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+6x^-2x + 3-(5x-l)Vx^+3=0
V-4x^+18x-20 + ^"'-^^^SV^
2x2-9x + 8

DVL.013496

4 1
Chung minh: — + — > 5 (1)
x 4y

NHA XUATBAN DAI HOC QUOC 6IA HA NOI

TAI LIEU O N T H I D A I H O C

Joqg boo VQ gioi
X

PHl/ONG T R I N H
BAT PHl/ONG T R I N H

' ;

HE PHUWNG TRINH
BAT D A N G

THUt

• Danh cho hoc sinh Idp 12 chiicrng t r i n h chuan va nang cao
• O n tap va nang cao k i nang lam bai
• Bien soan theo noi dung va cau true de t h i ciia Bo GD8fDT

-

I

THi;

VIEW T I f y ' H B I N H THUAN'

OVL

t

0[}{]

'4
/

-y. • • • -'


Phan 1:

Ldi noi dHu

PHL/ONG PHAP GlAl PHLTONG TRINH,

P h u o n g t r i n h , bat p h u o n g t r i n h , h f p h u o n g t r i n h , bat d 5 n g thuc la m ^ n g
kien thuc quan t r o n g t r o n g c h u o n g t r i n h toan pho t h o n g . Dac bi^t cac bai toan
ve p h u o n g t r i n h , bat p h u o n g t r i n h , h ^ p h u o n g t r i n h , bat d a n g thuc t h u o n g

BAT PHLfONG TRlNH VO TY

xuyen xua't hien t r o n g cac k y t h i chQn h p c s i n h gioi, cOng n h u T u y e n sinh dai
hQC va l u o n gay k h o k h a n cho hoc sinh.
N h S m g i i i p cac e m hpc sinh T H P T cung n h u cac e m hpc sinh chuyen Toan

NHOTNG KI^N T H L f C B6 T R O C H O GlAl P H l / O N G T R I N H V O i t

I.

CO m p t tai l i f u m a n g t i n h h f t h o n g de o n l u y f n , nang cao k i e n thuc k y nang

1. G i a i p h u a n g trinh bac 4:

giai toan de d a t ket qua cao nha't trong cac k y t h i H g c sinh g i o i , k y t h i T u y e n

a) PhuoiTg trinh dang: x^ = ax^ + bx + c

sinh dai hpc, c u n g n h u t h i vao cac l o p chuyen chpn, toi bien soan cuon: "Tai

P h u o n g phap: Ta t h e m bot vao 2 v e m p t l u p n g : 2mx^ + m ^ k h i d o p h u o n g

li^u on thi Dai hoc - sang tao va giai phuong trinh, bai phuong trinh, h$

t r i n h t r o t h a n h : (x^ + m)^ = (2m + a)x^ + bx + c +

phuong trinh, bai dang thitc".

Ta m o n g m u o n ve'phai c6 dang: ( A x + B)^

N p i d u n g cuon sach d u p e chia l a m 4 phan:

2m + a > 0

,Phan 1 : P h u o n g phap giai p h u o n g t r i n h , bat p h u o n g t r i n h v 6 t y

A = h^- 4(2m + a)(c + m^) = 0 ^

Phan 2: P h u o n g phap giai h^ p h u o n g t r i n h
Phan 3: P h u a n g phap h a m so'trong cac bai toan chua t h a m so'

Vi

Phan 4: P h u o n g phap h a m so trong c h i i n g m i n h bat dang thuc va t i m
a)

GTLN, G T N N

1: Giai cac phuang trinh:
x+V 5 W ^

=6

b)2x2-6x-l =V i ; ^

T r o n g m o i p h a n toi l u o n c6' gang h^ t h o n g p h u o n g phap, p h a n tich, d j n h
h u o n g each giai, cuoi m o i p h a n deu c6 bai tap ren l u y ^ n de cac e m hpc sinh
thusuc.

Giii:
a) D i e u k i ^ n : 1 < x < 6
Dat y =

T o i h y v p n g cuo'n sach se la tai li^u b o ich cho cac e m hpc sinh hpc tot m o n
Toan va dat ket qua cao t r o n g cac k y t h i .

' '

>0.

P h u o n g t r i n h t r o thanh: y2 + ^/^^^ ^ 5 ^ j y ' " l O y ' - y + 20 = 0
(0 < y < Vs

Mac d i i da co gang d a n h nhieu t a m huyet cho vi?c bien soan cuon sach

> ^f,,

song thie'u sot la d i e u k h o n g the tranh k h o i . Rat m o n g s u d o n g g o p phe b i n h

Xet p h u o n g t r i n h : y4 - I0y2 - y + 20 = 0 o y" = 10y2 + y - 20.

ciia ban dpc de Ian tai ban sau d u p e hoan thi^n h o n .

Ta t h e m vao 2 v e ' p h u o n g t r i n h m p t l u p n g : 2my^ + m^

C u o i cung toi x i n g u i l a i cam o n sau sac den ban be, d o n g n g h i f p, cac dien
dan toan da cung cap m p t so'tai l i ^ u quy gia de hoan thi^n cuon sach.

K h i d o p h u o n g t r i n h t r o thanh:

y'^ + 2my2 + m^ = (10 + 2m)y^ + y + m^ - 20

T a c gia
Nguyen Trung Kien

T a c o Ayp = l - 4 ( m 2 - 2 0 ) ( 1 0 + 2m) = 0 « > m = - 2
Ta Viet lai p h u o n g t r i n h thanh:
y ^ - 9 y 2 + (-]

=y

2

1
+y+7«>
4

y-2

=0


Tdt lieu

ou tin

h,u

tao vd ^fiat fi, oai fi, ne yrrmn^i

si'iii^

« ( y 2 - y - 5 ) ( y 2 + y - 4 ) = 0=*y =

-i +

Vi7 (TM) =>

-nguyen

X =

irung

IVICTI

II-V17

a) Dieu kifn: x > —
4
Binh phuong hai ve'ta thu duoc phuong trinh:

b) Dieu ki^n: x > —

o x ^ - 6 x ^ +8x2 + 2 x - l = 0 < » x ^ - 6 x ^ = - 8 x 2 - 2 x + l

Dat y = V4x + 5 > 0 thi phuong trinh da cho c6 dang:

Ta tao ra vetrai dang: (x^ - 3x + m)^ =

y ' ' - 2 2 y 2 - 8 y + 77 = 0 » y"* = 22y2 + 8 y - 7 7

- 6x^ + (9 + 2m)x2 - 6mx + m^

Tuc la them vao hai ve mpt lup-ng la:(9 + 2m)x2 - 6 m x + m2 phuong trinh

Ta them vao 2 ve phuang trinh mot luong: 2my^ + m^

tro thanh: (x^ - 3x + m)^ = (2m + l)x2 - (6m + 2)x + m^ +1

Khi do phuang trinh tro thanh:

Ta can A\p = ( 3 m + 1 ) - ( 2 m + I ) ( m 2 + 1 ) = 0 o m = 0

y^ + 2my^ + m^ = (22 + 2m)y^ + 8y +

- 77

Phuang trinh tro thanh: (x^ - 3x)2 = (x -1)^

Ta CO Ayp = 1 - 4(22 + 2m)(m2 - 77) = 0 o m = - 9 .

x = 2 + V3

Ta viet lai phuong trinh thanh:
y'* - 18y2 + 81 = 4y^ + 8y + 4 o (y^ - 9

- (2y + 2 ^ = 0

'y = -1 + 2^12

« ( y ^ + 2 y - 7 ) ( y ^ - 2 y - l l ) = 0:

<^(x2 - 4 x + l)(x2 - 2 x - l ) = 0
X =

(TM):

y = l + 2V3
b) Phuang trinh dang: x"* + ax^ = bx^ + cx + d

2
X

3
+—X

2

+m

Bang each khai trien bieu thuc:
2

I

X

a
+—x+m
2

,

a^^
= x'* + ax"' + 2m + — x^ +amx + m^
4

I

+

2

Bay gio ta can: •

2m+ •

61)

Ta tao ra phuang trinh: (y^ + y + mf =(2m + 6)y2 +(2m-6)y + m2 +3

A'vp = ( m - 3 ) 2 - ( 2 m + 6)(m2+3) = 0

x^ + amx + m^ khi do
« > ( y 2 + 3 y - 3 ) ( y 2 - y + l) = O o

f
—

2

o

m = -1

2m + — + b x^+(am + c)x + m + d
4

-3 + ^/2T

J

Vol V =

a2

2m + — + b > 0
4

y

=> m = ?
:

b) x^ = (1 - V5^)(2x - sVx + 3)

2 x 2 - 6 x - l = V4x + 5
Giai:

_ -3 + N/2T (Thoa man)
2
,_-3-V2T

^

Ayp =(am + c)2 - 4 2m + — + b (m^ +dj = 0
4
v
>

V i du 2: Giai cac phuang trinh:
a)

y^ = (1 - y)(2y2 - 3y + 3) < > y^ + 2y^ = Sy^ - 6y + 3
=

Phuong trinh tro thanh: (y^ + y -1)^ = (2y - if

2

X

11
thoa man phuong trinh.

Ta can:

phuong trinh tro thanh:
a
—X + m

= 1-N/2

2m + 6>0

(

Ta thay can them vao hai ve mpt luong:

2

*^^^V tat ca cac nghiem

b) Dat N/X = y > 0 thi phuong trinh da cho tro thanh:

Ta se tao ra 6 ve phai mot bieu thuc binh phuong dang:

2

X = 1 + N/2 ^^'^
X

1-N/2

x = 2 + V3

x = 2-V3

(Loai)

15-3^
r=>

X=

2

2

2. Phuang trinh v6 ty ca ban:

= g(x)
g ( x )= 0
> g2(x)
f(x)

7^

Vi du 1: Giai cac phuang trinh:
a) 7x2 +2x + 6 ^2x + l

b) V2x +1 +

= V4x + 9


GiAi:

+

( ^ - b ) [ ^ +^b +

^ = a-b^

+

( ^ / I - b ) ( V ^ + b) = a-b2

a) Phuong trinh tuong duang voi:
2
<>

=
+2x + 6 = (2x + l ) 2

x . - l
2

<=>x = l

i-,

3x2+2x-5 = 0
b) Dieu ki^n: x > 0. Binh phuong 2 ve ta dugc:
=
3x +1 + l^lx^ +x = 4x + 9 < > l^fhJTx
x>-8
<=> i
7x2-12x-64 = 0

=X +8 o

f /

+ Neu h(x) = 0 C O nghiem x = XQ thi ta luon phan tich dugc

, ...

h(x) = (x-Xo)g(x)

x>-8

Nhu vay sau buoc phan tich va riit nhan tu chung x - XQ thi phuong trinh

4(2x2+x) = (x +8)2

ban dau tro thanh: (x - Xo)A(x) = 0 < >
=

x=4

X - Xg =

0

A(x) = 0

Vi?c con lai la diing ham so', bat dang thuc hoac nhirng danh gia co ban de
ke't luan A(x) = 0 v6 nghiem.

16
x = —7

Doi chieu vai dieu kien ta tha'y chi c6 x = 4 la nghi^m cua phuong trinh.
II. MOT S 6 DANG PHUONG TRJNH VO

Neu phuong trinh c6 2 nghiem x ^ X j theo dinh ly viet dao ta c6 nhan tu

•

chung se la: x^ - (xj + X2 )x + Xj .Xj

THL/ONG GAP

Ta thuong lam nhu sau:

1. Giai phuong tiinh v6 ty bling phuong phap su dung bieu thuc lien hgp:
Dau hi^u:

+ Muon lam xua't hien nhan tu chung trong ^ ( ( x ) ta tru di mot lugng ax + b.

+ Khi ta gap cac bai toan giai phuong trinh dang: yfu^ + '^g(x) + h(x) = 0

Khi do nhan tu chung se la ke't qua sau khi nhan lien hgp ciia

Ma khong the dua ve mpt an, hoac khi dua ve mot an thi tao ra nhihig
phuong trinh bac cao dan den vi^c phan tich hoac giai true tiep kho khan.
+ Nham dugc nghi^m cua phuong trinh do: bang thii cong (hoac su dung
may tinh cam tay)
Phuong phap:
• Dat dieu kif n chat cua phuong trinh ( neii c6)
Vi du: Doi phuong trinh: Vx^ +3 + 3

^fly^+V + 2x.

+ Neu binh thuong nhin vao phuong trinh ta tha'y:
Phuong trinh xac dinh voi moi x € R. Nhung do chua phai la dieu ki^n
chat. De giai quyet triet de phuong trinh nay ta can den dieu ki^n chat do la:
+ Ta viet lai phuong trinh thanh: N / X ^ + 3 - ^ 2 x 2 + 7 = 2 x - 3
Dey rang: ylx^+S-^ll)^

+ 7 < 0 do do phuong trinh c6 nghiem khi

3
2x-3<0ox<2
• Neu phuong trinh chi c6 mot nghiem XQ :

# 0 - ( a x + b)

+ De tim a, b ta xet phuong trinh:

yfUx)

-

(ax + b) = 0. De phuong trinh c6

hai nghiem X j , X2 ta can tim a, b sao cho

axj + b = axj +

+ Hoan toan tuong tu cho cac bieu thuc con lai:
Ta xet cac vi du sau:
Vi d\ 1: Giai cac phuong trinh:
a)

V5x^-l+^2x-l+x-4 =0

b) V x ^ + V 4 - x = 2 x 2 - 5 x - 3

Giai:

a) Phan tich: Phuong trinh trong de bai gom nhieu bieu thuc chua can nhung
khong the quy ve 1 an. Neu ta liiy thua de tri^t tieu dau \/~, yT thi se tao ra
phuong trinh toi thieu la bac 6. Tit do ta nghi den huong giai: Su dung bieu
thuc lien hgp hoac dimg ham so :
Tu djnh huong tren ta c6 loi giai nhu sau:

Ta se phan tich phuong trinh nhu sau: Viet lai phuong trinh thanh:
# 0 -

+ - - ^ ^ g O ^ + h(x) - h(xo) = 0

Sau do nhan lien hgp cho tung cap so hang voi chii y:

Dieu ki^n x > ^—
Ta nham dugc nghiem cua phuong trinh la: x = 1.


2: G i a i cac phuong trinh:

K h i d o Vsx^-l = V 5 ^ = 2 ; ^ 2 x - l = V 2 ^ = l

Vi

V i vay ta phan tich p h u o n g t r i n h thanh:

a) x/x^ - 1 + x = Vx^ - 2

V5x3-l-2 +^ 2 x - l - l +x - l = 0

b) ^ - 2 ^ - ( x - 4 ) N A r ^ - 3 x + 28 = 0

5x^-5

2x-2

75x^-1+2-0

3|(2x-l)^ + ^ / 2 ^ + l

«(x-l)

a) D i e u kien: x >
Ta n h a m d u g c ng hiem x = 3 . N e n p h u o n g t r i n h dugc viet lai n h u sau:

2 •

5(x2 + x + l)
^(2x-lf

75x^-1+2

Giai:

+x - l =0 '

+1 = 0

\/x2 - 1 - 2 + x - 3 = 7 x ^ - 2

+^/2^ +l

N/X^ - 1 + 2 ^ x ^ - 1 + 4

+ 1>0
Voi dieu Itien x > 3i t h i f c l l ) +
V5
V5x3-l+2
^(2x-l)'+^^^ +l

.7x' - 1 +27x2 -1+4

= ^Jn = l

gia t r i X > 7 2 ta se thay

X - - 3 ^ ( x _ 3 ) ( 2 x + l)

1

1

Ta se c h u n g m i n h :

- ( 2 x + l) = 0

^

+

i = - ( 2 x + l) = 0

+

Taxet

Vx-2+1

< 0 (Bang each thay m p t

7x^-2+5

x+3

x2 + 3 x + 9
• +1 <0)
7x2 - 1 + 2 7 x 2 - 1 +4
7x^-2+5

'

x+3

, > x2+3x + 9 ^
< 1 va ,
>2
7x2 - 1 + 2 7 x 2 - 1 + 4
7x^-2+5

.

1

< 1;

- < li; , ^ A + l > 5 nen
^
2 x - r 1 !^ i j i i c i i

,

1+V 4 - X

1
,

Vx-2+1

1

1

N h a n xet: De d a n h gia p h u o n g t r i n h cuol cung v 6 n g h i e m ta t h u o n g d u n g
cac u o c l u g n g co ban: A + B > A v o i B > 0 tir d o suy ra
A + B>0
B>0


+27x2-1 > x - l

Dat 7 x 2 - 1 = t > 0 = > x = 7 t ^ + l .

.-(2x + l)<0
1 + V4-X
,

Bat p h u o n g t r i n h t u o n g d u o n g v o i :

T u d o suy ra: x = 3 la nghiem d u y nhat ciia p h u o n g t r i n h .

so A , B thoa m a n

x +3
^(x2-l)'+27^ +4

De y rang: V o i dieu kien x e 2; 4 t h i :
]

x^ + 3 x + 9

That vay:

x =3

,

+1 -

7x2-1+27x2-1+4

> A C ^ - 1 +1-N/4^ = 2X2-5X-3

.(x-3)

7x^-2+5

x+3

Ta d u doan:

P h u o n g t r i n h da cho t u o n g d u o n g v o i :

^

=0

7x^-2+5

x+3

<>
=

T u d o ta CO l o i giai n h u sau:

x-3

x^ + 3 x + 9

"x = 3

Ta n h a m d u g c n g h i e m cua p h u o n g t r i n h la: x = 3 .
= V,^[^^

+ 1-

. 7 x 2 - 1 + 2 7 x ^ - 1 +4

b) D i e u kien: x € 2; 4

yf^=^^f3^

7x^-2+5

x+3

«(x-3)

N e n p h u o n g t r i n h da cho c6 n g h i f m d u y nhat x = 2

^

^ '

+ x-3=

De t h a y :

Khi do

-5

A +B

<1 voi moi

t2 + 2t + 1 > 7t^ + 1 « t'^ + 3t^ + 6t2 + 4t > 0 . D i e u nay la hien nhien d i i n g .
+

Ta xet:
+ 3x 4- 9
I
I
> 2 <=> x2 + 3x - 1 > 27x^ - 2 » x^ + 2x^ + 7x2 - 6x + 9 > 0
7x^-2+5


<=> (x^ + x)^ + 6x^ - 6x + 9 > 0 . D i e u nay l u o n d u n g .

4
a =—
3
De CO dieu nay ta can:
<=> <
-2a + b = 4
20
b =
a+b=8

T u d o suy ra p h u a n g t r i n h c6 nghiem d u y nhat: x = 3

b) D i e u k i ^ n : x > 7 .
De d o n gian ta dat ^

=t>^

=> x = t^

+

P h u a n g t r i n h da cho t r o thanh:

t2 _ 2t - (t^ - 4)7t^-7 - 3t^ + 28 = 0

3t^ - 1 ^ + 2t - 28 + (t^ - 4)7t^-7 = 0

N h a m dug-c t = 2 . N e n ta phan tich p h u o n g t r i n h thanh:
c^4t^-t2+2t-32 + (t^-4)

o ( t - 2 ) (4t^ + 7 t + 16) + ( t ^ - 4 )

7-1

=0

T u o n g t y V l 9 - 3 x - (mx + n ) = 0 nh|n x = 1, x = - 2 la n g h i e m .
m = —
, fm + n = 4
3
Tuc la <
<=> <

!
-2m +n = 5
13
n = —
3
T u d o ta phan tich p h u o n g t r i n h thanh:

t^ + 2 t + 4

4Vx + 3 -

=0

4

D e y rang 4t^ + 7 t + 1 6 > 0 va t'' > 7 nen t a c o
4 t ^ + 7 t + 16 ) + ( t ^ - 4 )

t^+2t + 4

4
<> —
=
3

>0.

V i v a y p h u o n g t r i n h c6 nghi?m d u y nhat t = 2 < > x = 8 .
=
N h a n xet: Vi^c dat \/x = t t r o n g bai toan de g i a m so l u g n g da'u can da g i i i p

d a n gian h i n h thuc bai toan .
N g o a i ra k h i tao Hen h o p d o (t^ - 4) > 0 nen ta tach n o ra k h o i bieu thuc de
cac thao tac t i n h toan dup-c d o n gian h o n .
V i d y 3: G i a i cac p h u o t i g t r i n h :
b) V3x - 8 - Vx + 1 =

a) 4Vx + 3 + V l 9 - 3 x = x ^ + 2 x + 9

c)

x2+7
'^^x

d)

2(x + l )

x^+5x^+4x + 2
x^ + 2 x + 3

2x-ll

ri
= Vx

+X + 7
2

Giai:
19

a) D i e u k i ^ n : - 3 < x < —
3
Ta n h a m duoc 2 nghi em la x = l , x = - 2 nen ta p h a n tich de tao ra nhan t u
chung la: x^ + x - 2 . De l a m dug-c dieu nay ta th^c hien t h e m b o t nhan t u
n h u sau:
+

Ta tao ra 4\Jx + 3 - (ax + b) = 0 sao cho p h u o n g t r i n h nay nhan x = 1, x = - 2 la
nghiem.

o

4

20

+ Vl9-3x

—X +—

-

ri3_x
3

3

-(x2+x-2

3

3 , 3s/l9r^-(13-x)
3Vx + 3 - ( x + 5)

-X' - x + 2
3N/X + 3+(X + 5)

- x^ - x - 2

=0

/2
^
(x2-x-2) =0

-x-" - x + 2
3^3^19-3x + ( 1 3 - x )

•^'•^^^

x'=+x-2 = 0

1
3 3>/x + 3 + (x + 5)

3 3Vl9-3x +(13-x)

n

+

1 =0

1
De thay v o i - 3 < x < — t h i
> 0, — — - —
^>0
3
3Vx + 3 + ( x + 5)
3 3Vl9-3x +(13-x)
N e n —.
3 3N/XT3+{X

+ 5)

3\3^fl9^

+

{13-x)

+ 1.

P h u o n g t r i n h da cho t u o n g d u o n g v o i x^ + x - 2 = 0

x=l
x = -2

Vay p h u o n g t r i n h c6 2 nghiem la: x = 3, x = 8 .

I' •

N h a n xet: N e u da nha m dugc hai nghiem ciia p h u o n g t r i n h va d u doan
dugc p h u o n g t r i n h chi c6 2 nghiem t h i ta c6 the giai theo m p t each khac
n g i n g o n h o n n h u sau:
Xet ham so f(x) = 4Vx + 3 + V l 9 - 3 x - x^ - 2x - 9 tren
19
Ta thay x = - 3 ; x = - - k h o n g phai la nghiem,
i3

-3-^
'3


Cty TNHH

Tren

-3;

19

ta

-9

CO

5V3X-8 + (3x-4)
^
•-2x-2;f'(x) = -3x

'
^(x + 3 f

f'(x) =
^

-2<0

( X + 7) + 5>/X + 1

MTV DWH

Khattg

Viet

<0

c:>5V3x-8 + 3 x - 4 - 9 ( x + 7 + 5Vx + l ) < 0

^(19-3x)
< : : > 3 x - 8 - 5 V 3 x - 8 + —+ — + x + 4 5 V x T T > 0
4
4

nen f(x) = 0 c6 toi da hai nghiem tren

'3)

M a t khac: f ( - 2 ) = f(l) = 0 nen p h u o n g t r i n h c6 d i i n g hai n g h i f m la

V3x - 8 - —

x=l
x = -2

+

+ X + 45v'x + l > 0 . Dieu nay la hien nhien d u n g .

Vay p h u o n g t r i n h c6 2 nghiem la: x = 3, x = 8 .
Chuy:

b) D i e u k i ^ n : x > —.

N h u n g danh gia de ke't luan A ( x ) < 0 t h u d n g la n h i r n g bat dang thuc khong

3

P h u o n g t r i n h d u g c viet lai n h u sau:

5 V 3 X - 8 - 5 N / 5 ^

chat nen ta luon dua ve dugc tong cac bieu thuc b i n h p h u o n g .

= 2X-11

N g o a i ra neu t i n h y ta c6 the thay:

Ta n h a m d u o c 2 nghiem x = 3, x = 8 nen suy ra n h a n t u c h u n g la:

5\J3x - 8 + 3x - 4 < 9x + 63 + s V s i x + 81 N h u n g dieu nay la hien nhien d u n g

x^ - l l x + 24

do: 5 V 3 x - 8 < 5 7 8 ] x + 8 1 ; 3 x - 4 < 9 x + 63 v o i m g i x > -

Ta phan tich v o i n h a n t u 5\/3x-8 n h u sau:
+

,,,,

3

Tao ra sVSx - 8 - (ax + b) = 0 sao cho p h u o n g t r i n h nay n h a n x = 3,x = 8 la

N g o a i ra ta cung c6 the giai bai toan theo each d u n g h a m so 6 cau a)

nghiem.
Tuc la a,b can thoa m a n he:

+

5V3x - 8 + 3x - 4 - 9 ( x + 7 + sVx + 1) < 0

3a + b = 5

c) Dieu kien: x > 0

ja=:3

8a + b - 20 ^

Ta n h a m d u g c x = 1; x = 3 nen bie'n d o i p h u o n g t r i n h n h u sau:

b = -4

T u o n g t u v o i 5vx + 1 - ( m x + n) = 0 ta t h u dugc:

3m + n = 10
8m + n = 15

5 V 3 x - 8 - ( 3 x - 4 ) + (x + 7 ) - 5 V x + T = 0

x ^ - l l x + 24

573X-8 +(3x-4)

2(x + ] )

= 2,khi x=3

n=7

X

=0

-4x + 3

7

= 2 nen ta t r u 2 vao 2 ve

Vx^ + 3 - 2 V x

x^-4x +3

r

\

2(x + l )

2(x + l )

x''-4x + 3= 0

(1)

Vx^+3x + 2 x - 2 ( x + l )

x -4x + 3

Vx3 + 3 x + 2 x "

x2+7
2(x + l )

(2)

Giai (1) suy ra x = 1, x = 3

-9
5V3x-8+(3x-4)

(x + 7) + 5 V ^

=0

Giai (2) ta c6:
Vx-* + 3 x + 2x = 2(x + 1) c:> Vx^ + 3 x = 2 < o x ^ + 3 x - 4 = : 0 < = > x = l

x ^ - 1 1 x + 24 = 0

Ke't luan: P h u o n g t r i n h c6 nghiem la x = 1; x = 3

-9

1

5V3x-8+(3x-4)

(x + 7) + 5 V x + T

Ta xet A ( x ) =

Ta c6: k h i x = 1

7

{x + 7) + 5^[x + l

.(x^ - n x + 24J

<=>•

m=l

thi t h u dugc: J x + — - 2 = ^,
-2<»
X
2(x + l )

P h u o n g t r i n h da cho t r o thanh:

- 9 ( x ^ - l l x + 24)

j,

=0

-9

1

5x/3x-8+(3x-4)

( X + 7) + 5 N / ^

Ta c h u n g m i n h : A ( x ) < 0 tuc la:

N h a n xet: Ta cQng c6 the phan tich p h u o n g t r i n h n h u cau a,b .

' "

d) Ta c6: x^ + 5x^ + 4x + 2 = (x + 3)(x^ + 2x + 3) - 5x - 7 nen bat p h u o n g t r i n h
tuong duong voi
x^+5x^+4x + 2
;
= Vx^
x-+2x +3

+ X

7
,
n>
^
+ 2 « x + 3 - V x ' +2x + 3

5x + 7
x^+2x + 3

=0


Tai lifu 6n thi d^i hQC sang

t^o vd giai FT, bat l^l, WPT,

• (5x + 7)
,
^
[ ( x + 3) + V x 2 + x + 2

bal VI

- Nguyen

i rung

r^ien

Vang Viet

x^ - x - l = 0

=0
,
x2+2x + 3 j

x+1 +

'(5x + 7) = 0
Xet
= 0
(x + 3) + \/x^ + X + 2

t=2

t = -l(L)

•<=>x' + X - V x ^ + x + 2 = 0 .
x2+2x + 3

o ( x + l)2 +2(x + l ) V 8 - 3 x 2 + 8 - 3 x ^ + 3 = 0 .

o x + x - 2 = 0<=>

x=l

b) D i e u kien: 2x3 - 1 5 > 0 <=> x >

X +1

+ Vs -

-;2
3x

Dat t = (x^ - 7)

=> X =

- 4x^)72x3 - 1 5

+ 7 voi t > 2

Ta nham d u g c t = 1. N e n phan tich p h u o n g t r i n h n h u sau:

a) Dieu k i f n: I x 1 —
<
3

2 t ^ t T 7 - 4t + 5t - 5 + 4 t 7 2 t ^ T - 4t = 0

D i i n g may t i n h bo t u i ta t h u dugc 2 nghiem la: x^ - - 0 , 6 1 8 , X 2 =1,618
Xj.X2=-l

« 2 t f - V r f 7 - 2 ) + 5 ( t - l ) + 4t(V2t-l -1) = 0

nen n g h l d e n n h a n t u c h u n g la
»2t

x-^ - x - 1

t-1

+ 5 ( t - l ) + 4t

2t-2
U2t-l+lj

=0

Xet Vs-Sx^ - ( a x + b) = 0 sao cho p h u o n g t r i n h c6 2 n g h i e m X j =-0,618,
X 2 = 1,618 ta t h u duoc:
axj + b = y 8 -

3X]

axj + b = ^ 8 - 3x2

2t

«(t-l)
-0,618a + b = 2,61
l,618a + b = 0,38

7(t + 7)2 + 2 ^ / t 7 7 + 4

Ia = - 1

-4(x^ - x - 1 )
o(x^ - x - l ) ( x +l) = - ^
3x^ + ( 2 - x )
4
V8-3x2 +(2-x)^

+ 5-

4t

=0

721^ +1

2t
4t
De thay v o i t > — t h i

• + 5>0.
2
7(t + 7)2 + 2 ^ + 4
V2F^+1

' ^ ] b =2

P h u o n g t r i n h dug-c vie't lai n h u sau: x'' - 2x - 1 = \/8-3x^ - (2 - x)

«>(x^ - x - 1 ) x + 1 +

i + Vs

15

P h u o n g t r i n h da cho t r o thanh: 2t\/t + 7 - 3t - 5 + 4 t V 2 t - l = 0

Giai:

x^+X2=l,

+ 3=0

Ta vie't lai p h u o n g t r i n h thanh: 2x(x-' - 7) - 3{x^ - 7) - 5 = 4(7 - x^)^J2ix^ - 7 )

x^-3x + l = \/8-3x^

d o ta thay

+(2-X) + 4 = 0

Vay p h u o n g t r i n h ban dau t u o n g d u o n g v o i x^ - x - 1 = 0 <=> x =

x = -2

Vi du 4: Giai cac phuong trinh:

Tu

N/S-SX^

P h u o n g t r i n h nay v 6 nghiem.

Ket l u a n : P h u o n g t r i n h c6 3 nghiem: x = — r ; x = l ; x = - 2

b) 2x^ - 3x3 _ -,4^ ^ |g ^

= 0<:i>(x + l )

o ( x + l ) V 8 - 3 x ^ - X 2 + X + 6 = 0 O 2 ( X + 1 ) V 8 - 3 X 2 - 2 X 2 + 2 X + 12 = 0

D a t t = Vx^ +x + 2 > 0 . P h u o n g t r i n h t r o thanh:

a)

+1 +
•3x^ + ( 2 - X )

1

(x + 3) + \ / x 2 + x + 2

-t-2 = 0 ^

X

(1)

x 2 + 2 x + 3^

1

Giai (1):

t

=0
+(2-x)

VB-SX^

N h u vay p h u o n g t r i n h c6 nghiem d u y nhat t = 1 <=> x = 2
Vi du 5: Giai cac phuong trinh:
a)

2V4x' - X + 1 + 2x = 372x^ - x^ + 79x^ - 4x + 4

b) V3x + 1 - Vx + 3

- X

+1 =0


Tai l.cn 6nthUaihocshngtaovagtdtPl,batl'l,hel^l,bai

VI -N^yen

imngi^ien

Xetham so f(t) = Vt^ + 8 - Vt^ +15 - 3 t + 2 v a i t < - l

Giai:
a) Phuang trinh duoc viet lai nhu sau:
Vl6x2 - 4x + 4 + 2x = 3\/2x^ -

Tac6:f'(t) = -3 + 3t^
+ N/QX^ - 4x + 4

;.Vy "

I

« V l 6x2 - 4x + 4 - Vl 6x2 - 4x + 4 = 3\/2x2 - x^ - 2x > 0
Suy ra dieu kien de phuang trinh c6 nghiem la 3^/2x^ - x'' - 2x > 0
o 27(2x2 - x ^ ) > ^ x ^ » 35x3 - 5 4 x 2 < 0 o

X <

X ?i

Trucmg hop 1: x > 0 Chia hai ve phuang trinh cho x ta thu dugc:

X^

- - 1 = t ^ Phuong trinh da cho tro thanh:

Vt'^+15+2 = V t ^ + 8 + 3 t o V t ^ + 1 5 - V t ^ + 8 = 3 t - 2 .
2
De phuang trinh c6 nghiem ta can: 3 t - 2 > 0 o t > - . Nham duoc t = 1 nen
ta viet lai phuong trinh thanh: Vt^ +15 - 4 =
«(t^l)

't-Vl)(t2 +t +l)
15+4

De y rang:

(t^+l

^2 + t

( t 3 + l ) ( t 2 + t + l )- 3
Vt^78 + 3

+1

Vt" +15 + 4

^2

+t+l

- 3 + 3t - 3
=0

<0 nen phuong trinh c6

Vt^+8 + 3

X

ta nen dat an

sao cho v i f c bieu dien x theo an do la don

-'•'-1

Ta viet lai phuong trinh nhu sau: sfsx + l - Vx + 3 +1 - x = 0

x=l
V3x +1 + Vx + 3 = 2
Xet phuong trinh: >/3x + l + Vx + 3 = 2 . Binh phuang 2 ve ta thu dugc:
4x + 4 + 2J(3x + l)(x + 3) = 4 o J(3x + l)(x + 3) = - 2 x o i ' ^ f °
*
[x2-10x-3 = 0
O X = 5-2N/7
Ket luan: Phuang trinh c6 2 nghiem la x = 1, x = 5 - 2\l7
Nhan xet:
thanh V s x + T - 2 + 2 - V x + 3 + 4 - 4 x = 0 thi sau khi lien hgp phuang trinh

moi thu dugc se la:
3x-3

Truong hgp 2: x < 0 Chia hai ve phuong trinh cho x ta thu dugc:
X^

Dat 3 / - - I = t = ^ - - l = t 3 = > t < - l .
X

Nhan xet: Trong mpt phuong trinh c6 chiia nhieu dau

,' _

+ Ta thay phuang trinh c6 n g h i f m x = l . Neu ta phan tich phuong trinh

nghiem duy nhat t = 1 <=> x = 1

V

Tom lai: Phuang trinh ban dau c6 2 nghiem la x = 0, x = 1

2x-2
.
+l-x =0o(2x-2)
' ,
- ± =0
V3x + l + V x + 3
^
W 3 x +l+Vx +3 2j

;7X4.2=33/1:;.J9-i.4.

'

%' '

gian nhat. Vi^c la nay se giiip cau true phuang trinh mod d a phuc t^p hon.

54

0.

De don gian ta dat ^J- - 1 = t

< 0 do do f(t) > f ( - l ) = 2

Suy ra phuong trinh v6 nghiem.

b) Dieu ki^n x e

X

Vt^+15.

phu la mot bieu thuc chua

35

Xct X = 0 thoa man phuong trinh.
Xet

.Vt^+8

X

Phuong trinh da cho tro thanh:
- V t " +15+2 = -Vt^ + 8+3tc^Vt^ + 8 - V * ' ^15-3t + 2 = 0

+

1-x
^7"

jj
+4-4x = 0 c ^ ( x - l )

.

^

+

L _ _ 4 = 0.
V3X + 1+2 2 + Vx + 3
\ V 3 x + l + 2 2 + Vx + 3
3
1
'
R6 rang phuong trinh h? qua ,
+

.
- 4 = 0 phuc t^p hon
V3X + 1+2 2 + VX + 3
phuong trinh ban dau rat nhieu.
+ De y rang khi x 1 thi VSx + l = Vx + 3 nen ta se lien hgp tr\rc tiep bieu
thuc: Tsx + l - Vx + 3 .

Tm V!r:N TifvHBINH THUA;\


2. Dat an p h y d y a vao tinh d i n g cap ciia p h u o n g trinh:
Ta t h u a n g gap p h u a n g trinh dang nay 6 cac dang bien the nhu:
+ a x ^ + b x + c = d^/px^+qx + r (1)

t

+ ax^ + bx + c = d-Jpx"* + qx^ + rx^ + ex + h (2)

.j;.,.^,

-r,,,

iq ,

+ AVax^ + bx + c + B^ex^ + gx + h = C^/rx^ + px + q (*)
•''
Thuc chat p h u o n g trinh (*) khi binh p h u o n g 2 ve thi xuat hi^n theo dang
(1) hoac (2).
De giai cac p h u o n g trinh (1), (2).
P h u o n g p h a p chung la:

+ Phan tich bieu thuc trong dau V~ thanh tich ciia 2 da thuc P(x),Q(x)
+ Ta bien doi ax^ + bx + c = mP(x) + nQ(x) bang each dong nhat hai ve.
Khi do p h u o n g trinh tro thanh: mP(x) + nQ(x) = d^/P(x)JQ(x)
Chia hai ve cho bieu thuc Q(x) > 0 ta thu dup'c p h u o n g trinh:
P(x) , „ ^ —
m — ^ + n = d J fPix)^ . Dat t = f P W > 0 thi thu dugc p h u o n g trinh:

Q(x)

VQ(X)

•

Wi^)

mt^ - dt + n = 0.
Mpt each tong quat:
Voi nriQi p h u o n g trinh c6 dang:
aP"(x) + bQ"(x) + C P " - ' ' ( X ) Q ' ' ( X ) + d^^?{x).Qi\) = 0 thi ta luon giai dugc
theo each tren.
Mpt so VI d\i:
Vi dy 1: Giai cac p h u o n g trinh:
a) 2(x^ - 3 x + 2) = 3\/x^ + 8
b) x + l + 7x^ - 4 x + l =3N/X
c) 4 x ^ + 3 ( x ^ - x j V ^ = 2 ( x ^ + l )

n=l
m = -l
m - 2n = - 3 <=> n = l
2m + 4n = 2

P huong trinh da cho c6 dang:
-2(x + 2) + 2(x^ - 2x + 4) - 3yJ{\ 2)(x'^ - 2x + 4) = 0 . Chia p h u o n g trinh cho
- 2x + 4 > 0 ta thu dugc: - 2

x+2

-3

(x + 2)
+ 2 =0
(x^ - 2 x + 4)

Dat t = — —
> 0 ta thu duoc p h u o n g trinh: -2t^ - 3t + 2 = 0
^ ( x ^ - 2 x + 4)
t = -2
(^ + 2)
1
2
^
4
'^S
1 d o t > 0 => t = 1t=2 y ( x 2 - 2 x + 4) 2
^
' = — <=> X - 2x + 4 = 4(x + 2).
X = 3 + N/I3
x' - 6 x - 4 = 0 o x = 3-Vl3
A,

0

fx>0
<=>
x>2 + S
b) Dieu kien: x^ - 4x + l > 0
Binh p h u o n g 2 ve'ciia p h u o n g trinh ta thu dugc:
<^
x^ + 2x + 1 + 2(x + l)7x^ - 4 x + l + x^ - 4 x +1 = 9x
c^2x-

- n x + 2 ^ 2 j ( x 2 + 2x + l)(x^ - 4 x + l) = 0

Gia su
m= —
m +n =2
2
2x^ -11x + 2 = m(x^ + 2x + l) + n(x^ - 4 x + l)=> 2m - 4n -11 <=>
5
m +n=2
n =—
2
P huong trinh tro thanh:
- i ( x ^ f2x + l) + | ( x ^ - 4 x + l) + 2j(x^ +2x + l)(x^ - 4 x + l) = 0

Giai:
a) Dieu kien: x > - 2 .
Ta Viet lai p h u o n g trinh thanh: 2(x^ - 3x + 2) = 37(x + 2)(x^ - 2 x + 4)
Gia su x^ - 3 x + 2 = m(x + 2) + n(x^ - 2 x + 4). Suy ra m , n phai t h o a m a n

Ji

Chia p h u o n g trinh cho x ' + 2x +1 > 0 ta thu dugc:
x^ - 4 x + l = 0 . Dat t - 'x^ - 4 x + l '
x^ - 4 x + l
> 0 ta CO
-1 + 5
^x^ + 2x + l ,
x^ + 2x + l^
^x2+2x + l ^
\
•


Cty TNHH MTV DWH

t=- l
-4x + l
1
1 <=> t = — »
t =5
x^ +2x + l
5

Phuong trinh 5t^ + 4t - 1 = 0

25

Giai:
a) Dieu kien x > -

^1

<=>24x^ -102x + 24 = 0<=>

Phuong trinh da cho dugc viet lai nhu sau:
2x^ - 8x2 + 1 0 x - 4 - 3 x ( x - 2 ) V 2 x - l = 0

x =4

o (X - 2)(2x2 - 4x + 2) - 3x(x - 2 ) V 2 x - l = 0

Ket luan: Phuong ^rinh c6 2 nghi^m -x = - , x = 4

c > ( x - 2 ) (2x2 _ 4 x + 2 ) - 3 x V 2 x - l = 0

Nhan xet: Trong lai giai ta da bien doi:

x-2 = 0

(x + l)Vx^ - 4 x + l = J(x^ +2x + l)(x^ - 4 x + l ) la vi X +1 >0

(2x2 _4x + 2 ) - 3 x V 2 x - l =0

c) Dieu ki?n: x > -1

Xet phuong trinh:

Ta Viet lai phuong trinh thanh: (x - 1 ) 2x^ - 2x - 2 - 3xVx + l = 0

2 x 2 - 4 x + 2 - 3 x V 2 x - l =0<=>2x2 - 4 x + 2 - 3 V x 2 ( 2 x - l ) = 0

x=l

Ta gia su: 2x2 - 4x + 2 = ^^^2 ^ ^^2x -1):

2x2 - 2 X - 2 - 3 X N / X + T = 0

De thay x = -1 khong phai la nghi^m.

Ta c6: 2 - 2 .

Xet X > -1 ta chia cho x +1 thi thu dugc phuong trinh:
X

2^^
'x + 1

3 - ^ ^ - 2 = 0<=>
Vx + 1

Giai (1):

-2C:>

Vx+T

=- l o

TxTT
x

-2

-4x-4 = 0

-4x-4 = 0

'2x-l'

-2t2 - 3 t + 2 = 0 < »

Voi t = i

o x = 2 + 2^/2

x<0

n = -2

2 1
J'—x--— = 0 . Dat t =

'2 x - l
/ — - — > 0 phuong trinh moi

la:

(1)

= -1 (2)

X>0

Jm = 2

Phuong trinh tro thanh: 2x2 _ 2(2x -1) - 3 ^ x 2 ( 2 x - l ) = 0 . Chia cho x^ > 0

Xet phuong trinh: 2x^ - 2 x - 2 - 3 x V x +1 = 0 < > 2x2 - 3xVx +1 - 2 ( x +1) = 0 .
=

Giai (2):

Khang Viet

t = -2
.=1
2

taco: . F ^ = i c : > x 2 - 8 x + 4 = 0 o

x=4+2V3
x = 4-2>/3

Nhan xet:
<»x = 2-2V2

+

chia nhu tren thi bai toan van dugc giai quyet. Vif c dua vao

x = l;x = 2±2V2

a) 4(2x2 +1) + 3(x2 - 2X)N/2X - 1 = 2(x^ + 5x)
b) Vsx^ +4x -N/X2 - 3 X - 1 8 ^5y/x
c) V5X2-14X + 9 - V X 2 - X - 2 0 = 5 N / X T I

^ 2 - 3 x V 2 x - l =0 ta c6 the khong can dua

X vao trong dau V ~ khi do ta phan tich: 2x2 _ 4^ + 2 = mx2 + n(2x -1) va

Ket hg-p dieu ki^n ta suy ra cac nghi^m ciia phuong trinh la:

Vi dvi 2: Giai cac phuong trinh:

D61 voi phuong trinh 2x2 _

la giiip cac

em hgc sinh nhin ro hon ban chat bai toan.
+

Ngoai ra can iuu y rang: Khi dua mgt bieu thuc P(x) vao trong dau
thi dieu kien la P(x) > 0 . Day la mgt sai lam hgc sinh thuong mac phai khi
giai toan.


T n i lieu on thi iliii hoc •.mix tao vii giai PT, hat PT, he PT, boT DT- Nguyen

TrungKien

-3x-18>0
b) D i e u ki^n:

x>0

Cty TNHH MTV DWH Khang Viet

T o m lai: P h u o n g t r i n h c6 2 n g h i f m la: x = ^

va x = 9

<=>x>6.
c) D i e u ki?n x > 5 .

5x^ + 4 x > 0

C h u y e n ve b i n h p h u o n g ta dugc: 2x2 - 5x + 2 = S^^x^ - x - 20J(x + 1 )
P h u o n g t r i n h da cho dvtqc viet lai thanh: Vsx^ + 4x = Vx^ - 3x - 1 8 + 5^/x
B i n h p h u o n g 2 v e v a t h u gpn ta dugc: 2x^ - 9x + 9 - 5yjx{x^ - 3 x - 1 8 ) = 0

Gia sir: 2x2 _ 5 x + 2 = m | x 2 _ x - 2 0 J + n ( x + 1 )
m-2

N e u ta gia s u 2 x ^ - 9 x + 9 = mx + n(x^ - 3 x - 1 8 ) t h i m , n phai thoa m a n
n =2

K h i do ta

CO :

'

- m + n = -5 k h o n g ton tai m , n thoa m a n h$.
-20m + n = 2

m - 3n = - 9 dieu nay la hoan toan v 6 l y .
-18n = 9

N h u n g ta c6 : (x^ - x - 20J(x +1) - (x + 4 ) ( x - 5 ) ( x + 1 ) = (x + 4)(x2 - 4x - 5

De khac phuc van de nay ta c6 chii y sau : x^ - 3x - 1 8 = (x - 6)(x + 3 k h i do
m =2
A / X ( X ^ - 3X -18)

= 7x(x

- 6)(x

+ 3)

= 7(x^

- 6x)(x

+

Gia sir: 2x2 - 5x + 2 = a^x^ - 4x - sj + p(x + 4 ) . c,^^y

3)

,m = 2
^

^

3^.

m =2

- 6 m + n = - 9 <=>

Ta viet lai p h u o n g t r i n h : 2^x2 - 4x - sj + 3 ( x + 4) = 57(x2 - 4 x - 5 ) ( x + 4) .

n =3

n =3

Chia hai v e c h o x + 4 > 0 ta t h u duoc: 2

N h u vay p h u o n g t r i n h t r o thanh:
2{x^ - 6x) + 3(x + 3) - 5^{x^ - 6x)(x + 3) = 0
Dat t =
Chia cho x + 3 > 0 ta t h u dugc: 2

x^ - 6 x
x+3

Dat t = ,

x^ - 6 x
x+3

x^-6x

-5

x2-4x-5^

f

x+ 4

> 0 ta thu d u o c p h u o n g t r i n h : 2t2 - 5t + 3 = 0 <=>

= 1 o x 2 - 5 x - 9 = 0 <=>
x+4

5 + 761
x=•
2
5-761

X = •

2
X =•

2

= l » x ^ - 7 x - 3 = 0c^

7-761

3
T r u o n g hgp 2: t = - <=>
^
2

X

-4x-5
x+4

X = •

x=8
9
7
= - o 4x^ - 25x - 56 = 0 <=> _ _ 7
4
^~ 4

Ket hgp dieu kien ta suy ra cac n g h i ^ m ciia p h u o n g t r i n h la:
c
7 + v'6T
.
_
,
b u y ra X = —
thoa m a n dieu k i ^ n .

X =

T r u o n g h o p 2:

8;

X =

5 + 761

V i du 3: Giai cac phuong trinh:

x^ - 6 x
x+3

x=9
= -<:>4x2-33x-27 = 0 ^
2

J

x+4

T r u o n g h ( ? p l : t = l<=>^^

7W6T

2

+ 3= 0

t=l

2

x+3

x+ 4

_Jrx2-4x-5]

2

> 0 = > 2 t ^ - 5 t + 3 = 0<:^

T r u o n g h o p 1: t = 1 <=>

x2-4x-5

+3=0

x+3
t=l

x^-6x

n =3

- 5 m + 4n = 2

Bay g i o ta viet lai p h u o n g t r i n h thanh: 2x^ - 9x + 9 - 5yj(x^ - 6 x ) ( x + 3) = 0

Gia sir: 2x2 - 9x + 9 = ^^^^2 _

m =2

-4m + n = -5:

_

a) 7x2 + 2 x + 7 2 x - l = 7 3 x 2 + 4 ^ + 1
3:r>x = 9
b) x ^ - 3 x 2 + 2 ^ / ( x + 2 ) ^ - 6 x = 0


Giki:
a) Dieu ki?n:

l _ 3 ( ^ . 2 i ^

—•

Binh phuong 2 ve phuong trinh ta thu dirge:

= 0.

Dat t = ^ ^ ^ ^ ta CO phuong trinh: 2t^ - 31^ +1 = 0 »

+ 4x - 1 + 27(x^ +2x)(2x-l) = Sx^ + 4x +1 o x^ +1 - ^(x^ +2x)(2x-l) = 0 Ta
m =l
gia sit: x^ +1 = m(x^ + 2x) + n(2x -1) o < n = - l

<>
=

2m + 2n = 0

(x^ + 2x) - (2x -1) - ^(x^ + 2x)(2x -1) = 0

Dat t =

2x-l
Ix^ +2x

Truong hgp 1: * =

m =l
n =- l

-

,x^ +2xJ

J

i

( 2x-l

1+ 1 = 0

<>

=

viec tinh toan se gap kho khan.

<=>

x>0
^
I
= !<:=> \/x + 2= x<=> x^ - x - 2 = 0 «>x = 2

Vx + 2

V i du 4: Giai cat phuong trinh:
a)

De khac phuc ta c6 the xu ly theo huong khac nhu sau:
Ta viet lai: ^(x^ + 2 x ) ( 2 x - l ) = 7(x + 2)(2x^ - x ) liic nay bang each phan tich
nhu tren ta thu dugc phuong trinh:

b) 5 N / X ' ' + 8 X = 4 X 2 + 8

2 X ^ - X 2 - 3 X + 1 = N/X^ + X ' * + 1

Giai:
a) Hinh thuc bai toan de lam cho nguoi giai bo'i roi nhung de y that ky ta thay:
Chia khoa bai toan nam 6 van de phan tich bieu thuc: x^ + x"* +1

2 . p ^ ^+1=0
V x+2

Dat t = J—
^ > 0 = > t 2 - 2 t + l = 0 < » t = lci>2x2-x = x + 2<=>x2-x-l = 0
V x+2
. Kiem tra dieu ki?n ta thay chi c6 gia tri x = ^ ' ^ ^
2
2
man dieu ki^n.
<=> X =

la thoa

b) Dieu kien: x > -2 .

Ta thay do ve trai la bieu thuc bac 3 nen ta nghi den huong phan rich:
x^ + x"* + 1 = (x^ + ax + l)(x'' + bx^ + cx +1). Dong nhat hai ve ta thu dugc:
a = 1; b = 0; c = - 1 . Nen ta viet lai phuong trinh da cho thanh:
2(x^ - X + 1) - (x^ + X + 1) - ^(x^ - X + ]).(x2 + X +1) = 0

2.

x^-x + 1

=0

Neu ta dat y = Vx + 2 thi phuong trinh tro thanh: x^ - 3xy^ + 2y'^ = 0 . Day
la mot phuong trinh d i n g cap bac 3. T u djnh huong tren ta c6 loi giai cho

1

•m<;

Chia cho x^ + x +1 > 0 ta thu dugc:

X^ + X +

De y rang:

bai toan nhu sau:

—

Ke't luan: Phuong trinh c6 2 nghiem: x = 2;x = 2 - 2^3

Ve CO ban den day ta hoan toan tim dugc x. Nhung voi gia tri t nhu vay

Ta viet lai phuong trinh thanh: x^ - 3x(x + 2) + lyjix + lf

=

X
2
Truong hgp 1: t = 1

>0=>-t^-t +l =O o t =

1
1(2x2 - X) + i ( x + 2) - J(x + 2)(2x2-x) = 0 « ^ ^ ^ ^ 2^
' 2^
'

'
x+2

x<0
„ I
r
<=>x = 2-2^3
2\lx + 2 = - x <=> x^ - 4 x - 8 = 0

1

Jx + 2
<r>

Phuong trinh tro thanh:

t= - i
2
t=l

x-' - x + 1

1

X^ + X +

X +

1

X^ + X +

Dat t =

x-" -

1

X - X

+

1

- 1 = 0.

> 0 ta CO phuong trinh: 2t - 1 - 1 = 0 <=>
x=0
1 ,
-i
2 ^
x=- l
= 1 <=> x-^ - x^ - 2x = 0 «
x=2
1

t=l
t = -f(L)

+ Xet truong hop: x = 0 khong thoa man phuong trinh:

Giai t = 1 <=> —
x^

+ Xet X 0. Ta chia phuong trinh cho x'^ thi thu dugc:

Ke't luan: Thu lai ta thay 3 nghiem: x = 0,x = - l ; x = 2 deu thoa man.

+X +

1!

V


ml im on tm a^t

sang t4d

H^J

b) D i e u ki?n:

+ 8x > 0

tagtat

»

X

m.

bat l^l,

Vi,

uai

trr^

Nguyen

Cty TNHH

TnmgKien

+ 4x2 + 4 - (4^2 _ 8x + 4) = (x^ + if

- (2x -

<=> (a - 2b)(a - b)(a + b) - (a - b)(a - 2b) = 0 c:> (a - 2b)(a - b)(a + b - 1 ) = 0

if

m +n =4

N/2X + 3 - Vx + 1 = 0

• '

Ta q u y bai toan ve giai 3 p h u o n g t r i n h co ban la: 2N/2X + 3 -N/X + 1 = 0

- 2 m + 2n = 0<=>m = n = 2

Gia s u 4x2 + 8 = ^ ^ ^ 2 _ 2x + 4) + n{x^ + 2x)

N/2X + 3 + V X + 1 - 1 = 0

4m = 8

V o i dieu kien: X > - 1 => a > l , b > 0

P h u o n g t r i n h tro thanh:
2(x2 - 2x + 4) + 2(x2 + 2x) - 5^{x^ - 2x + 4)(x2 + 2x) = 0 . Chia
x^+2x
-2x + 4

-5

hai

ve

cho

- '

T r u o n g hop 1: N/2X + 3 - Vx + 1 = 0<=>2x + 3 = x + l<=>x = -2(L)
T r u o n g hop 2: lyJlx

x^ + 2 x

+ 3 - Vx + 1 = 0<=:>8x + 12 = x + l<=>x = - y

T r u o n g hop 3: N/2X + 3 + %/x7T - 1 = 0 . V i N/2X + 3 > 1,N/X + 1 > 0 => VT > O
t= 2

x^ + 2 x
^ ' ^" = 4 o
X -2x + 4

«

Dau bang xay ra k h i va chi k h i x = - I .

,.1

T o m lai p h u o n g t r i n h c6 n g h i f m d u y nha't x = - 1
b) Dieu kien x > - 2

2

Ta thay rang ne'u b i n h p h u o n g true tie'p se dan den p h u o n g t r i n h bac 5

Sx^ - lOx +16 = 0 v 6 n g h i ^ m

De khSc phuc ta se t i m each tach x2 + 4 ra k h o i V2x + 4
-5-V37

T r u o n g h g p 2: t =

i « - i ^ - t ^ =lo3x2+10x-4
-2x + 4

X=•

= 0<^

4

-5
x=-

Ket luan: P h u o n g t r i n h c6 hai n g h i ^ m la:

+ V37

+8x = J x

X

« - (x2 + 4)(V2x + 4 + 1) = 2x(2x + 3) o (x2 + 4)(V2x + 4 + 1) =

x2 + 4 = 2x(V2x + 4 - 1 )

o

X =

V2x + 4

V i dv 5: G i a i cac p h u o n g t r i n h :
a) (x + 2)(V2x + 3 - 2Vx + 1 ) + ^2x2 + 5x + 3 - 1 = 0
b) (x^ + 4 ) V 2 x + 4 =3x2 + 6 x - 4

chung.

o

x2 + 2x + 4 - 2xV2x + 4 = 0 o

x>0

x^ - 2 x - 4 = 0

Chu y r5ng: T r o n g m p t so p h u o n g t r i n h : Ta can dua vao tinh ddng cap cua
detit do phan tich tao thanh nhan tu

- 1)

(x2 + 4)(V2x + 4 + 1) = 2x(V2x + 4 + l)(V2x + 4 - 1 )

o

^ + 8) = 7x(x + 2)(x2 - 2x + 4) = 7(x2 + 2x)(x2 - 2x + 4)

timg nhom so'hang

2x(7(2x + 4)2

D o 72x + 4 + 1 > 0 . P h u o n g t r i n h da cho t u o n g d u o n g v o l

N h a n xet: Ta c6 the phan tich:
X

T u do ta viet lai p h u o n g t r i n h n h u sau: (x2 + 4)\/2x + 4 + x2 + 4 = 4x2 ^

o

3

X =•

i

tj •

3

-5->/37

X = •

(L)

+ 2 = 0.

Vx2-2x + 4

^ ^"^^
> 0 ta CO p h u o n g t r i n h : 2 t 2 - 5 t + 2 = 0
x^ - 2 x + 4

T r u o n g h(?p 1: t = 2 o

Vift

P h u o n g t r i n h da cho t r o thanh: (a2 - b2 )(a - 2b) - (a2 - ab - 2b2) = 0

= ( x 2 - 2 x + 4)(x2+2x)

Dat t =

Kltattg

a) Dat N/2X + 3 = a , V x + l = b = > a , b > 0

x<-2

- 2 x + 4 > 0 ta t h u dug-c: 2-

DWH

Giii:

>0

Ta thay chia khoa bai toan nSm a v i f c phan tich bieu thuc:

+ 8x =

MTV

(x - V2x + 4 )

=0

o x = l + V5

Ket luan: P h u o n g t r i n h c6 n g h i ^ m d u y nha't x = 1 + N/S
c) D i e u ki?n: x > 2
fm = l
Gia sir x2 - 6 x + l l = m(x2 - x + l ) + n ( x - 2 )

- m + n = - 6 <=> m = 1, n = - 5
m-2n = l l


Tai lieu oil thi dai hoc sang tao va giai PT, bat PT, hf PT, boT DT - NguyettTrung

Kien

+ De giai cac phuong trinh dang nay ta thuong lam theo each:

p= l
- p + q = -4 <=> p = 1, q = -3
p-2q = 7

x2 - 4 x + 7 = p ( x 2 - x + l ) + q ( x - 2 ) :

-

1-2 (x^ - x + l ) - 3 ( x - 2 )

N/X^

=

0

1-5.D3t t =

x^-x + l

^

I

• -21—

x-2

Vx^-x + l

+ 6,

x-2

V ,x - x + 1

A(x)]^ thi dieu ki§n can va dii la A^j, = [ g i ( m ) ] ^ -4£j(m).gj(m) = 01=> m
Ta xet cac vi du sau:

Chia phuong trinh cho ^(x^ - x +1)^ ta thu duQc:
x-2

Ta t^o ra phuong trinh: mt^+g(x)t + h(x) = 0

Ta CO A = [ g ( x ) ] ^ - 4 m . h ( x ) = fj(m)x^+gj(m)x + h j ( m ) . De A eo dang

Phuong trinh da cho tro thanh:
(x^ - x + l ) - 5 ( x - 2 ) V x ^ - x +

Dat 7f(x) = t => t^ = f(x)

Vi
=0

1: Giai cac phuong trinh:

a) x ^ + l - ( x + l)\/x^-2x + 3 = 0

= 79x^+16

Giai:

^ >0
Vx^-x + l

a) Dat t = V x ^ - 2 x + 3 > 0 = > t ^ - x 2 - 2 x + 3
t= l

Ta thu duoc phuong trinh: 6t^ - 5t^ - 2t +1 = 0 <=> t = l
3

1
t= ~(L)

Phuong trinh da cho tro thanh: x^ +1 - (x + l)t = 0

''

Ta se tgo ra phuong trinh: mt^ - (x + l)t + x^ +1 - m(x^ - 2x + 3) = 0
(Ta da them vao mt^ nen phai bot di mpt lupng mt^ = m(x^ - 2x + 3))
Phuong trinh dupe vie't lai nhu sau:
mt^ - (x + l ) t + (1 - m)x^ + 2mx +1 - 3m = 0

+ Neu t = l « a = b o x 2 - 2 x + 3 = 0(VN)

A = (x +1)^ - 4m (1 - m)x^ + 2mx +1 - 3m

+ Neu t = -<=>x^-10X + 19 = 0 < » X = 5±N/6
3

= (4m^ - 4m + l)x2 + (2 - Sm^)x + 12m2 - 4m +1
Ta mong muon

Ketluan: X = 5±N/6
2. Giai phuong trinh v6 ty bang phuong phap dat an phy khong hoan toan.

+ Dat an phu khong hoan toan la phuong phap chpn mpt so hang trong
phuong trinh de dat lam an sau do ta quy phuong trinh ban dau ve dang
mpt phuong trinh bac 2: mt^ +g(x)t + h(x) = 0 (phuong trinh nay van con
an x )
+ Van de ciia bai toan la phai chpn gia trj m bang bao nhieu de phuong trinh
bac 2 theo an t c6 gia trj A chan

b) 2V2x + 4 + 4 ^ 2 ^

A = A(x)

nhu the viec tinh t theo x

se dupe de dang.
+ Thong thuong khi gap cac phuong trinh dang:
ax^ + bx + c + (dx + e)^jp\^ + qx + r = 0 hay

A = (Ax + B)2 o A ^ = (1 -4m2)2 - { U m ^ - 4 m + l){4m^ - 4 m + l ) = 0c:>m = l
Phuong trinh moi dupe tao ra la: t^ - (x + l)t + 2x - 2 = 0
Ta

CO

A = x^ - 6x + 9 = (x - 3)^

t_x + l-(x-3)_,
2
Tu do ta c6:
t . i i ± i ± i ^ = x-l
+ Truonghppl: t = l o > / x ^ - 2 x + 3 - 2<=>x^-2x-l = 0 o x = l ± r y 2

^ ,
fx>i
+ Truong hpp 2: t = x - 1 o Vx -2x + 3 = x - l<t:>
-2x + 3 = x^-2x + l

ax^ + bx + c + (dx + e)^px + q = 0 thi phuong phap dat an phu khong hoan

Phuong trinh v6 nghiem.

toan to ra rat hi^u qua:

Tom lai: Phuong trinh c6 2 nghiem la: x = 1 ± %/2


Dat t = Vl - x2 ta tao ra p h u o n g t r i n h :

b) D i e u k i # n : - 2 < x < 2
B i n h p h u o n g 2 ve p h u o n g t r i n h va t h u gpn ta dugc:

mt2 + (6x + 18)t + (8 + m)x2 - 6x - 18 - m = 0

_ 1 6 7 8 - 2 x 2 + 8x - 32 = 0 .
Dat t = yjs-lx^

Co A' = (3x + 9)^ - m (8 + m)x'^ - 6 x - 1 8 - m

ta tao ra p h u o n g t r i n h la:

A = ( A x + B)2

+8x-8m-32

Ta

CO

A.;^ =16m2 - ( - 2 m 2 - 9 m ) ( 8 m 2 + 32m + 64) = 0 o m = - 4

+

t=

+ ^4 = (2x + 8)^

/

9~

X

x>0

Suv ra

8 - ( 2 x + 8) _ x

-4
~2
8 + (2x + 8)
x

t=
-=
-4
2

,
4

4 o

4 ( 8 - 2 x 2 ) = (x + 8)2

T o m lai p h u o n g t r i n h c6 n g h i e m d u y nha't x =

VN

4V2

+ Vl - x2

Giai:
a) D i e u kien: - 1 < x < 1 . Ta vie't p h u o n g t r i n h t h a n h :
- X =

3x + 1 + \ / l - x2 .

B i n h p h u o n g 2 ve ta t h u d u o c p h u o n g t r i n h m o i :
16(x +1) + 4(1 - X ) - 1 6 \ / l - x 2 = 9x2 + 6x + 1 + 2(3^ + l ) 7 l - x2 '+1 - x2
»

o

4 ^ 1 - x 2 = 3x + 5 « 16(1 - x2) = 9x2 +

^ 25

3
Ket l u a n : P h u o n g t r i n h c6 2 n g h i e m x = 0, x = —
1
5
Chii y : O b u o c cuoi cung k h i giai ra n g h i e m ta phai t h u lai v i phep b i n h
p h u o n g liic dau k h i ta giai la k h o n g t u o n g d u o n g .

b) 2x2 ^ 7 x ^ 1 0 - ( 3 x + 2)(2Vx72 - V2x + 5 ) - 4 7 2 x 2 + 7x + 1 0 = 0

4Vx + 1 - 2V1

T r u o n g h o p 2: t =

16(1 - x 2 ) = 9x2 + 3 0 x + 2 5 0 2 5 x 2 + 30x + 9 = 0 o x = - -

V i d\ 2: G i a i cac phucmg trinh:

a) 47x + 1 - 1 = 3x + IsfT^

T r u o n g h g p 1: t = 1 < > Vl - x2 = 1 <=> x = 0 thoa m a n d i e u k i ^ n
=

T h u lai ta thay: ^ = ~ ~ thoa m a n p h u o n g t r i n h :

x<-8

2

4

-8
t - - 3 x - 9 + (3x + l ) _ ^

4V2

="^<=>" 4 ( 8 - 2 x 2 ) = x

T r u o n g h o p 2:
t =

A' = (3x + 9f - 8(6x + 10) = (3x + 1)2
^_ - 3 x - 9 - ( 3 x + l ) _ 3x + 5

T u d o suy ra p h u o n g t r i n h m o i la: -4t2 - 16t + x2 + 8x = 0

X

A ' ^ = ( 3 m + 27)2 - ( 9 - 8 m - m 2 ) ( m 2 + 1 8 m + 81) = 0

P h u o n g t r i n h da cho t r o thanh: -8t2 + (6x + 18)t - 6x - 1 0 = 0

Ta m o n g m u o n A' = ( A x + B)2 < > A = 0 phai c6 n g h i e m kep . Tuc la:
=

T r u o n g h o p 1: t = —<=> v 8 - 2 x

'' ^

T u do tinh dugc m = - 8

= ( - 2 m 2 - 9m)x2 + 8mx + 8r
8m2 + 32m + 64

T i n h dugc: A' = 4x2 +

v

Ta m o n g m u o n

mt2 - 16t + (9 + 2m)x2 + 8x - 8 m - 32 = 0

A' = 64 - m (9 + 2m)x2

5 !

= ( 9 - 8 m - m 2 ) x 2 + ( 5 4 + 6m)x + m 2 + 18m + 81

mt^ - 1 6 t - m ( 8 - 2 x 2 ) + 9x2 + 8 x - 3 2 = 0
o

v£

8x2 _ 6x - 1 8 + (6x + 1 8 ) \ / l - x 2 = 0

T u do ta can l u u y ; Khi gidi mot phuong

trinh

md cdc phep dat dieu

phuc tap ta c6 the bo qua buoc nay nhung khi gidi xong phuang
phdi thu lai vdo phuong trinh ban ddu detim

nghiem chinh xdc.

b) D i e u ki?n: x > - 2 .

^

Dat t = 2 ^ x 7 2 - V2x + 5 t h i t2 = 6 x + 13 - 4^2x2 + 7x +10
P h u o n g t r i n h da cho t r o thanh: t2 - (3x + 2)t + 2x2 + x - 3 = 0
T a c o A = 9x2 + 12x + 4 - 8 x 2 - 4 x +12 = (x + 4)2

t=2x+3
t= x - l

T r u o n g Rgp 1: t = 2x + 3 <=> Isjx + l - sJlx + B = 2x + 3 .

trinh

kien
ta

De y r3ng: 2x + 3 = 4(x + 2) - ( 2 x + 5) = (2Vx + 2 - V 2 X + 5)(2N/X + 2 + 7 2 x + 5J
N e n ta c6:

D | t t = 72x^ - 3x + 1 ta tao ra p h u o n g t r i n h :
m t ^ - 8xt + (10 - 2m)x^ + (3m - 9)x + 3 - m = 0

(2VX + 2 - V2x + 5 ) ( 2 V x + 2 + V2X + 5 ) = (2VX + 2 - V2X + 5 )
2Vx + 2 - V2x + 5 = 0

Ta CO A = 1 6 x ^ - m (10 - 2m)x^ + (3m - 9)x + 3 - m

• T
= 16x2 _ ^ ( 1 0 - 2 m ) x 2 + ( 3 m - 9 ) x + 3 - m

2Vx + 2 + V2x + 5 = 1
2Vx + 2 - V 2 x + 5 = 0
2Vx + 2 + V2x + 5 = 1

3
—
2
x = -2

= ( 2 m 2 - 1 0 m + 16)x2 + ( 9 m - 3 m 2 ) x + m ^ - 3 m

X=

Ta can :

T n r o n g hgrp 2 : t = x - 1 <=> 2\fx + 2 - V2x + 5 = x - 1

(2Vx + 2 - V2x + 5)(2Vx + 2 + V2x + 5) = (x - l){2yfx + 2 + V2x + 5 )

= ( 9 m - 3m^f

- 4(2m2 - 10m + 16){m^ - 3 m ) = 0 => m = 3

P h u o n g t r i n h da cho t r o thanh: 3t^ - 8xt + 4x2 = 0 <=>
t = 2x

<» (2x + 3) = X - l ) ( 2 V x + 2 + V2x + 5)

T r u o n g h o p 1:

De thay x = 1 k h o n g p h a i la n g h i ^ m nen p h u o n g t r i n h t u o n g d u o n g v a i :

x>0
2
I
2
t = - x c t . V 2 x 2 - 3 x + l = - x < = > - 9 ( 2 x 2 - 3 x + l ) = 4x2
3
3
3
x =—
2
o
3
X= —
7

^ ^ ^
x-1

= 2 V ^ + V2x + 5 < = > 2 V ^ + V2x + 5 - ^ ^ = 0
x-1

N h a n thay tren m o i khoang ( - 2 ; ! ) va (l;+oo)
h a m so f ( x ) = 2\/x + 2 + -Jlx + S _ x - 1 ^

lien tyc va c6

1
1
f(x) = .
> 0 nen h a m so' d o n g bien. Suy ra tren m o i
• +yJ2x= = = + - {x-lf
— +5
s/x + 2
khoang d o p h u o n g t r i n h c6 nhieu nhat m o t n g h i ? m .
Xet tren khoang ( - 2 ; ! ) ta c6 f ( - 2 ) > 0 nen p h u o n g t r i n h v 6 n g h i f m
Xet tren k h o a n g (l;+oo) c6 f(2) = 0 n e n p h u o n g t r i n h c6 n g h i ^ m d u y nhat
x =2
T o m lai p h u o n g t r i n h da cho c6 d u n g 3 n g h i f m : x =
Vi
a)

3

lOx^ - 9 x - 8 x V 2 x 2 - 3 x + 1 + 3 = 0

3
3
1
Ke't l u a n : P h u o n g t r i n h c6 3 n g h i f m : ' ' ^ ^ ' ' ^ ^ y ' ' ^ " ^
b) D i e u k i e n : x > 1 .

2,x = - 2

Dat t = 7 x ^ + 3 > 0 <=> x^ = t2 - 3 . D o h? so cua x^ t r o n g p h u o n g t r i n h la: 1

A = (5x-1)2 -4(6x2 -2x) = x 2 - 2 x + l = ( x - l ) 2 .
^_(5x-l)-(x-l)_

b) x^ + 6 x ^ - 2 x + 3 - ( 5 x - l ) \ / x ^ + 3 = 0
Giai:
x>l
a) D i e u k i f n:

x . l
2

14x2 - 27x + 9 = 0

T r u o n g h o p 2: t = 2x < » ^j2x^ -3x + l =2x<=>i'^^^
<=>x = [-3x + l = 0
3

P h u o n g t r i n h da cho t r o thanh: t2 - (5x - l ) t + 6x2 - 2x = 0
=

3: G i a i cac p h u o n g trinh:

x>0

Suy ra:
2

2x


b) Ta viet 1 ^ phuang trinh thanh: 3 - (2x2 _^

x=l
Tmong hgp 1:

+ 3 = 2x •

x>0
x ^ - 4 x ^ + 3 = o'

Ta coi day la phuang trinh bac 2 cua Vs ta c6:

3 + 721
2
3-^/2T
(L)
x=X=•

A = ( 2 x 2 + 1 ) 2 - 4 ( x + x^) = 4 x 2 - 4 x + l = (2x + l)2
> / 3 = - ( 2 x 2 + l + 2 x - l ) = x2 + x

Tir do suy ra

x= l
Truong hg-p 2: 7 x ^ + 3 = 3x - 1 <=>

X

x^ -9x^ +6x + 2 = 0
Tom lai phuong trinh c6 3 nghi^m: x = 1, x =

3+ >^

V3=-i(2x2+l-2x +l)= x2-x + l

= 4 + 2N/3

V i dv 4: Giai cac phuong trinh:
a) V s - x =x^ - 5

. \

«

x2 + X - Vs = 0
x2 - x + l - V s = 0

Giai 2 phuang trinh tr§n ta thu dugc cac nghi?m ciia phuong trinh da cho

X = 4-2N/3(L)

,x = 4 + 3V2

+ x + x'* = 0

- l ± V l + 4V3 . „
-1±V4N/3-3
^
hoac x =
^
2
2
c) Dieu ki?n x > - 4
Ta viet lai phuang trinh thanh: x + 4 + (4x2 ^^_2|7x + 4 + 8x2 + 2 x - 8 = 0.
la: x =

b) X ' ' - 2 N / 3 X 2 + X + 3->/3=0

Coi day la phuang trinh bac 2 an Vx + 4 thi

c) 8X2+3X + ( 4 X 2 + X - 2 ) N / X + 4 = 4

A = (4x2+x-2)
Gidi:

a) Dieu ki^n:

Tu do suy ra

x<5
x^ >5

^
- 4 ( 8 x 2 + 2 x - 8 j = 4x2 _ ^
-4-

Vx + 4 = -2x
Vx + 4 =2x + l

Giai 2 truong hgp ta thu dugc cac nghi^m cua phuang trinh la:

Binh phuong 2 ve ta thu dug-c: 5^ - (2x^ +1).5 + x + x"* = 0

1-V65
x=8

Ta coi day la phuong trinh bac 2 cua 5 ta c6:

-3 + V57
x=8

A = (2x2 +1)2 - 4(x + X * ) = 4x2 - 4x.+1 = (2x +1)2
5 = i ( 2 x 2 + l + 2 x - l ) = x2 + x

Vi d\ 5: Giai cac phuong trinh:

5 = -i(2x2+l-2x +l)= x2-x + l

a) 3(V2x2 + 1 -1) = x ( l + 3x + 8V2x2 +1)

T u do suy ra

X =•

Truonghgp 1: x 2 + x - 5 = 0<=>

-1-N/2T

Truong hgp 2: x - x - 4

= 0o

Vx2 +3x + 6 + 7 2 x 2 _ i =3x + i
Giii:

~*"2

x =-

X =-

b)

a) Ta viet lai phuang trinh thanh: 3x2 + x + 3 + (8x - 3 ) ^ 2 x 2 + 1 = 0 .
2

D|t t = V 2 x 2 + l > 0 suyra

'

t^=2x^+\.

Ta tao ra phuang trinh: mt2 + (8x - 3)t + (3 - 2m)x2 + x + 3 - m = 0 .
2

X =•

Doi chieu voi dieu ki^n ta c6 4 nghi^m deu thoa man phuong trinh.

Ta CO A = ( 8 x - 3 ) 2 - 4 m ( 3 - 2 m ) x 2 + x + 3 - m
= (8m2 - 12m + 64)x2 - (48 + 4m)x + 4m2 - 12m + 9.


Ta can A'

= (24 + 2mf

- (8m^ - 12m + 64)(4m2 - 1 2 m + 9 ) = 0 = > m = 3 .

E)6'i chie'u v o l dieu k i ^ n ban dau ta tha'y chi c6 x = -^''"^TlS

P h u o n g t r i n h t r o thanh: 3t^ + (8x - 3)t - 3x^ + x = 0 .
Ta c6: A = (8x - 3)^ - \2.{-3x^ + x) = lOOx^ - 60x + 9 = (lOx - 3)^ .

1

6

T r u o n g h o p 2: ^2x2 - 1 =

^_ 3 - 8 x + ( 1 0 x - 3 ) _

6

<=>

3

1

T r u o n g hgip 1: V2x^ + 1 = - 3 x + 1 <=> x < 3
9x2 _ 6 x = 0
x<0
T r u o n g hp-p 2; V2x^ +1 =-—<::>
17x2
3

2x-l

X

~

^g^Q

<::>X =

man

dieu kien .

.*

^^3-8x-(10x-3)_3^^^
Tir d o t i n h dvtqc :

^j^^^

x.l

X

2
o
4x2 + 4 x - 5 = 0

Do'i chieu v o i dieu k i ^ n ban dau ta thay chi c6 x =

0

=•

1

X = •

^

la thoa m a n dieu

ki#n .

VN

Vay p h u o n g t r m h co 2 n g h i ^ m la: x =

va x = —

Vay p h u o n g t r i n h c6 n g h i f m d a y nhat: x = 0
b) D i e u k i ^ n : x >

1

PHLTONG P H A P HAM

Ta viet lai p h u o n g t r i n h thanh: N / X 2 + 3 X

+

6=3X

+

Dau hi?u:

1-\/2X2-1.

+

Binh p h u o n g 2 ve'va thu gon ta dugc p h u o n g t r i n h m o i :
lOx^ + 3 x - 6 - 2 ( 3 x +1)72x2
Dat t = \/2x2 - 1 > 0 suy ra

S6

Bai toan p h u o n g t r i n h giai bang p h u o n g phap h a m so' t h u o n g c6 dac d i e m
la l u o n d u a ve d u g c dang: f u(x) = f v(x)

-1=0

trong d o h a m so' dac t r u n g

t h u o n g la ham d o n d i e u tang hoac d o n d i ^ u g i a m tren m i e n xac d i n h D

=2x2-1.

+

Dac diem noi bat nhat ta c6 the de phat h i f n la: Trong p h u o n g t r i n h c6 nhieu

Ta tao ra p h u o n g t r i n h : mt2 - 2(3x + l ) t + (10 - 2m)x2 + 3x - 6 + m = 0 . Ta c6

bieu thuc chua can, hoac da thuc bac cao ma ta k hong the q u y ve m p t an.

A' = (3x + 1)2 - m ( 1 0 - 2 m ) x 2 + 3 x - 6 + m

Ta t h u o n g giai cac p h u o n g t r i n h dang nay theo each:
Cach 1 :

= (2m2 - 1 0 m + 9 ) x 2 + (6 - 3m)x - m2 + 6 m + 1 .

+

D u a p h u o n g t r i n h ve dang f(x) = 0 v o i x 6 D

+

T a c a n A ^ = ( 6 - 3 m ) 2 - 4 ( 2 m 2 - 1 0 m + 9)(-m2 + 6 m + l ) = 0 = > m = 4 .

Xet ham so' y = f(x) tren D .

P h u o n g t r i n h t r o thanh: 4t2 - 2(3x + l ) t + 2x2 + 3x - 2 = 0
Ta c6: A' = (3x + 1)2 - 4.(2x2 + 3x - 2) = x^

Tir do t i n h dugc:

C h u n g m i n h f'(x) > OVx € D hoac f "(x) < OVx € D

6x + 9 = (x - 3)2 .

+

3x + l - ( x - 3 )
x+ 2
t=4
2
^_3x + l + ( x - 3 ) ^ 2 x - l

T r u o n g h g p 1: V2x2 - 1 =

<^

n g h i e m d o la d u y nhat.
*

T a xet cac vi dvi sau:
V i dy 1 : G i a i cac p h u c m g t r i n h sau:

2 +

x>-2
7x2-4x-8 = 0

N h a m m o t n g h i f m x = Xg . D u a vao t i n h chat ciia h a m so' d o n di?u ta suy ra

X =•

<=>

a) V x - l + x 2 - 7 + ^ x + 6 = 0

2V15
7

b)

^(x-l)2 - 2 ^ x - l - ( X - 5 ) N / X - 8 - 3 X

2-27T5 '

+ 31 = 0

Giai:

X = •

a) D i e u k i ^ n x > 1 .
^

Ta tha'y x = 1 k h o n g phai la n g h i e m ciia p h u o n g tririh:


Xet h a m so f ( x ) = V x - 1 + x^ - 7 + ^ x + 6 = 0 tren (1; +oo) ta c6

2t

T a c o f('t) = 2 ^ / t 7 7 + -

- - 3 + 4>/2t-l +

5^(t + 7)^

f '(x) = — /
+ 2x + — , ^
> 0 . V i v|iy h a m so d o n g bien t r e n (1; +oo)
2Vx-l
3^(x + 6)2

De

y

rang:

Voi

1

te

thi

4t
V2t-1

'

2\/t + 7 > 23/^ > 3

n e n suy ra

M a t khac ta c6: f(2) = 0 => x = 2 la n g h i ^ m d u y nhat cua p h u a n g t r i n h .
b) D i e u k i ? n : x > 8 .

'

1
f ( t ) > O V t e -;+oo

' -i"'

De d a n gian h i n h thuc ciia p h u a n g t r i n h ta dat ^ x - 1 = t

t>

, n h u v ^ y h a m so' f ( t ) d o n g b i e n t r e n

1
-;+oo

. Lai

CO

^
f ( l ) = 0 => t = 1 la n g h i f m d u y nha't cua p h u a n g t r i n h . Tu d o ta c6 : P h u a n g

P h u a n g t r i n h d a cho t r o thanh: t^ - 2t - (t^ - A)yjt^-7

- 3t^ + 28 = 0

t r i n h ban dau c6 n g h i ^ m d u y nhat x = 8

<=> 3t^ -1^ + 2t - 28+(t^ - 4)7t^-7 = 0. N h a n thay t = y/? khong phai la nghi^m.
Xet h a m so f (t) = 3t^ - 1 ^ + 2t - 28 + (t^ - 4)ylt^-7
T a c o f('t) = 9 t ^ - 2 t + 2 + 3 t ^ ^ t ^ - 7 +

tren

b) Ta thay: x = 0 thoa m a n p h u a n g t r i n h da cho

+oo).

Xet t r u a n g hg-p x > 0 . Chia h a i ve p h u a n g t r i n h cho x va d a t t = — ta
du(?c: 2Vt^ - 1 + 4 + 2 = 3^2t - 1 + ^41^ - 4t + 9

"^^'^^ . V i 9 t ^ - 2 t + 2 > 0 nen suy

W-7f
ra

Dat a = ^ 2 t - l ta c6 p h u o n g t r i n h m a i : 3a + yja^ +8 - Va^ +15 - 2 = 0

f ' ( t ) > 0 V t e ( \ / 7 ; + o o ) . N h u vay h a m so f ( t ) l u o n d o n g

^\/7; +00j. Ta c6 f(2) = 0

Xet h a m so: f(a) = 3a + Va^ + 8 - Va^+15 - 1 . De thay a < 0 t h i f(a) = 0 v 6

bien tren

nghi^m

t = 2 la n g h i f m d u y nhat cua p h u a n g t r i n h :

3a=
Ta can xet k h i a > 0 c6: f (a) = 3 + •
Va^+8

T u d o suy ra p h u o n g t r i n h ban dau c6 n g h i f m d u y nhat: x = 9
Vi

2: G i i i c a c p h u a n g t r i n h s a u :

M a t khac: f ( l ) = 0

a) 2x'' - 3x^ - 14x + 16 = (28 - i\^)\l2x^ -15
*

b) 2\/4x^-x + l + 2 x = 3 ^ 2 x 2 - x ^ + V 9 x ^ - 4 x + 4

a = 1 la n g h i ^ m d u y nhat

Dat a = v 2 t ^ 1 => a < - 1 ta c6 p h u a n g t r i n h m a i :

lai p h u o n g t r i n h thanh: 2x(x^ - 7) - 3(x^ - 7) - 5 = 4(7 - x^

Va^+8-3a-Va^+15 + 2 = 0

)^j2{x^-7)

Xet h a m so f(a) = Va^ + 8 - 3a - Va^ +15 + 2
D|t t = ( x ^ - 7 ) = > x = ^/t + 7 v o i t > i
P h u a n g t r i n h da cho t r o thanh: 2 t ^ t + 7 - 3t - 5 + 4 t 7 2 t - l = 0
k h o n g phai la n g h i ^ m cua p h u o n g t r i n h , n e n ta chi giai

1

1

_Va^ + 8

T a c o : f'(a) = - 3 + 3a'

De thay t = ^

x=1

-2\/t^ - t + 4 + 2 = 3 ^ 2 t - l - V 4 t 2 - 4 t + 9

a) D i e u k i f n : 2x^ - 1 5 > 0 o x > 3 / ^
Viet

>0

Va^+15

Xet t r u o n g h ^ p x < 0 . Chia hai ve cho x v a dat t = — ta duQc:

Giai:

Ta

3a^

Va^+15.

p h u a n g t r i n h v o i nhiang gia t r i t e - ; + = o
Xet h a m so f(t) = 2t^/t + 7 - 3t - 5 + 4 t V 2 t - l tren

a) 2(x-3)(^/x + 4 + 2 V 2 x - 7 ) = 3 x - 4
1

-;+oo

< 0 do do

Vay p h u a n g t r i n h c6 2 n g h i f m: x = 0; x = 1
V i d\ 3: G i a i cac p h u a n g t r i n h s a u :

ri

b)

^I2^^1^3x^T6x

+ U-yf4^

= 2•J3

vaia<-l

f(a) > f ( - l ) = 2


Giai:

Giii:

7
,
7
a) D i e u k i ^ n : x > —. De y rang x = — k h o n g p h a i la n g h i ^ m nen ta xet p h u a n g

Hoac: ax^ + bx^ + cx + d = e^px'^ H-qx^ + r x + h (2)

trinh tren

ta t h u o n g g i a i theo each:

Chia 2 v e c h o 2(x - 3) ta t h u d u o c p h u a n g t r i n h :
2x-6

*

D o i v o i (1): D a t ^ p x + q = y k h i d o x =

2x-6

-l^L_i

Xet ham so f(x) = \Jx + 4 + 2\J2x-7

tren

dua

7

Taco f ( x ) =

—7=1—+V 2 ; ^+ 2 x - 6 ) 2^
^=^= ( — ^

3^(x + 4)2

N h u vay h a m so' d o n g bien tren

7

> 0

tren

ax'^ + bx^ + c x + d = s.y
g ^px + qx

x<4

x<4

2x^+3x2 +6x + 1 6 > 0

(x + 2 ) ( 2 x 2 - x + 8 ) > 0

<> - 2 < X < 4 .
=

Ta thay x = -2;x = 4 khong phai la nghiem nen ta chi xet tren (-2;4)

T

'

f . / X

+6X + 6

1

Ta CO f (x) =

3 cpng hai p h u a n g t r i n h ta t h u dup-c:
+ rx + h I = y^

A x ' ' + Bx^ + Cx + D = s.y + y^ sau d o d u a p h u a n g t r i n h ve d a n g :

nen p h u a n g t r i n h f(x) = 0 c6 n g h i ^ m d u y nha't x = 4

_ 2^

v e d a n g : ax"^ + bx^ + cx + d = A y ' ' + By . Sau d o bien d o i p h u a n g t r i n h

D o i v o i (2): D a t gyjpx^ +qx^ + r x + h = y sau do tao ra h ^ t a m :

7

. Mat khac f (4) = 0

Xet ham so f(x) = V 2 x ^ + 3 x 2 + 6 x + 16 _ ^J^Z^

thay vao p h u a n g t r i n h ta

thanh: A . [ u ( x ) ] ^ + B.u(x) = A y ^ + By

2x-6

b) D i e u k i ^ n

N h u ' n g p h u a n g t r i n h c6 dang: ax"' + bx^ + cx + d = (ex + h ) ^px + q (1)

j^en (-2;4)

„
.
r
r>
> 0 vol mpi x e (-2; 4)

.
+ —.
V2x^+3x2+2V4-X
Suy ra ham so dong bien tren (-2; 4). Mat khac f(l) = 0 nen x = 1 la nghiem
duy nha't ciia phuong trinh.
C a c V I du sau se tap trung vao lop cac bai toan s u dung p h u a n g phap ham
so bling each 2:

u(x)

l3

+ S.

u(x) = y + s . y

Ta xet cac v i d u sau:
X r^^. 3/;
7
,
f8x^-36x2+53x-25 =y
a) Dat ^
(I)
3x-5=:y^
C p n g hai p h u a n g t r i n h ciia he v o i nhau ta t h u dupe:
8x^ - 36x2 +

- 30 = y^ + y (*). Ta n g h i den v i f c bien d o i ve'trai thanh:

A ( x ) ] ' ' + A ( x ) de p h u a n g t r i n h c6 d ^ n g : [ A ( X ) ] ^ + A ( x ) = y^ + y
G i a s u : 8 x ^ - 3 6 x 2 + 5 6 x - 3 0 = (2x + a)^+(2x + a ) .

^ •'

D o n g nha't h f so ciia x2 => a = - 3

+

Bien doi phuong trinh ve dang: f u(x) =f v(y) thong qua he tarn.

+

Tu do dya vao tinh chat. Neu f(t) dan di|u tang hoac dan di^u giam tren

Xet h a m so f ( t ) = t^ + 1 ta c6 f ' ( t ) = 3t2 + 1 > 0

D ma f ru(x)l = f rv(y)1 <> u(x) = v(y)
=

T u p h u a n g t r i n h ta suy ra

V i dvi 4: G i a i cac p h u a n g trinh sau:

N h u vay p h u a n g t r i n h (*) c6 dang: (2x - 3)'' + (2x - 3) = y ' ' + y (1)
f ( t ) d o n g bien tren R .

f(2x-3) = f ( y ) « . y = 2 x - 3 o 8 x ^ - 3 6 x 2 + 5 3 x - 2 5 = 2 x - 3

a) 8x^ - 36x2 + 53x - 25 = ^ 3 x - 5

b) 8x^ - 13x^ + 7 x = 2 \ / x 2 + 3 x - 3

c) ^ 2 4 x - l l - 1 6 x V 2 x - l - 1 = 0

c) x^ - ^ x + 2 1 n x - | l n ( x + 21nx) = 0

o 8 x ^ - 3 6 x 2 +51x-22 = 0 o

X =

5±V3

x =2

0

'


Tai

lieu

on thidai

UQC sang

tao va gidi

PT, bat PT,

FT, barUT-Nguyen

Cty

ThmglOen

Nen ham so' f(t) dong bien tren

Qua v i dy tren ta thay vi^c chuyen qua h^ tarn (I) giup ta hinh dung bai
toan dudu

x 2 + 3 x - 3 = y^

Theo (*) ta c6:
f(2x -1) = f(y) » y = 2x - 1 c:> 8x^ - 13x2 + 7x = 4x - 2
<=> 8x^-13x2 +3x + 2 = 0 »

5 + >/89

X = •

c) Dieu ki^n: x > ^ . Ta dat V 2 x - 1 = a >0 thi phuong trinh da cho tro thanh:
^123^ +1 - 8 a ^ - 8 a - l = 0 o 8 a ^ + 8 a + l = \/l2a2+l
Dat \^12a^n = y ta thu dugc h$ sau: .

^ y Cgng hai phuong

(0;+oo).

=

3x^-x-2

( x - l ) ( x 2 + x + 2)

X

X

V i dv 5: G i a i cac phuong trinh sau:

a)

d)

- 13x + 8 = 2x2^x(l + 3x - 3x^)
3 X ^ + 4 X 2 - 1 = N / X ^ + 2 X 3 + X2

Vx^ - x + 2
l-V-x^ + X +2

4 ^ :

= x2-l

1 +V-x^ - x + 4

2014^(7x2 +1 - x ) = l
Giai:

a) Nhan thay x = 0 khong phai la nghiem ciia phuong trinh:
3 . .,
*
7
Chia hai ve phuong trinh cho x ta thu dugc:

Xet ham so f(t) = t^ +1 ta c6 f'(t) = 31^ +1 > 0 => f(t) dong bien tren R .
Theo (*) ta c6
f(2a +1) = f(y) <=>y = 2a + l=>8a^+8a + l = 2a + l o a = 0=>x = ^

7

1
3
Dat y = 3 / — + — 3 ta thu duoc hf sau:
Vx^
X

Dat y = \/x + 21nx > 0 ta thu dugc h§ sau:
x-y-21ny =0 _
y = x + 21nx

Xethamso f(t) = t^ + t + 21nt tren

(0;+oo)

ta c6 f'(t) = 3 t 2 + l + - > 0 .

12

10
+

3

.

8

^

3

f2
^
3
^
3 = y+2y<::> Ix - 1

3

J

f2
^
+ 2 - - 1 = y ' + 2y (*)
Ix
J

Xet ham so f(t) = t^ + 2t ta c6 f'(t) = 3t2 + 2 > 0 => f(t) la ham so dong bien
Theo (*) ta c6:
f

= y + 21ny _ _ 3 _ , ^,
x'^ + x + 21nx = y'^ + y + 21ny
y^ = x + 21nx

13

Cpng hai phuong trinh cua h^ ta c6:
8

Ke't luan: x = i la nghif m duy nha't ciia phuong trinh:
x>0
d) Dieu ki^n:
x + 21nx>0

1

^,[1

13 8
3
- + — z= 2 3 | — + — 3
-2
X
x2 x3

X

\ld} + l = y^

trinh ciia h? voi nhau ta thu dugc: (2a +1)^ + (2a +1) = y^ + y (*)

. Ham so f(x) c6 dao ham

Tu do suy phuong trinh c6 nghi^m duy nha't x = 1 .

c)

16

Vig/

f(x)Sf(l)-0.

b)

'x = l

Khang

doi dau tir (-) qua (+) khi din qua x = 1 . L^p bang bien thien ta thay

= y ^ + 2 y (*)

= 3t^ + 2 > 0 => f(t) la ham so dong bien

2
X

. Cpng hai phuong trinh ciia h? ta thu du^c:

Xet ham so f(t) = t^ + 2t ta c6

DWH

trinh: x'^ - x - 2 1 n x = 0 . Xet ham so f(x) = x ' ' - x - 2 1 n x tren ( 0 ; + o o ) . T a c 6
f(x) = 3x^-1

8x^-12x2+10x-3 = y ^ + 2 y < » ( 2 x - l ) ^ + 2 ( 2 x - l )

MTV

Tu do suy ra f(x) = f(y) <» x = y .Ta quy bai toan ban dau ve giai phuong

b) Dat ^x^ + 3x - 3 = y ta thu duQ-c h^ phuong trinh sau:
8 x ^ - 1 3 x 2 + 7 x = 2y

TNHH

,

2 .

= f(y)oy =- - l o
-5±V89
X
Suy ra x = l , x =

7

13

8

4

^

- + — = --2<^
X

x^

X

8

13

3 ^ „

— — - + _ + 2 = 0.
x^

x^

X


fdi

lifu on thidai hoc sang tao va giai PT, bat PT, h? PT, baT DT- Nguyin Trung KietT

T H 1 : Neu: 1 < x <

b) Nh|n thay x = 0 khong phai la nghi#m cua phuong trinh nen ta chia hai ve
phuang trinh cho x thi thu duQC phuang trinh tuang duang
la:

< » x ^ - x + 2 < x ^ + x = > f ( x ^ - x + 2J<f^x^ + x) dong thoi x ^ - l > 0

3x2+4x--!- = 3 x 3 + 2 + i .
X

V

+2+—

Dat y =

3 x ^ + 4 x - —= y
^
. Cpng hai phuong trinh
ta c6 h§ sau:
, ^ + 2 - f i = y^

TH

Neu

khi do

x-" +1

d) De y r a n g

X

- X

+

1

Lay
X

logarit theo co so e
V

x=±

y

V

-x^+x + 2>0

^
- l + >/l7
<=>-l—
-x^ + x + 2 ^ 1
2

t

-x^ - x + 4 > 0

Vi In2014>l,

Mat khac f(0) = 0

7

X

+ 1 + x = 0.

f (x) = ln2014-

l + ^4-(x^+xj

b)

c)

J^

=:2x + :

x^+3x

Giai:

-4=
a) Dieu ki?n: x + 7x^ + 1 > 0 .
Ta thay

dong bieh tren [0;4
Phuong da cho CO dang: f x - x + 2

Ta xet hai truong hg-p:

-f(x2+x)-(x2-ij=on

x^ +1 >

trinh:

Xet ham so'

x = 0 la nghifm duy nhat ciia phuang trinh.

-2x + 6\n(x + /x^+1 = 0

Xethamso f ( t ) = —^^^,t€r0;4l ta c6

phuang

< 1 nen f'(x) > 0 => f(x) la ham so dong bieh tren R.

a) x^ +2Vx^ + l - 2xlnj^x + 7x^ +1 j - 2 - 0

=x^-l.

l ^ V l ± > 0 V t E r 0 ; 4 l suy ra ham so f ( t ) =

:;

x^+l

V i du 6: Giai cac phuong trinh sau:

Phuong trinh da cho tuong duang vai

f ( t ) = 2X/L^

suy ra

= 1 nen phuang trinh da cho tuong

l n v x " +1 + x
In 2014 = i n f 7 x ^ + l +: <=>xln2014-ln

x'^ + x > 0

l + ^ 4 - ( x 2 - x + 2]

+X

j

ca hai ve ta thu dugc

f(x) = x l n 2 0 1 4 - l n ^ x'^ +1 + x

Vx^ - x + ;

nen

duang voi: 2014"

x= -l

c) Dieu ki^n xac djnh:

x^ - x + 2 < x^ + x

Vay phuang trinh da cho CO nghiem duy nhat la x=l.

f(t) dong bieh tren R .

+ 1 < > 3x-^ + 3x^ - X - 1 = 0 o
=

-1 < x < 1

Thu true tiep ta thay x = 1 thoa man (*).

f(x +1) = f(y) <n>y = x + l < » 3 x ^ + 4 x - - = x + l

- - = X
X

2:

f ( x 2 - x + 2 ) > f ( x 2 + x ) , mat khac x^ - 1 < 0 nen phuong trinh v6 nghiem

T u phuong trinh ta suy ra

3x^ + 4x

k h i do

f^x^ - x + 2 J - f ^x^ + x j - ^ x ^ - l j > 0 . Nen phuong trinh v6 nghi^m

X

cua he ta c6: (x +1)^+(x +1) = y^ + y . Xet ham so f ( t ) - t ^ + t ta c6
f'(t) = 3t^ +1 > 0

- I + N/T?

x^ +1

+ X

> OVx

Xet ham so f(x) = x^ + 2Vx^ +1 - 2 x l n f x + 7x^ +1 - 2. Ta

CO


V i d v 7: G i a i cac p h u o n g t r i n h sau:
a) ^6x + 1 = 8x^ - 4x - 1
Xet h a m so' g(x) = x - I n

x +V?+l

ta CO gXx) = l - -

Vx^+1

X >

J

V

Vx^+l

g'(x) = O o x = 0 .
Khi

I'

(2x + l ) 2 + V 4 x 2 + 4 x + 4 + 3 x 2 + 7 9 x ^ + 3

c)

\

2x + l
i l o g 2 ( x + 2) + x + 3 = l o g 2 ^ +
Giai:

(-) qua (+) k h i d i qua x = 0. L a p bang bien thien ta suy ra f ( x ) > f(0) = 0 .

a) P h u o n g t r i n h t u o n g d u o n g v o i 6x + l + \/6x + l = (2x)"' + 2x

D o d o p h u o n g t r i n h da cho c6 n g h i f m d u y nhat x = 0

Vx^ + 1 + x > OVx .

Xet h a m so f ( x ) = x^ - 2x + 61n

x +Vx

Neu x > l = > V T > 2 . N e u x < - l thi V T < - 2 .

^+l

Suy ra m p i n g h i ^ m p h u o n g t r i n h deu thupc [ - 1 ; ! ] .
Dat t = cos

.-3.

,

3

+

3

>3sl9

suy

ra

e 0; t

M a t khac ta c6: f(0) = 0 suy ra phuong trinh c6 nghi^m d u y nhat x = 0
c) Ta Viet l^i phuong trinh thanh:

( x ^ - x ) ^ - 1 6 x = 4\^4x^Tl2x

D ^ t y = V 4 x ^ 12x ta c6 h ^ t ? m sau:

^3-x

- 1 6 x = 4y

4 x ^ + 1 2 x , = y^
Cong hai ve h? p h u o n g t r i n h ta t h u du<7c: {x^ " ''j

P h u o n g t r i n h t r o thanh cos3t =

(2x + l ) f 2 + ^ ( 2 x + l f + 3 ] = ( - 3 x ) f 2 +

+ '^{^^ - xj = y ^ + 4y .

= 0

o

x2=3

V g y p h u o n g t r i n h da cho c6 3 n g h i fm la: x = 0; x =

3t = ± - + k27t. V i x e [O; 7t] nen
3

^

yj{-3xf+3

t a c o i'{t) = 2 + ^t^+2

+

> 0 , l a ham

d o n g bien tren R ,
Ta CO £(2x +1) = £ ( - 3 x ) » x = -1

x^ - x = y < » (x^ - x)^ = 4x^ + 12x

x^^x^-1)^-4x^-12

= cos - «
3

V ^ y p h u o n g t r i n h c6 3 n g h i ? m x = cos - ; x = c o s — ; x = c o s —
9
9
9
b) P h u o n g t r i n h da cho c6 dang:

Xet h a m so f { t ) = t f 2 + V t ^ ]

Xet h a m so f ( t ) = t^ + 4t ta c6 f ' ( t ) = St^ + 4 > 0 nen h a m so f(t) d o n g bien .

x= 0

2

ta t i m d u g c 3 gia t r j t t u o n g u n g thoa m a n d i e u k i f n 1^: t = - ; t = — ; t = —
9
9
9

f (x) > 0 . N h u vay h a m so f(x) d o n g bien tren R .

X

X, X

x^+l

Theo bat dSng t h i i c Co si ta c 6 : ( x 2 + l ) +

o

'

t a c o f ( ^ 6 x + l ) = f(2x) <=>2x = ^6x + l o 8 x 3 - 6 x - l = 0 o 2 x ( 4 x 2 - 3 ) = 0

Taco f ( x ) = 3 x ^ - 2 +

Ta c6: f (x^ - x ) = f ( y ) »

' '

Xet f(t) = t^ + 1 ta CO f ' ( t ) = 3t2 + 1 suy ra h a m so f ( t ) d o n g bien. Theo de bai

b) D i e u k i f n : x + V x ^ + 1 > 0 .

Ta thay V x ^ T l > x

X

X

0 t h i g'(x) > 0 , k h i x < 0 t h i g'(x) < 0 t i r d o suy ra f (x) d o i d a u tir

=0

x eK
i
c) D i e u k i ? n :

-2;H

Xe

-00;--

t h u l a i ta thay thoa m a n d i e u k i $ n .

1

j(0;+co)

ôX

-'ãã-2

u(0;+oo)

a

K h i d o p h u o n g t r i n h viet l ^ i la:
l o g j Vx + 2 - 2yjx + 2 + X + 2 = l o g j

1^

2 + - -2
X

X

2.1

\