M =

<=> X n = 1 ±

x„-l
1+

1 ;1+

2

l-4=;l-i-^

B a i 7. Cho ham so y =

gnu

J V

(x-l)-^+a + l
X

1^'
a) T i m cdc gia t r i cua a de
do thi cua ham so c6 ba ciTc t r i va chrfng minh rSng

^'i^'"'

suy ra

J)

vdi cac gia t r i do thi cac cifc t r i nay se n ^ m tren mot parabol c6 dinh.
b) Chtfng m i n h rang v d i m o i a ^ K , do thi cua h a m so y =
X

g(x)>g(0) = i .

+X +1

l u o n cd ba

d i e m uon thdng hang.
HiTdng d§n giai

1
V a y gia t r i nho nha't can t i m la - , dat diTdc tai x = 0
a) Ta cd y = x^ - 3x + 3 +

1
B a i 6, Cho ham so y = 2x - 1 + •
^
x-1
a) Khao sat va ve do thi h a m so'.

• y' = 2x - 3

,

-

2x^-3x^ - a
x^

b) T i m toa do d i e m M ba't k i thuoc do thi sao cho tong khoang each tit M den
hai du'dng t i e m can nho nha't.

Ta cd do thi cua ham ^ g " " " " ""^"^

Hifdng d§n giai
,(iOttteriq;iliji3!i%
a) Hoc sinh tuf g i a i .
'
•"
YiT .
b) Ta tha'y t i e m can duTng cua do thi la x = 1 va t i ? m can x i e n la x = 2x - 1.'

DiTa theo do thj cua ham so nay thi ta tha'y

X6t mot d i e m M n k m tren do thi c6 toa dp la M X();2X(,-1 +

',,1'

2

< Q nen day la ham nghich bien, suy ra

h^"^

!

Suy ra gia u-i nho nha't do c h i n h la d = - ^ ^ , dat diTdc k h i

V a y d nho nhal k h i M =

A = x + i ^ = ^ ^ ' + ^ ^ + ^=g(x)
2x + 3
2x + 3

'^'^ ^ ^ " " ^ " " T ;
(•^x +

Xo-1

x„-l

= f(x).

i.

Ne'uOrj,

-2x^-4x + l

3r

(2x +

2

1

-1

x„-l

-X

!'

Ta c a n tim gia trj nho nha't c u a b i e u thuTc n a y iJng v d i m o i gia tri x,, ^ 1.

\

- 1 < x < 1 . K h i do, do 2x + 3 > 0

2x + 3

1

x» - I
N/5

nen ta chi can xet 2 tru'dng hdp sau:

Ne'u-1 < x < O t h i

1
X,) - 1

til

5 " Ta tha'y rang v d i x = 1 thi A = 1 nen ta chi can x6t cac gia trj cua x khac i
sao cho A < 1, turc la chi can xet |x| < 1 o

2x„-l +

d = |xo-l

f':'~; mil \

vf;"'!'?'-

2xn-

'.],.,.'}

* (*,,**'

b) Theo de bai, ta can t i m gia t r i nho nha't cua bieu thiJc sau

T o n g khoang each tiT M t d i hai diTcJng t i e m can ciia (C) la

(.,• t,\m-'

\

a) Hoc sinh tU" khao sat.
x-1

HU

,•>',,, ;

M I V uvVH'Khang Vii

Xn - 1

, X„

I

^^f"''

y = 2x^ - 3x^ nhi/ dirdi day:

;^

^[~7
i.

/,

rhng cac gid t r i a can t i m la - 1 < a < 0.
V d i d i e u k i e n tren, theo cong thuTc ve toa

dp edc d i e m ciTc trj t h i de tha'y cac ciTc t r i
ciia do thi h a m so da cho se n l i m t r e n
parabol y = 3x^ - 6x + 3 c^ dinh.
, W^*t,

/
'
-3

at>n m f o ^tuMA

x -


b) Ta

CO

( x ^ + x + l ) - ( x + a)(2x + l ) _
y' =
:i
5
—
(x^+x + i r
f
2(xU3ax2+3(a-l)x-l

' •
^

J ) T i m tham so Ihifc m dc tpa dp cifc dai va ciTc l i e u cua ( C m ) n a m ve hai phia

x^ + 2ax + a - 1
5
5—
(x^+x + i r
/

^

cua dirdng t h i n g 3x + 4y - 7 = 0.

ii;

Hufdng d§n g i a i

,

a) Hoc sinh tif g i a i

(x2+x)+l)'

m

:uiT>

•:i)n

(

v

,

:• j

'<.

b) Ta C O y' = 3x^ + 2 ( m - l ) x - (m + 3). Phi/dng trinh y' = 0 cd

Ta se chtfng minh r^ng phi/dng trinh tiTdng tfng 1^:

' i j ^ ' i^'i (fi'' i '

x' + 3ax' + 3 ( a - l ) x - 1 = 0

(•)

j , " ^ ^ ',1 .

C O dung ba nghiem phan biet.

, cv'

i

,

~ -.

D a t f ( x ) = x ' + 3ax^ + 3 ( a - l ) x - l , x €

>» i
,

j;

39

A' = ( m - 1)^ + 3 ( m + 3) = m^ + m + 10 =

^
jj 1

>0

nan luon cd 2 n g h i e m phan biet. Suy ra v d i m p i m , h a m so c6 ciTc d a i , ci/c

i l - ,,/i'lV

ticu. Thi;c h i c n phep chia y cho y', ta cd.

R.

Ta C O f(0) = - 1 < 0, f ( - l ) = 1 > 0, l i m f ( x ) = - o o ,

l i m f ( x ) = +oo

dong

thdi h^m so nay l i e n tuc tren tap so thifc nen phiTdng trinh f ( x ) = 0 c6 ba

m-1
y = x ' + ( m - l ) x ' - ( m + 3 ) x - 1 = ( 3 x ' + 2 ( m - l ) x - ( m + 3)) — + •
3

nghiem phan biet thuoc cac khoang ( - o o ; - 1 ) , ( - 1 ; 0), (0; + o o ) .

2(m-l)^

Do do, phi/dng trinh (•) c6 ba nghi?m phSn biet hay do t h i da cho c6 ba d i e m

2 ( m + 3)

m^+2m-12
x +-

uon.
G'd sur ho^nh do cua mot trong cac d i e m uon cua do t h i h ^ m so da cho 1^ x„
va day cung la nghiem cua phiTcfng trinh (*) hay X Q + 3axo + 3(a - 1)X(, - 1 = 0 ;
khi d6, tung do trfdng uTng cua d i e m n ^ y chinh 1^ yo = —5-^
Xo + Xo

. Ta se t i m

X Q + 3axf, + 3 a x „ + 3a - 1 = 3 x „ + 3a o

+ x„ +1)

(x,) + 3a - l)(xo + x,, +1) = 3 ( X Q + a ) .

^

XQ +

3(Xo + x„ +1)

'

:d

3a - 1 ^

ttfc Ik chung t h i n g hang. Du'dng t h i n g di qua cdc d i e m uon tifdng tfng chinh
x+ 3a-l
y =

.

S .

+

9

3

2(m - 1 ) ^

2 ( m + 3)

:ntfd

,

r

m^ + 2 m - 1 2
XcT

+

DO TH| HAM S6 NANG CAP

^

2(m-l)^

2(m + 3 ) 1 ^ ^ m ^ + 2 m - 1 2

y =

c) N c u d i e m A cd tpa dp lii (x; y) thi d i e m d o i xtfug cua A qua diTdng t h i n g
y = X cd tpa dp la (x; y). V i the y c u cau cua bai toan tiTdng diTdng v d i viec
tim nghiem nguyen (x; y) v d i x^y

r

cua he phi/dng trinh I
[x = y " ^ - 4 y - l

MQT S6 DANG TONG H0P TOAN LIEN QUAN DEN ^
'

Trir hai phi/dng irinh ve thco ve, sau do chia hai ve' cho x - y ;>t 0, ta dU'dc
X'

+ xy + y^ = 3.

Dc dang t i m dU'dc nghiem nguyen v d i x^y
Bk\. Cho ( C „ ) c6 phifdng trinh y = x ' + ( m - l ) x ' - ( m + 3)x - I .
a) Khao sdt

ve do t h i (C) ciia hkm so k h i m = 1.

b) Chtfng m i n h r^ng v d i m p i m, hkm so' c6 ci/c d^ai, ci/c t i € u . Vie't phifdng trinh
dirdng t h i n g d i qua c i c d i e m cifc d a i

1'

va do chinh la phUdng trinh diTdng t h i n g di qua hai d i e m ciTc tri.

.3

Do d6, cdc d i e m uon cua do thi cung thoa man quan h$ tuyen tinh nhif tren,

la

X c D + -

Suy ra cac d i e m cifc dai va ciTc tieu n^m tren di/dng t h i n g cd phifdng trinh

Tir dieu k i ^ n cua Xo, ta thafy r^ng
2

m^+2m-12

2 ( m + 3).

ycT =

+1

mot quan he tuyen tinh giffa Xo, y o .

S „ y ra y„ -

2(m-l)^
Do y c D =

ciJc tieu cua do t h i .

c) T i m nhffng cap d i l m nguyen tren (C) do'i xtfng v d i nhau qua diTdng t h i n g
y = X v£l khong nKm t r e n di/dng t h i n g d6.

cua phu'dng trinh x ' + xy + y^ = 3

ia(2;-l),(-l;2),(-2; l),(l;-2).
Thur lai vko he, ta nhan bp nghiem ( 2 ; - 1 ) . ( - 1 ; 2 ) . '

"^'^

'

'

V a y la t i m dirpc cap d i e m nguyen duy nha't d o i xuTng v d i nhau qua di/dng
t h i n g y = X va khong nam tren di/dng t h i n g do la (2; - 1 ) , ( - 1 ; 2).

, ......


Lu^n

gUu di trUOC

thi DH 3 m,.'n / U

/'

\.,:m /

B a i 2. Cho ho diTcfng cong (C J : y =
a) Khao sat

>i h

\ . , • 7 , ',,;/( Thdng

CdngtyTNHH

-x^ + mx x-m

MTVl)\> / /

.

c6 diing hai nghiem phan bipt (an so la m).
Viet lai (1) diTdi dang m^ - (x,, + y„)m + X(,(x,i + y„) = 0, ta suy ra tat ca nhuTng
diem (xo; yo) thoa man yeu cau bai toan la nhCTng diem c6 toa do thoa man

ve do thj ( C ) cua ham so khi m = 1.

b) Xac dinh m de ham so c6 ciTc dai, ci/c tieu. Vie't phiTctng trinh diTdng thing di
qua cac diem ci/c dai va cifc tieu cua do thj ham so'.

Bai 3. Cho ham so y = x ' - 3(m + l)x^ + 2(m^ + 4m + l)x - 4m(m + 1).

c) Tim cac diem trong mat phlng sao cho c6 dung hai di/cfng cua ho ( C J di qua.

a) ChiJng minh rang ( C ^ ) luon di qua mot diem co djnh khi m thay doi.

H i f d n g dSn g i a i

—x^ + X — 1

a) Khi m = 1, ham so trd thanh y

x-1

•
1

= -x ' x-1

'^^te

U

Taco y = - 1 +

x(2-x)
= —^

^,

y' = 0 k h i x = 0 h o a c x = 2.

(x,) + y())(yo -

3x()) > 0.

H i M n g dSn g i a i

X->-oo

-X --

yo =

= +00;

- 3(m + l)xf) + 2(m^ + 4m + l)x„ - 4m(m + 1) diing vdi moi m.

(2X() - 4)m^ + (-3 xf, + 8x0 - 4)m + xf, -

-X

= -00;

-

x-1

x->l-'

1

lim y = lim
X-++00

-X

-

x-lj

X->+00

lim y = lim
x-*i~

x->r

-X

- -00:

x-1

Dieu nay xay ra khi va chi khi

'

-3xf) + 8x„ - 4 = 0

Tur day giai ra dU'pc Xo = 2, yo = 0.
Vay ( C m ) luon di qua diem M(2; 0) CO djnh.
x ' - 3 ( m + l ) x ^ + 2(m^ + 4 m + l ) x - 4 m ( m + 1 ) = 0

—00

0

+00

1

0
+00-

0

CO 3 nghiem phan biet. TiT cau a) ta thay ring x = 2 la nghiem cua phiTdng
(x - 2)(x^ - (3m + l)x + 2(m^ + m)) = 0

-00

-00

n fn»t
^^^^ ^ . ^ x ( 2 m - x )

x-m
+ mx - m

(1)

trinh tren. Nhd do, bie'n doi da thufc ve trai, ta diTdc phiTdng trinh tiTdng diTdng

r+00

D 6 thi ham so: (hoc sinh tif ve Hnh)
2
2
b) Ham so' y =
^ " ^ ^ - m ^,

niiid

t») ( C m ) c i t true hoanh tai 3 diem phan biet khi va chi khi phtfdng trinh

Bang bie'n thien

-X

+ 2xo - y,, = 0 vdi mpi m.

xf,-3xf,+2xo-yo =0

= +00

-

3XQ

'2xo-4 = 0

H^m so CO tiem can xien y = - x va ti^m can dtfng x = 1.

so y =

f"^' tJ'' f^

Vict dang thuTc tren nhu'da thiJc theo m, ta difdc

j

x-lj

X-+-00

lim y = lim

1

J

* ^ ^ x (>: nM si ui il / f

a) Gia suTdiem co djnh la (x„; y„). Khi do ta c6

Ham so dat ciTc tieu bing 1 khi x = 0 v^ dat ciTc dai bllng -3 khi x = 2.
Um y = lim

C

c) Khao sat ve ve do thj ham so' khi m = 1

H^m so tang tren (0; 1) va (1; 2), giam tren (-00; 0) va (2; +00).

• Gidi han:

(

b) Tim m sao cho ( C m ) c i t true hoanh tai 3 diem phan biet.

Mien xac d j n h D = R \ { 1 } .
1

, Vi(t_

(x-m^

2

Tilf day ta thay r i n g (1) co 3 nghiem phan biet khi va chi khi phi/dng trinh (2)
sau day CO 2 nghiem phan bietkhac 2:
- (3m + l)x + 2(m^ + m) = 0
Dieu nay xay ra khi va chi khi
Giaira ta diTdc m

Khi 66 hai diem cifc tri c6 Ipa dp ti/dng ufng la (0; m) va (2m, -3m). Suy ra

|A = (3m + l ) ^ - 8 ( m ^ + m ) > 0
l 2 ^ - ( 3 m + l).2 + 2 ( m 2 + m ) ^ 0

c6 ciTc dai v^ ciTc tieu khi vS chi khi m^O.

x-m

^,
(2)

1.

Vay vdi m ; t 1 thi ( C n , ) c i t true hoSnh tai 3 d i l m phan bi$t.

dudng thing di qua hai diem cifc tri c6 phiTdng trinh l i y = m - 2x.
c) Gia sur (x,), y„) la mot diem trong mSt phing ma c6 dung hai diTdng cong ( C J
di qua. Khi do phifdng trinh y„ =

'"^o ~ "^^

Xo-m

(1)

B^i 4. Cho ham so y = — .
x-1

( C )

3) Khdo sdt sir bie'n thien v^ ve do thi (C).
^) Tim hai diem A, B thupc (C) v^ doi xtfng nhau qua difdng thing y = x - 1-


Luyfn giat ae truac Kytnttitijmi

g

Hifdng dSn giai
b) N e u M ( x , y)

pai 6. Cho ham so y =

!. if, i . f \ ^.r t>i|V

M ' ( x ' , y') 1^ hai d i e m doi xtfng v d i nhau qua du"dng thSng

y = X - 1 Ihi ta C O

(x-ir

'

a) Hoc sinh tiT khao sal.
. ,„., ,,„,,,,,.,,

v=,.

a) Khao sat sir b i c n thien va ve do thi ham so da cho.

•

(x-1)^
b) B i e n luan theo m so nghiem cua phi/dng trinh

i)

- — - = - 1 (do dirdng th^ng M M ' vuong goc v d i diTdng t h i n g y = x - 1).
x'-x
,

Tir i ) va i i ) ta suy ra x ' = y + 1, y ' =

Hifdng d§n giai

X -

^

=—

1

a) Hoc sinh t i f g i a i .
b) + m < 0: phiTdng trinh v6 nghipm

1.

Gia suf M va M ' la hai d i e m thupc (C) doi xufng v d i nhau qua diTdng thang
-hav f

y=

,
.
x'^
(y + l)2
x - l = y ' = — — = ••'
x'-l
y + 1-1

1
U

1

va cap

^ I

sflj

. D o chinh la nhi?ng d i e m A , B

0 < m < 12: phi/dng trinh co 2 nghiem;

+

m = 12: phiTdng trinh c6 3 nghiem;

+

m > 12: phiTdng Irinh co 4 nghiem.

1.:.

Bai 5. Cho h ^ m so y =

v = ^^~'^
x + 2

nhir sau: V d i x > - 2 , G' chinh la G. V d i x < - 2 , G' la anh doi

xiJng cua G qua true hoanh.
TCr do thi G' ta cd ket qua bien luan neu tren.

- 3 x - l

2x^+(6-m)x + 4
^
mx + 2

C O dinh duy nha't. Xic dinh tpa dp cua d i e m do.

Dap an. 1 < m <

6N/3-9.

Cdch I. Chu y phiTdng trinh luon cd nghiem x = 1 v d i m p i m . Khiio sat ham so

b) Khao sat va ve do thi cua ham s6' k h i m = 5.

-3x

Hifctng dan giai
y =-

Gia sur (x,,; y,,) la d i e m co djnh ciia do thj ham so.

-1

(x ;^ 1). Ta can t i m m sao cho difdng thang y = m cat do

x-1

thi tai 3 d i e m phan biet.

L'u- 1 ' .
'
2x,^)+(6-m)x„+4
^.

K h i do ta C O y;, = — — ^
'—il
vdi moi m.
mx,, + 2
ra mXd.yi, + 2y„ = 2 xf, + (6 - m ) X ( , + 4 v d i m p i

= mx-m

Hi/(tng d§n giai

a) Chtfng minh r^ng v d i m o i gi4 trj cua m , do thi ham so luon di qua mot diem

Suy

'l:'.

Bai 7. T i m m dc phi/dng trinh sau cd 4 nghiem phan biet:

can t i m .
I > ^ • ^ / ^ l .

+

2)
••••X' ,

V2J'

m = 0: phiTdng trinh co 1 nghiem duy nhat;

Sd liTdc each g i a i : TiT do thi G d cau a), ta suy ra do thj G' cua ham so

G i a i he nay ta diTdc c5p d i e m triing nhau M = M ' c6 toa dp

'

+

','•'•1:

y = X - 1 thi ta C O

72'

= m.

V + Y ' X + X'

V

ii) Trung d i e m cua M M ' nSm tren y = x - 1, suy ra

diem M

x +2

Cdch 2. Del thj ham so' y = 4|x|-^ - 3 | x | - 1 g6m hai nhanh, nhanh 1 la d6 thi ham
so' y = - 4 x ' + 3x - 1 v d i X < 0 va nhanh 2 la d6 thj ham so' y = 4x- - 3x - 1

m.

vdi X > 0. T r o n g khi do y = mx - m la diTdng thdng quay quanh d i e m A ( l ; 0).

V i e t phirong Irinh tren diTdi dang nhj thiJc iheo m , ta diTdc

Hay vie'l phi/dng trinh lie'p tuye'n ke tiT A den nhanh 1 cua d6 thj ham so'.

(xoyo + x,i)m + 2y() - 2 xf, - 6x1, - 4 = 0 v d i m p i x
Suy ra Xoy„ + x„ = 0 vii 2y„ - 2 x,^, - fix,, - 4 = 0.

h

G i a i he nay ta diTdc nghicm duy nhal x„ = 0, y„ = 2. Tit do suy ra do thi ham
so luon di qua d i c n i co djnh duy nha't c6 tpa dp (0; 2).

Jo.


BAI TAP T O N G H 0 P L U Y g N THI DAI H Q C V A L U Y g N H Q C SINH G I O I

Cho ham so y = -x^ + 3x^ - 2. (C)
' '^
a) Khao sat sir bic'n thien va ve do thj (C). Tif do vc do thj (C) y = + 3x^-2
b) Tim tren (C) nhffng diem ma qua do chi ke diTdc mot ticp tuyen vdi (C).
H\i6ng dSn giai
a) Hoc sinh tiT khao sat.
b) Gia sur M(x„; y,,) la mot diem tren (C). Ta giai bai loan vict phiTOng trinh ticp
tuyen cua (C) qua M. Gia su" ticp tuyen (t) kc lit M den (C) tiep xiic vdi (Cj
tai N(x,; y,). Khi do phiTdng trinh cua (t) co dang
y - y , =(-3x| +6xi)(x-xi)

Vi (t) di qua M ncn ta CO
6>
: '
+
y o - y i =(-3xf +6x,)(Xo-x,).
(1)
Ngoai ra, do N thuoc (C) nen ta CO ^
y, = - x ) ' + 3x^ - 2
• (2)
Nhir vay toa do tiep diem la nghiem cua he (1), (2). Ycu cau bai toan o he
(1), (2) v('Ji an la (x,; y,) co nghiem duy nhat..
Thay y, tif phiTdng trinh (2) vao phi/dng trinh (1), ta di/dc

Phi/dng trinh hoanh do giao diem cua (C) va tiep tuyen co dang
(3x^ - 3 ) ( x - X A ) + yA = x ' - 3 x + 2

B a i 1.

2x-| - 3(x,) + l)xf + 6x„x, - 1 -y,) = 0

Lai thay

2xi'-3x„x?+xil-3xf+6x„x,-3xi^-0 ^ •
o(x,-x„)2(2x,+x,)-3) = 0

(3)

Ta thay he (1), (2) co nghiem duy nhat khi va chi khi ~ ^ = x,, o x,, - 1 .
Ttrdo tinh difdc y„ = 0.
Vay M(l; 0) la diem duy nha't tren (C) ma qua 66 co the kc dung mpt Uep

tuyen vdi (C).
Bai 2. Cho ham so y = - 3x + 2. (C)
a) Khao sat su" bien thien va ve do thj (C).
b) Xet 3 diem A , B, C ihdng hang va thuoc (C). Gpi A', B', C' la giao diem cua
(C) vdi tiep tuyen cua (C) lai A, B, C. ChiJng minh rang A', B', C th^ng hang
c) Tim tren do thj (C) cac diem doi xuTng nhau qua 1(0; 2)
Hi^cJng d§n giai

a) Hocsinhtirkhiiosal.
^
b) Phirdng trinh tiep tuyc'n cOa (C) tai diC'm A(XA; y^) c6 dang
y = (xx-3)(x-XA) + yA.

'

Thay yA =

- 3 X A +2 vao phUOng trinh, ta di/dc
(3x^ - 3)(x - x A ) + X ' A - 3 X A = x-^ - 3x
< : > ( X - X A ) ^ ( X + 2 X A ) = 0.

Nhi/ vay tie'p tuyc'n cua (C) tai A c^t (C) tai diem co hoanh dp XA (chinh la
A) va diem co hoanh dp - 2 X A (la diem A'), tufc la X A = - 2 X A
TiTdng tU" X g . = - 2 X B , X c = - 2 x c .
(1)
Bay gicf ke't luan cua b^i toan sc dtfdc chufng minh nhd nhan xet sau:
Nhgn xet: X6t 3 diem A, B, C thuoc (C) co hoanh dp liTdng uTng la XA, XR, XC.
Khi do A, B, C thang hang khi va chi khi XA + XB + Xc = 0.
ChuTng minh. Gi a suT A, B, C nam tren difdng thang co phU'dng trinh y = ax +b.
Khi do XA, XB, XC la nghiem cua phU'dng trinh

x' - 3x + 2 = ax + b o x' - (3 + a)x + (2 - b) = 0
Ap dung djnh li Vi-ct, ta suy ra XA + XB + Xc = 0.
NgiTdc lai, gia suT XA + XB + Xc = 0. Viet phifdng trinh difdng thang di qua A, B
c^t (C) tai C thi theo phan thuan ta co XA + XB + Xc = 0 suy ra Xc- = Xc suy ra
C' trilng C va co nghla la A, B, C thang hang. Nhan xet diTdc chuTng minh.
Quay trd lai bai toan, do A, B, C th^ng hang ncn theo nhan xet, ta co
XA + XB + Xc = 0. Theo (1), ta co XA- + XB- + X c = -2(XA + XB + XC ) = 0.
Tiep tuc ap dung nhan xet ta suy ra A', B', C th^ng hang (dpcm).
Bai 3. Cho ham so' y - — - 3x - - CO d6 thj (C).
2
X
a) Khao sat va ve do thj (C).
b) ChuTng minh rSng ham so co ba diem cifc tri phan biet A, B, C. , rir;.
c) Tinh dien ti'ch tarn giac ABC.
d) Tim tam va ban kinh di/dng tron ngoai tiep tam giac ABC.
Htf(?ng d i n giai

a)

TFa CO' y . = x - 30+ —1 = x-^-3x^+1
;

X

Tird6y' = 0 o x ' - 3 x ^ + 1 = 0 . (1)
•
vi,;.V
Dat f(x) = x' - 3x' + 1 thi f(-l) = - 3 , f(0) = 1, f(l) = - I , f(3) = 1 nen theo tinh
chat ham lien tuc, phiTdng Irinh y' = 0 c6 3 nghiem XA, XB, XC thoa man dieu
ki^n -1 < XA < 0 < XB < 1 < Xc < 3. Tir do suy ra dpcm.

LuySn gUU dS trade

thi DH 3 miin Bdc, Trung. Nam ToOn hoc - NguySn Van ThOng

C6ng ty TNHH MTVDWH

b) DiOn tich tarn gidc ABC c6 the tinh theo cong thtfc

c) Hi^(^ng dan. Hay tim cac hang so a, b, R sao cho (x^ - a)^ + ( y A - b)^ =
XA la nghiem cua ( 1 ) va yA = •

S = | | ( X A -xeKyA - y c ) - ( X A -XcKyA - Y B )
..2
Ta CO yA - y b

-Xy)

-(XA

..2

/• .

XA+XB

.

1

=^ ^ ^ - 3 ( X A - X B ) -

1

\

1

-3 +

Thay vao

(2),

la difde

•

yA

-y,j

1

o . - J < i v , ,.,,5

Gia suT phu-dng trinh du^dng tron c6 dang (x - A)^ + (y - B)^ = R I
Thay y = x' vao va bien ddi, ta du"dc phu"ctng trinh hoanh do giao diem cua

-1.

(P) va dirdng tron la x* - (2B - l)x^ - 2Ax + A^ +

- R^ = 0

(1)

Theo gia thiet a, b, c, d la 4 nghiem ciia phu'dng trinh (1).

= - - ( X A - X ^ K X C + 1).

Ap dung dinh l i Vi-et cho phu'dng trinh bac 4, ta c6 a + b + c + d = 0.

;fi,:. !M

3

Tirdng lir y ^ - y y

XAXBXC =

--

Hrf(?ngdSn giai

iih

Do XA, XH Xf IJi 7> nghiem ciia (1) nen theo djnh li Vi-ct, ta co
x.v + X|, + Xf = 3, X A X | , + x „ X c + X C X A = 0 va

•-3XA

vdi

jj^i 4. Xet parabol (P) c6 phiTdng trinh y = x l M6t diTdng tr6n c^t (P) tai cac
diem A, B, C, D c6 hoanh do tUdng u'ng la a, b, c, d. Chtfng minh r^ng a + b +
c + d = 0.
(2)

••'ji

Khang Vi?t

Bai 5. Trong mat phang vdi he toa do De-cac vuong goc Oxy, cho di/dng cong
(C): y = 2x'' - 3x^ + 2x + 1 va difdng thing (d): y = 2x - 1.

=--(XA-XCKXB+1)

a) Chilng minh rang diTdng cong (C) va diTdng thang (d) khong cat nhau. u
Turdo suy ra

S-^|(XA -X^KXA

-XCKX^

X^Xfj + X ^ X c + X ^ X A

b) Tim tren (C) diem A c6 khoang each den (d) la nho nhat.

- X C )

-(XBXA

+X^XB

HUiing, dSn giai

+X^XC)

a) PhiTdng trinh hoanh do giao diem cua (C) va (d) co dang
' D a t

X = X ^ X B + x ^ X c + X ^ X A , Y = XgXA + x^x^

+ x\\^,

ta c6

2x^ - 3x^ + 2x + 1 = 2x - 1
o 2 x ' ' - 3x^ + 2 = 0

X + Y = X A X „ ( X A + Xu) + XtXn(Xc + X,)) + X A X C ( X A + Xc) = - 3 X A X I
(X^XB+X^XC + X^XAKX^X^

X.Y=
=

Dat t = x^ thi ta duTdc 2t^ - 3t + 2 = 0. Phu'dng trinh cuoi cung nay c6:

+X^Xi3+X^Xc)

XAXB + X ^ X ^ + X^X^. + 3 - ( X ^ X A + X ^ X B

Siir dung hang dfmg thii-c \\ x^ + x^ -

(1)

A = 9 - 16 < 0, do do CO nghiem. Tif do (1) v6 nghiem va nhu" vay ta co dpcm.

+X^Xc)

b) Xet diem

A(X();

yo) thuoc (C). Ap dung cong thtfc tinh khoang each, ta tim

SXAX^XC

duTdc khoang each tiT A den (d) b^ng d =
= ( X A +Xi3 + X e ) [ ( x A + X B +xc)^

Ta tinh diTdc x^ + Xy +

- 3 ( X A X B + x^Xc + X C X A )

Tirctng tir x ^ x ^ + x'^x^'- + x;^x^ = 3 X ^ X - B X ^

+(XAXB +XBXC +XCXA)[(XAXB +XBXC +XCXA)^ -3XAXBXC(XA

Suyra

XAXB+XBXJ +

\2

Do 2x,'5-3xf,+2 = 2

+

+XB

+XC)'

Thay vao ta tinh di/ac X Y = 3 + 3 - 24 = -18.
Tir day suy ra (X - Y ) ' = (X + Y ) ' - 4XY = 9 - 4(-18) = 81
Suy ra | X - Y | = 9
'
4

7 7
+ - > - .
8 8

* V

^t'-

2xo - 3xo + 2

•'.'J'.

Dau bing xay ra khi va chi khi XQ = — O X „ = ± —
4
2

Tur day ta tim diTdc hai diem A Ih A,

X^X^=3

• 2x0 + 1
75

.

= 3 . ( - i ) + (3)(3- - 3.0) = 24.

yo

^.V3-i^
2

8

Chu y: Co the kicm tra lai r^ng tiep tuyen ciia (C) tai A,, A2 trung nhau va
song song vdi (d). Sir kien nay khong ngau nhien. Co the chiJng minh diTdc
r^ng neu A(x„; y„) thuoc (C) la diem co khoang each den (d) la nho nhat thi
•Jircfng thing qua A va song song vdi (d) se tiep xuc vdi (C). Day cung la mot
ttnh cha't co the diing de giai bai loan khoang each. .(Ou!, 11 'ii! n ':i> I . v

Luy?n glai d? trwfr

i-f tWi

lui

B a i 6. Cho ham so y =

' misn ttac. imng, nam wan

x'

+3X

L^wig ijr 1111111 mt

niM- - lyguyen vurrrmmg

+3

y,
^'

x+1

= - X i

'

+m =

m-3
4

+m =

r uf

rii Tvnang

ri^

3m+ 3
4

a) Khao sat va ve do Ihi (C) cua ham so.

Tir day, khur m ta diTdc y, = 3x, + 3. Suy ra phiTdng trinh quy tich trung d i e m I

b) Xac dinh hai d i e m A , B Ian liTdl d tren hai nhanh cua (C) sao cho A B ngrii,

cua doan M N la di/dng th^ng c6 phi/dng trinh y = 3x + 3.
m^x^ + 1

nhat.

B a i 8. Cho hp (C J : y
Hifdng d§n giai

CO

a) Hoc sinh tiT khao sat.

do thj ( C J nao di qua.
Hifdng d i n giai

b) X e t d i e m A ( X A ; YA) ihtipc nhanh phai (tfng v d i X A > - 1 ) va B ( X B ; ya) thuot
iii

nhanh trai (tfng v d i XQ < - 1 ) .
Ta

CO

the dat

XA +

1 = a v^

. r oia

XB +

. T i m tren dU'dng thang y = 1 d i e m ma khong

0^b+.

a

Gpi (a; 1) la d i e m ma khong c6 do thj (Cm) nao di qua, ttfc la khong ton tai m

%. tnii i

1 = - b v d i a, b > 0. Ta c6

sao cho 1 =

m^a^+1

XR ~ X ;
AB'

=

(XA -

X B ) ' + (yA + y s )

= (XA -

XB)'

+

XA - X B

D i e u nay tifdng dU'dng v d i phu'dng trinh m ' a ' + 1 = a khong cd nghipm m

+
(XA

+1)(XB+I)

o
1

= (a + b ) ' + (a + b + - + - ) ' = ( a + b ) '

2

a - 1 < 0.

V a y tat ca nhCTng d i e m can l i m la nhffng d i e m cd tpa dp (a; 1) v d i a < 1.
Bai 9. T i m tat ca cac gia tri m sao cho (x + l ) ( x + 3 ) ( x ' + 4x + 6) > m v d i m p i x.

2

Jli/if! or

+ •

2

a^b'

ab

a = b =

^

HiiTdng d§n giai

= 8(v^ + l).

X6t hhm so y = f(x) = (x + l ) ( x + 3 ) ( x ' + 4x + 6) thi ba't dang thifc da cho
dung v d i m p i x k h i va chi k h i m < min f(x). Dat t = x ' + 4x + 4 = (x + 2 ) ' > 0

D a u bang xay ra k h i va chi k h i a = b va 2 =

o

thi f(x) =

a^b^
Vay khi

1
A -

(X-

+ 4x + 3 ) ( x ' + 4x + 6) = (t - l ) ( t + 2) = t ' + t - 2 = g(t).

Khao sat ham so' g(t) v d i t > 0, dc dang t i m di/dc m i n g ( t ) = - 2 .
thi AB

t>()

.IT

Tiir do suy ra m i n I"(x) = - 2 va nhU' vay dap so cua bai toan la m < - 2 .

ngan nhat.
B a i 7. Cho ham so y -

Bai 10. Bien luan theo m so nghiem cua phu'dng trinh x ' - (4 + m ) | x | + 5 + I m = 0.

x^ + 2 x + 5

(C).

x+1

' '

. not) A "u*

a) Khao sat sir bien thien va ve do thi (C).

''

> i

Hrfdng d§n giai

*

+

m < - ^ : phi/dng trinh cd 2 nghiem;

+

m = - — : phu'dng trinh c6 3 nghiem;

b) Dirdng thang (d): y = - x + m c^t do thj (C) tai hai d i e m phan bipt M , N . Tim
phu'dng trinh quy tich trung d i e m I cua doan M N .

,a L-

HUdng dSn giai
a) Hoc sinh tiT khao sat.

^

*J<

b) Phu'dng trinh hoanh dp giao d i e m cija (d) va (C) c6 dang
x^ + 2 x + 5
= - X

+m o

2x + (3 - m)x + 5 - m =: 0

x +1

Theo cong thiJc tinh tpa dp trung d i e m va dinh l i V i - e t , ta co
x,

_ X M

+ X N

2

' ^ A ' A C H

+

—^ < m < - 2 : phu'dng trinh cd 4 nghiem;

+

m = - 2 : phu'dng trinh cd 2 nghiem;

+

- 2 < m < 2: phu'dng trinh vd nghiem;

+

m = 2 : phu'dng trinh cd 2 nghiem;

+

m > 2: phu'dng trinh cd 4 nghiem.

_m-3
4

V i I n ^ m tren diTdng thdng y = - x + m nen ta c6

x2-4
Htfdng dan: V i c t phUdng trinh lai du'di dang
X

X

-2

+ 5
•=m

< y •


Luy^n gidi dS truOc

Do

thi h a m

thi DH 3 miSn Bdc, Trung, Nam Todn hoc - ISguySn van i nong

so'd vc'

trai difOc

suy

ra tiT

do

thi

4^ + 5

hiim so' y =
X

c d c h gii? nguyen
nay true tung.

*

nhanh phai

(u'ng vdi x > 0)

\d\h doi xu'ng

hOp

bang

-2

Bai 12. Cho f(x) = x' - 3x. ChiJng minh rSng neu x > y thi f(x) - f(y) > - 4 . Dan
bang xay ra khi va chi khi nao?
'
Hi/(?ng d§n gial
*») " ^ n ( . P ) ^ l ift!)OD
Khao sat do thj ham so f(x) = x' - 3x, ta thay f(x) dat cUc dai bkng 2 khi x =
- 1 va dat ciTc ticu b^ng -2 khi x = 1.
-»> r ' M * *
\ r«J
"

y

i

PHlJdNG T I ^ B X T PHl/dNG TRiNH M C vA L6GARIT
Hfe PHl/dNG TRiNH, Hfe B X T PHIJCJNG TRiNH VA NfU L6GAWT

c u a nhanh

I

T 6 M TAT LI THUYET

jjuw-'u.] u,u,

1. phi/dng trinh cd ban

a phuang tiinh mu CO ban a" = m(0< a ^ 1).
Neu m < 0 thi phifdng trinh v6 nghiem.

' « •

,

• .7 . !

.....

Neu m > 0 thi phu'dng trinh co nghiem duy nhat x = logam.

• ' '

b Phuang trinh Idgarit ca ban

i

log„x = m (0 < a ^ 1) o X = a'"
3

3"^^^^ ^''^

2. Mpt so phu'dng phap giai phu'dng trinh mu va logarit.

'

8'>Mi>

FhiTdng phap du'a vc cung cd so, phi/dng phap dat an phu, phifdng phap
logarit hoa, phu'dng phap sijf dung tinh dong bicn, nghjch bicn cua ham so.

\

Chiiy:
-3

-2/

/

-1

"

-2

V

I I

1/2

1

3

•

w
"

Cho 0 < a

"
?t

.

• i

1. Khi d6

^

,

' ^

.

\

+ a"^'= a ' ' * ^ ' f ( x ) = g(x)
+, loga f(x) = log, g(x) « f(x) = g(x) > 0.
•
3.

•'

Cho ham so y = f(x) dong bie'n tren D va x,, e D.
Khi do, phu'dng trinh l"(x) =

.0 =

<

f(X())

co nghiem duy nhat x =

Xo

tren D.

phu'dng trinh mu va Idgarit.

Phrfdng phap chung
•

luy thiTa, mu v^ logarit va dilng cac phiTdng phap the, phiTdng phap cong dai

//i/i/i 15. Doth f ham soy = x' - 3x
•

Ncu y < - 1 thi i"(y) < 2. Ta c6 2 triTimg hdp:

so, phifdng phap dat an phu,...

+ Neu X < - 1 thi do l"(x) tang tren -oo; -1) nen la c6 l(x) - f(y) > 0 > - 4 .
•+ N e ' u x > - 1 t h i l ( x ) > - 2 d o d 6 f ( x ) - r ( y ) > - 4 .
•

Khi giai cac he phi/dng trinh mu va logarit, ta difa vao cdc phep bicn doi ve

•

Can chu y dat dieu kien dc he phifdng trinh c6 nghiem, dac biet la cac bieu
thiJc n^m trong logarit.

^

,

Ncu y > - 1 ihi X > - 1 va ta c6 f(x) > -2. Ta lai xct 2 triTtJng hdp:

4. Bat phifdng trinh mu va Idgarit.

+ Ncu y < 1 thi f(y) < 2 va ta c6 r(x) - l(y) > -4.

a- Bat phuang trinh mu ca ban.

+ Ncu y > 1 ihi do h i m so f(x) tang len (1; +oo), la c6 f(x) - l"(y) > 0 > -4

•

Ba't phifdng trinh a" > m ( 0 < a 7i 1).

'

Neu m < 0 thi ba't phifdng trinh nghiem dung vdi moi gia tri cua x.
Neu m > 0 va a > 1 Ihi nghiem cua bat phifdng trinh la x > log„m.

Vay trong moi irifdng hdp ta dcu co f(x) - f(y) > -4 (dpcm).

Neu m > 0 va a < I l h i nghiem cua bat phifdng trinh la x < log^m.
•

Bat phifdng trinh a" < m (0 < a

1)

Ne'u m < 0 thi ba't phifdng Irinh v6 nghiem.

•/

,

,

'
45


Liiy?n gidi di trade

COng ty TNHH MTVDWH

thi DH 3 miSn Bdc. Trung, Nam Todn hoc - NguySn Van ThOng

Ne'u m > 0 va a > 1 Ihi nghicm cua bat phtfdng trinh la x < logam.

log, M(logy z - l )

Neu m > 0 va a < 1 thi nghicm ciia bat phUdng trinh la x > iogam.

logy M (logy Z - l )

b. Bdt phuang trinh Idgarit ca ban
•

Bat phiWng trinh logaX > m (0 < a

1).

"^^^ k ^ y M - logy M

j Ji-, j- fj

Neu 0 < a < 1 thi nghiem cua bat phU'dng trinh la a < x < a"". ,5^,;.,^
•

log, M - logy M ^ log, M

!•;

Neu a > 1 thi nghiem cua ba't phU'dng trinh la x > a'".

Khang Vt(t

logy M

nai 2. Giai cac phiTdng trinh:
log,(x' + x + l ) - i o g 3 X = 2 x - x

Bat phU'dng trinh logaX < m (0 < a ^ 1)

• ,og,(x' + X+ 1) + log2(x' - X + 1) = logjCx" + X' + 1) + l0g2(x' - x ' + 1)

Ne'u a > 1 thi nghiem cua bat phU'dng trinh la 0 < x < a'".

3« + 5" = 6x + 2

Ne'u 0 < a < 1 thi nghicm cua phU'dng bat phU'dng trinh la x > a'".
c. Phuang phdp giai bdt phuang trinh mu vd. logarit.

' X ' d i u i l gnburirt^ •'

HiMngdSngiai

Cung nhU'phi/(Jng trinh mu va logarit, dc giai bat phi/dng trinh mu va logarit ta

^ Dieu kien: x > 0. PhiTdng trinh tiTdng diTdng vdi

cung siif dung cac phifdng phap: di/a ve cung cd so, dat an phu, phiTdng phap

x^+x + 1

logarit hoa, phiTdng phap sOc dung tinh dong bicn, nghjch bie'n cua ham so,...

log3

= 2x-x^ o

logj

+3 + (x-l)'=l

Vx N/X >

Chii y: Khi giai cac bat phU'dng trinh logarit, ta phai dac biet chu y dieu kien
N2

xac dinh cua bat phifdng trinh.

+ 3 >logj 3 = 1

+ 3>3:

De thay:

II. C A C B A I T O A N M I N H H Q A

I

Vx

Bai 1.
a. Cho X, y la hai so du'dng ihoa man x^ + y^ = 7xy. Chiirng minh r^ng

V 1 ( X - 1)^>0=:> log.

Vx

x+y
1 ^,
,
'og3 — ^ = - (logix + logiy)
1

+ 3 + (x-l)^>l

- i

b. Cho X, y, z > 0; X, y, z

1 va M > 0; M

>/x

1.

= 0

Ding thuTc xay ra khi va chi khi:

ChuTng minh rkng ne'u x, y, z lap nen cap so'nhan thi

o x = l (Thoa dieu kien)

log^ M - logy M _ log^ M
logy M - log^ M

Vay phiTdng innh c6 nghi$m x = 1

logy M

b. PhiTdng trinh tiTdng diTdng vdi:

Hi^i^ng dSn giai
1
a. Taco: log3-^^-!^ = -log3
3
2

1,

--\ogT,

l0g2 [(x^ + X + l ) ( x 2 - X + 1)] = log2 [ ( x ^ + X^ + l ) ( x ^ - X^ + l )

x^+y^+2xy
1
9xy
=^ = - l o g - , — -

o(x^^lf-x^=(x^ +

1

= 2 ('^Ssx + log3 y )
b. X, y, z lap nen cap so nhan

xz = y^ => logyX + logyZ = 2

c- PhiTdng trinh tiTdng diTdng vdi: 3" + 5* - 6x - 2 = 0 (1)
De thay x = 0; x = 1 l i hai nghi?m ciia phiTdng irinh (1).

Ta c6:

log^M-logyM

log^M

logx M 1 -

logy M - logy M ~ logy M
logy M

X6t ham so f(x) = 3 ' + 5 ' - 6x - 2 tren R-

logyM^

f (x) = 3Mn3 + S'lnS - 6

l o g , M j _ logxM(-logyx)

logy M
jOgyM

-1

lf-x^ox«-x^==Oo^^:J'

f"(x) = 3"(ln3)' + 5x(ln5)' > 0; Vx e R

lOgyM(lOgyZ-l)

=> f ( x ) dong bien tren R.
A

Matkhdc:

.
J

'
' ~ - i.

.: id.'tt 3" ^

' '


LuySn mi d? trade ky thiPH 5 mmBUc. Trung, Nam loan /

tySn van 'IhO

f(()) = ln3 + ln5 - 6 = lnl5 - 6 < 0
r ( l ) = 31n3 + 5 1 n 5 - 6 > 0

t = 2cos(p

Suy ra ton tai duy nhat a e (0; 1) sao cho f (a) = 0

2(4cos'(p - 3cos(p) = 1 <=> cos3(() = ^ <=> cp = ^

0
JJ

Bang bien thien

-

f(x)

I

'•'a

0

—00

X

-

0

+00

+

+

Ta c6: 2008^l''"''U4l'-''''''' >4'''''"''U-.4l'-''''''U-4'''''''''U-.4

= x'' - 3x^ -1
x'' + x^ +1

. 1
2 :)t«t»>^*'»'>^*:i'f!,a
sinx + cosx > s i n ' x + cos x = l )

Hifdng din giai
a = 4x^ + 2 > 0

=> Phifdng trinh xac dinh vdi mpi x e R.

b-x^ + x^+l>0

D a i t = 2008^'^'"''l+4i'-'"''l-5>0
PhiTdng trinh (1) trd thanh

PhU'ctng trinh trd th^nh \og2(m ~^

T

a . 2008*'
Neu a > b thi - > 1 >
b
2008"

;

Neu a < b thl - < 1 < 2008^
b
2008"

"f4 M'iVil

'u = 2 0 0 8 ' - l

:

[2008'-u + l

[2008'-2008" = u - t

[2008'=t + l

(2)

u =t

u=t

Xet ham so 1(1) = 2008' - t - 1, t e [0; +oo)

r(t) = 2008'.ln2008 - 1 > 0; Vt > 0
Bang bien thien
t

Dat t = x ^ t > 0. Phi/cfng trinh trd thanh: t^ - 3t - 1 = 0

0

+00

m
m

Xet ham so f(t) = t^ - 3t - 1 tren [0;+00)

f (t) = 3t^ - 3
o

,^

[2008'-u + i

t + 1-2008"

Vay a = b =^ x"" - 3x^ -- 11 == 0

f'(t) = 0

'"Sr

Hif(}ng dSn giai

Vay phiTdng trinh c6 hai nghiSm x = 0, x = 1.
Bai 3. Giai phu'dng trinh: logjoos

'^^^

= 2008^'""* '

l0g2(KW

Dat

2cos —
9

Vay phiTdng trinh c6 dung hai nghiem x = ^2cos-^ \x=
Bai 4. Giai phiTdng trinh

+00~.^_____

f(x)

•

+
—
0

*• +°o

•

1

t=l

2QQg2|sinx|^44|sinx|

t =- l

Khi do, (2) suy ra u = t = 0

,.

I

,

I

,

o x = kn (k e Z)

^4|sinx| _ J _ ^|cosx|

X6t bang bie'n thien

~4"

0
f(t)
f(t)

+C0

0

' •

•

Vay phi/cJng trinh c6 nghiem x = kn (k e Z)
^ a l 5. Gi^i phircjng trinh 2'" + 3'" = 2" + 3"^' + x + 1

+00

HUdng dSn giai
Phifdng trinh liTcfng difdng vdi
PhiTcfng trinh chi c6 dung mot nghiem t e (1; 2) nen ta dat

2^'+3'"+2''=2''+'+3''^'+x + l

(1)

'•
AO