nhannguyen95 - 9 hướng giải phương trình không mẫu mực

Phng trnh khng mu mc Trn Xun Bang- GV Trng THPT Chuyn Qung Bnh

Trn Xun Bang- GV Trng THPT Chuyn Qung Bnh Phng trnh khng mu mc

1

PHNG TRNH KHNG MU MC

Ta xem phng trnh khng mu mc nhng phng trnh khng th bin i tng tng, hoc bin i h qu t u cho n khi kt thc. Mt s phn loi nh th ch c tnh tng i.

I. PHNG TRNH GII BNG PHNG PHP T N PH. 1. Mc ch t n ph. 1.1. H bc mt s phng trnh bc cao. a mt s phng trnh bc 4 v phng trnh trng phng. Phng trnh bc bn: ax4 + bx3 + cx2 + dx + e = 0 ( a 0 ) a v c phng trnh trng phng ch khi th hm s: f(x) = ax4 + bx3 + cx2 + dx + e c trc i xng. Gi x = x0 l trc i xng. Php t n ph x = x0 + X s a phng trnh ax4 + bx3 + cx2 + dx + e = 0 v phng trnh trng phng. V d 1: Gii phng trnh x4 - 4x3 - 2x2 + 12x - 1 = 0 Gii. t y = x4 - 4x3 - 2x2 + 12x - 1 Gi s ng thng x = x0 l trc i xng ca th hm s.

Khi qua php bin i: 0x x Xy Y

= +

=hm s cho tr thnh:

Y = (x0 + X)4 - 4(x0 + X)3 - 2(x0 + X)2 + 12(x0 + X) - 1 =

4 3 2 2 3 40 04 6 4o ox x X x X x X X+ + + + -

-

3 2 2 30 0 04 12 12 4x x X x X X -

-

2 20 02 4 2x x X X +

012 121

x X+ +

Y l hm s chn ca X 03 20 0 0

4 4 04 12 4 12 0

x

x x x

=

+ =

Suy ra: x0 = 1 v Y = X4 - 8X2 + 6 Phng trnh cho tng ng vi: X4 - 8X2 + 6 = 0 X2 = 4 10 X = 4 10 , X = 4 10 + Suy ra phng trnh c 4 nghim: x = 1 4 10 , x = 1 4 10 +

V d 2: Gii phng trnh x4 + 8x3 + 12x2 - 16x + 3 = 0 Gii. t y = x4 + 8x3 + 12x2 - 16x + 3. Gi s ng thng x = x0 l trc i xng ca th hm s.

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2

Khi qua php bin i: 0x x Xy Y

= +

=hm s cho tr thnh:

Y = (x0 + X)4 + 8(x0 + X)3 + 12(x0 + X)2 - 16(x0 + X) + 3 = =

4 3 2 2 3 40 04 6 4o ox x X x X x X X+ + + + -

3 2 2 30 0 08 24 24 8x x X x X X+ + + + +

2 20 012 24 12x x X X+ + + +

016 163

x X ++

Y l hm s chn, suy ra: x0 = - 2 Y = X4 - 12X2 + 35 Y = 0 X2 = 5, X2 = 7 X = 5 , X = 7 Suy ra bn nghim X = - 2 5 , X = - 2 7

Bi tp tng t: BT1. Gii phng trnh 2x4 - 16x3 + 43x2 - 44x + 14 = 0 S: x = 2 1

2 , x = 2 2 .

BT2. Gii phng trnh 6x4 + 24x3 + 23x2 - 2x - 1 = 0

S: x = - 1 23

, x = - 1 32

.

a phng trnh bc bn dng: (x - a)(x - b)(x - c)(x - d) = m, trong a + d = b + c v phng trnh bc hai. Do a + d = b + c nn phng trnh cho tng ng: (x - a)(x - d)(x - b)(x - c) = m [x2 - (a+d)x + ad] [x2 - (b+c)x + bc] = m

2 2

( )( )( ) ( )

X ad X bc mx a d x X x b c x

+ + =

+ = = +

Phng trnh cho chuyn c chuyn v: (X + ad)(X + bc) = m X2 + (ad + bc)X + abcd - m = 0 V d 1: Gii phng trnh (x - 1)(x - 2)(x + 3)(x + 4) = 14. Gii. Phng trnh cho tng ng vi: (x - 1)(x + 3)(x - 2)(x + 4) = 14 (x2 + 2x - 3)(x2 + 2x - 8) = 14

2

( 3)( 8) 142

X Xx x X

=

+ = 2

2

11 10 02

X Xx x X

+ =

+ = 2

1, 102

X Xx x X

= =

+ =

x = - 1 2 , x = - 1 11 .

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3

V d 2: Gii phng trnh (x2 - 1)(x + 2)(x + 4) = 7 Gii. Phng trnh cho tng ng vi: (x - 1)(x + 4)(x + 1)(x + 2) = 7 (x2 + 3x - 4)(x2 + 3x + 2) = 7

2

( 4)( 2) 73

X Xx x X

+ =

+ = 2

2

2 15 03

X Xx x X

=

+ = 2

3, 53

X Xx x X

= =

+ = x =

3 292

V d 3: Tm tt c cc gi tr ca tham s m phng trnh sau: (x2 - 1)(x + 3)(x + 5) = m a) C nghim. b) C bn nghim phn bit. Gii. Phng trnh cho tng ng vi: (x - 1)(x + 5)(x + 1)(x + 3) = m (x2 + 4x - 5)(x2 + 4x + 3) = m

2

( 5)( 3)4

X X mx x X

+ =

+ = 2

2

2 15 (1)4 (2)

X X mx x X

=

+ =

a) Phng trnh (2) c nghim X - 4 Phng trnh cho c nghim ch khi phng trnh (1) c nghim X - 4.

Cch 1: Phng trnh (1) c nghim X - 4

( 4) 0

' 0( 4) 0

42

f

fba

m - 16

Cch 2: Hm s f(X) = X2 - 2X - 15 , X - 4 c f '(X) = 2X - 2. f(X) lin tc trn [- 4; + ) v c cc tiu duy nht trn ti X = 1. Suy ra, trn [- 4; + ) ta c min f(X) = f(1) = - 16. Vy phng trnh (1) c nghim X - 4 khi m - 16. b) 4 nghim phn bit ? Thy ngay l cc phng trnh x2 + 4x = X1, x2 + 4x = X2 c nghim trng nhau khi v ch khi X1 = X2. Do vy phng trnh cho c 4 nghim phn bit khi v ch khi phng trnh (1) c hai nghim phn bit X1 > X2 - 4.

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4

Cch 1. Ta phi c: ' 0( 4) 0

42

fba

>

>

- 16 < m 9

Cch 2: Hm s f(X) = X2 - 2X - 15 , X - 4 c f '(X) = 2X - 2.

X - 4 1 +

f '(X) - 0 +

f(X)

9 +

- 16 Bi tp tng t: BT1. Gii phng trnh x4 - 2x3 - 7x2 + 8x + 7 = 0. HD. Tm a, b: (x2 - x + a)(x2 - x + b) = x4 - 2x3 - 7x2 + 8x + 7. t x2 - x = t BT2. Cho phng trnh (x + 1)(x + 2)(x + 3)( x + 4) = m.

a phng trnh bc bn dng: ax4 + bx3+ cx2 + bx + a = 0(a 0) Thy ngay x = 0 khng tho phng trnh. Chia hai v ca phng trnh cho x2: Phng trnh cho tng ng : ax2 + bx + c + b 1

x + a 2

1x

= 0

22

1 1( ) 0a x b x cx x

+ + + + =

( )2 2 0a X bX c + + = ,

trong X = x + 1x

hay x2 - Xx + 1 = 0, 2X

VD1. Gii phng trnh 2x4 + 3x3 - 10x2 + 3x + 2 = 0. 2

21 12 3( ) 10 0x xx x

+ + + =

( )22 2 3 10 0X X + = 22 3 14 0X X + =

72,2

X X = = , trong X = x + 1x

hay x2 - Xx + 1 = 0, 2X

i) X = 2: x2 - 2x + 1 = 0 x = 1 ii) X = - 7

2: 2x2 + 7x + 2 = 0 7 33

4

VD2. Cho phng trnh x4 + hx3 - x2 + hx + 1 = 0. Tm h phng trnh c khng t hn hai nghim m phn bit.

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5

Gii. 2 21 1( ) 1 0x h xx x

+ + + =

( )2 2 1 0X hX + = 2 3 0X hX + = (1), trong

X = x + 1x

hay x2 - Xx + 1 = 0 (2) , 2X . Cch 1. Phng trnh (2) nu 2X th c hai nghim cng du. Nn mun c nghim m th - b/a = X < 0. Suy ra X - 2. Nhng (1) lun lun c hai nghim X1 < 0 < X2 nn ch mang v cho (2) c X1. Vy X1 < - 2 < 0 < X2. Khi f(- 2) < 0, f(X) =

2 3X hX+

1 2 0h < 12

h > .

Cch 2. (1) 23 Xh

X

= , 2X

t 23( ) Xf X

X

= , 2X 2 2

23 3

'( ) 0,X Xf XX X

= = < 2X

X - - 2 2 +

f '(X) - -

f(X) + - 12

12

-

Phng trnh (2) nu 2X th c hai nghim cng du. Nn mun c nghim m th - b/a = X < 0. Suy ra X - 2. Nhng (1) lun lun c hai nghim X1 < 0 < X2 nn ch mang v cho (2) c X1. Vy X1 < - 2 < 0 < X2. Theo trn: 12h > . Bi tp tng t: BT1. Gii phng trnh 2x4 - 5x3 + 2x2 - 5x + 2 = 0. BT2. Cho phng trnh x4 + mx3 - 2x2 + mx + 1 = 0. Tm m phng trnh c khng t hn hai nghim dng phn bit.

1.2. Lm mt cn thc. VD1. Gii phng trnh x(x + 5) = 2 3 2 5 2 2x x+ Gii. t 3 2 5 2x x+ = X 3 22 5X x x+ = + Phng trnh cho 3 2 4 0X X + = X = - 2 2 5 6 0x x+ + = x = - 2, x = - 3 VD2. Cho phng trnh 3 6 (3 )(6 )x x x x m+ + + = (1)

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6

1) Gii phng trnh khi m = 3 2) Tm tt c cc gi tr m phng trnh (1) c nghim. Gii. t 3 6 , 3 6x x t x+ + = 1 1' , 3 6

2 3 2 6t x

x x= < 3 (2) Phng trnh t2 + 4t = m (3) 1) m = - 3: Phng trnh (3) t2 + 4t + 3 = 0 t = - 1, t = - 3. Thay vo (1): * t = - 1: 2

3 03 01( 3) 1 ( 3)( 1) 13 2 4 0xxx

xx xx x x

< < +

= + = =

1 5x =

1 5x = tho iu kin x - 1.

* t = - 3: 23 03 01( 3) 3 ( 3)( 1) 93 2 12 0

xxxx

x xx x x

< < +

= + = =

1 13x =

1 13x = tho iu kin x - 1. 2) (3) c nghim t m - 4.

Xt phng trnh 2( 3)( 1)x x t + = , x - 1 hoc x > 3 x

2 - 2x - 3 = t2, x - 1 hoc x > 3

t f(x) = x2 - 2x - 3, x - 1 hoc x > 3

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7

f '(x) = 2x - 2 x - - 1 3 + f '(x) - +

f(x)

+ +

0 0

v t2 0 nn (2) lun lun c nghim. Cch 2. Nu dng nh l o v du ca tam thc bc hai th vi m - 4. Xt 3 trng hp khi thay vo (1): i) t = 0: 1( 3) 0

3x

xx

+ =

: Phng trnh c nghim x = - 1.

ii) t > 0: (1) 2 2 23 0 3

( 3)( 1) ( ) 2 3 0x x

x x t F x x x t > >

+ = = =

Thy ngay F(3) = - t2 < 0 nn F(x) c nghim x > 3. 3i) t < 0: (1) 2 2 2

1 0 1( 3)( 1) ( ) 2 3 0x x

x x t F x x x t+

+ = = =

Thy ngay F(- 1) = - t2 < 0 nn F(x) c nghim x - 1.

VD4. Gii phng trnh 2 2 2( 1) 3 ( 1) 2 1, 2nn nx x x n+ = HD. Thy ngay x = 1 khng tho phng trnh. Vi x 1:

Chia hai v ca phng trnh cho 2 1n x , ta c: 1 13 21 1

n nx x

x x

+ =

+ (1)

t 11

nx

tx

+=

, khi (1) t - 31t + 2 = 0 t2 + 2t - 3 = 0 t = 1, t = - 3

i) t = 1 : 1 11 11 1

nx x

x x

+ += =

: V nghim

ii) t = - 3: 1 31

nx

x

+=

(2) + n chn: (2) v nghim + n l: (2) ( )1 3 13 1 ( 1)( 3) (3 1) 3 1

1 3 1

nn n n n

n

xx x x x

x

+ = + = + = =

+

1.3. Lm mt gi tr tuyt i. VD1. Tm m phng trnh sau c nghim

2 22 1 0x x m x m + =

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8

HD. t 1 0x t = 2 22 1x x t = Phng trnh cho tng ng t2 - mt + m2 - 1 = 0 (1) Phng trnh cho c nghim khi ch khi phng trnh (1) c nghim t 0. = m2 - 4m2 + 4 = 4 - 3m2 i) = 0 4 - 3m2 = 0 m = 2

3 : Pt(1) c nghim kp t =

2m

m = 23

tho

ii) > 0 - 23

< m < 23

:

+ (1) c 2 nghim dng P > 0, S > 0 m > 1. Suy ra 1 < m < 23

tho

+ (1) c hai nghim tri du P < 0 - 1 < m < 1 + (1) c 1 nghim bng 0 m = 1 . Khi nghim kia t = m nn m = 1 tho KL: - 1 < m 2

3

VD2. Cho phng trnh 2 2 1x x m x + = (1) 1) Gii phng trnh khi m = 0. 2) Tm m phng trnh (1) c 4 nghim phn bit. HD. t x - 1 = t 2 22 1x x t =

Pt(1) 2 1t m t + = 2

2

01 0

01 0

t

t t m

t

t t m

+ = + + =

2

2

0( ) 1

0( ) 1

t

f t t t mt

g t t t m

= = = + =

f '(t) = 2t - 1, g'(t) = 2t + 1

V x = 1 + t nn mi nghim t cho (1) mt nghim x. Suy ra khng c m tho 1.4. Lng gic ho cc phng trnh. VD. Gii phng trnh 3 2 3 2(1 ) 2(1 )x x x x+ = HD. Do 1 - x2 0 - 1 x 1. t x = cost, [ ]0;t pi Ptrnh cho 3 3cos sin 2 sin cost t t t+ =

x 0 +

g '(x) +

g(x)

+ - 1

x 0 1/2 +

f '(x) - 0 +

f(x)

- 1 + - 5/4

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9

3(cos sin ) 3sin cos (sin cos ) 2 sin cost t t t t t t t+ + = (1) t sint + cost = X

2 1cos , 2,sin cos

4 22X X

x X t tpi = =

.

(1) 2 2

3 1 13 22 2

X XX X = 3 22 3 2 0X X X + = 2( 2)( 2 2 1) 0X X X + + = 2, 2 1X X = .

Nhng 2 2, 1 2X X X = = . i) X = 2 : sint + cost = 2 21 2x x + =

21 2x x = 2 21 2 2 2

2 0x x x

x

= +

22 2 2 1 02

x x

x

+ =

12

x = .

i) X = 1- 2 : sint + cost = 1 - 2 21 1 2x x + = 21 1 2x x =

2 21 2 2 2 2(1 2)1 2 0

x x x

x

= +

2 (1 2) 1 2 01 2

x x

x

+ =

1 2 2 2 12

x

= .

1.5. i s ho cc phng trnh lng gic, m, loga. VD1. Gii phng trnh ( ) ( )2 3 2 3 4x x + + = HD. t ( )2 3 0x t+ = > ( ) 12 3 x t = Pt 1 4t

t+ = t2 - 4t + 1 = 0 2 3t =

( )( )

2 3 2 3

2 3 2 3

x

x

+ = +

+ =

( ) ( ) 22

2 3 2 3 2 3x

x

= + = = +

2, 2.x x = =

VD2. Cho phng trnh ( ) ( )tan5 2 6 5 2 6x tanx m+ + = 1) Gii phng trnh khi m = 4 2) Gii v bin lun phng trnh (1) theo m. HD. t ( )tan5 2 6 0x t+ = > ( )tan 15 2 6 x

t =

Pt cho tng ng 21 1 0t m t mtt

+ = + = (1)

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10

1) m = 4: 2 3t = ( ) ( )tan 5 2 65 2 6 2 3 log 2 3x tanx + + = = ( )5 2 6log 2 3x arctan kpi+ = + 2) Ptrnh cho c nghim khi v ch khi Pt(1) c nghim t > 0 Thy ngay rng, nu (1) c nghim th c hai nghim cng du. Do vy nu pt (1) c nghim dng th c hai nghim dng. Suy ra, cn v l:

2 4 02

0m

mS m

=

= >. Khi t =

2 42

m m ( ) 2tan 45 2 6 2

x m m + =

2 2

5 2 6 5 2 64 4

tan log arctan log2 2

m m m mx x kpi

+ +

= = +

.

2. Cc kiu t n ph. 1.1. t mt n ph chuyn phng trnh v phng trnh ca n ph. VD. Gii v bin lun phng trnh 243 1 1 2 1x m x x + + = HD. Thy rng x = - 1 khng tho ptrnh.

Pt cho tng ng vi 41 13 21 1

x xm

x x

+ =+ +

(1)

t 4 1 01

xt

x

= +

. Khi (1) 23 2 0t t m + = (2) Ptrnh cho c nghim khi v ch khi (2) c nghim khng m Cch 1: Phng trnh (2) c 2 nghim tri du m < 0

Phng trnh (2) c 2 nghim khng m ' 0

00

PS

103

m

Hai nghim ca (2) l 1 1 33

mt

=

Nh th, khi m < 0:

1 1 33

mt

+ = 4

1 1 1 31 3

x m

x

+ =

+

4

11

1

11 1 1 31 3 1

Mx m M xx M

+

= = = + +

khi 0 m 13

: 41 1 1 31 3

x m

x

=

+

4

11

1

11 1 1 31 3 1

Mx m M xx M

+

= = = + +

hoc 4

22

2

11 1 1 31 3 1

Mx m M xx M

= = = + +

1.2. t mt n ph v duy tr n c trong cng mt phng trnh. VD1. Gii phng trnh 2(1 - x) 2 22 1 2 1x x x x+ =

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11

HD. Cch 1: t 2 2 1x x t+ = 0 2 2 2 22 1 2 1 4x x t x x t x + = = Pt 22(1 ) 4x t t x = 2 2(1 ) 4 0t x t x =

2' ( 1)x = + (1 ) ( 1) 2, 2t x x t t x= + = =

2 0 0t x x= : 2 2 2 22 1 2 2 1 4 3 2 1 0x x x x x x x x+ = + = + = : VN 2t = : 2 22 1 2 2 5 0 1 5x x x x x+ = + = =

Cch 2: Pt (x - 1)2 - 2(x - 1) 2 2 1x x+ - 2 = 0 VD2. Gii phng trnh (4x - 1) 2 21 2 2 1x x x+ = + + Cch 1: t 2 1x t+ = Cch 2: Bnh phng hai v 1.3. t mt n ph v duy tr n c trong mt h phng trnh. VD1. Gii phng trnh x2 + 5 5x + = HD. t 25 0 5x y y x+ = = + (1) T Pt cho x2 = 5 - y (2) Tr tng v (1) v (2) ta c: y2 - x2 = x + y x + y = 0 hoc y - x - 1 = 0 i) x = y = 0 y = - x 0 x 0: (1) x2 - x - 5 = 0 x = 1 21

2

Nhng x 0 nn 1 212

x

=

ii) y - x - 1 = 0 y = x + 1 0 x - 1: (2) x2 - x - 4 = 0 x =

1 172

Nhng x - 1 nn 1 172

+

Cch 2.(Bin i Pt v dng tch) x

2 + 5 5x + = 2 ( 5) ( 5) 0x x x x + + + + = ( 5)( 5 1) 0x x x x + + + + =

VD2. Gii phng trnh x3 + 1 = 32 2 1x HD. t 33 2 1 2 1x y y x+ = = + (1) T Pt cho x3 = 2y - 1 (2) H (1)&(2) l mt h i xng loi 2. Cch 2.(Dng tnh cht th ca hai hm ngc nhau) Pt cho tng ng

331 2 1

2x

x+

= (1)

Cc hm s 3

31, y 2 1

2xy x+= = l cc hm s ngc ca nhau. Vy nn phng

trnh (1) tng ng 3 12

xx

+=

3 2 1 0x x + = -1 51, x = 2

x

=

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12

VD3. Gii phng trnh (x2 - 3x - 4)2 - 3x2 + 8x + 8 = 0 HD. Ptrnh cho tng ng (x2 - 3x - 4)2 - 3(x2 - 3x - 4) - 4 - x = 0 (x2 - 3x - 4)2 - 3(x2 - 3x - 4) = 4 + x t x2 - 3x - 4 = y x2 - 3x = 4 + y (1) T phng trnh cho suy ra y2 - yx = 4 + x (2) H (1)&(2) l mt h i xng loi 2. VD4. Gii phng trnh 7x2 + 7x = 4 9

28x +

PP chuyn v h i xng loi 2: - VT bc hai, VP cn hai

- Nn t 4 928x +

= at + b (bc nht ca t khi bnh phng th thnh bc hai) - Khi t ta c ngay : 7x2 + 7x = at + b Ta phi c mt pt mi: 7t2 + 7t = ax + b

4 928x +

= at + b x = 7a2t2 + 14abt + 7b2 - 9/4

ax + b = 7a3t2 + 14a2bt + 7ab2 - 94

a

7t2 + 7t

Ta phi c:

3

2

2

7 114 7

97 04

a

a b

ab a b

=

=

+ =

a = 1, b = 12

Bi tp tng t: BT1. Gii phng trnh 2x2 - 6x - 1 = 4 5x + (Thi chn T12QB 21/12/2004) BT2. Gii v bin lun theo a phng trnh 3 2 23(2 ) 2 2 ( 2)x a a x a a+ = +

1.4. t hai n ph v a phng trnh v phng trnh hai n ph. VD1. Gii phng trnh 2 2 4 3 23 5 1 8 3 15x x x x x x x + + = + + + a phng trnh v dng u + v = 1 + uv VD2. Gii phng trnh 2 2 23 2 2 15 2 5 132 2 1 2x x x x x x + + = + a phng trnh v dng u + v = 1 + uv 1.5. t hai n ph v a phng trnh v h phng trnh hai n. VD1. Gii phng trnh 4 5 3x x+ + = HD. t 4 0, 5 0x u x v+ = = 2 2 9u v + =

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13

Ta c h phng trnh 2 2 9

3u v

u v

+ =

+ =

Cch 2. Bnh phng hai v. Cch 3. t f(x) = 4 5 0x x+ + 2 ( ) 9 2 (4 )(5 ) 9 ( ) 3f x x x f x= + + Du ng thc xy ra khi ch khi x = - 4 hoc x = 5. Cch 4. t f(x) = [ ]4 5 , x -4;5x x+ + . Kho st, lp bng bin thin. VD2. Gii phng trnh 3 6 (3 )(6 ) 3x x x x+ + + = . HD. t 3 0, 6 0x u x v+ = = 2 2 9u v + =

Ta c h phng trnh 2 2 9

3u v

u v uv

+ =

+ =

Cch 2. t 2 93 6 0 (3 )(6 )2

Xx x X x x + + = + =

Phng trnh cho tng ng 2 9 32

XX =

VD3. Gii phng trnh 21 2( 1) 1 1 3 1x x x x x+ + + = + + (TS 10 Chuyn Ton HSPHNI, 97 - 98) a phng trnh v h c mt phng trnh tch : u + 2u2 = - v2 + v + 3uv u - v + v2 - 3uv + 2v2 = 0 u - v + (v - u)(v - 2u) = 0

1.6. t hai v ca phng trnh cho cng mt n ph. VD1. Gii phng trnh 3 22log cotgx log cosx= HD. t 3 22log cotgx log cosx= = t ta c:

2 2 2

22 2

2

2

cos 4 cos 4 cos 4cos 2

cos 4 4cot 3 3 sin 4 1

sin 3 3cos 0,cot 0

cos 0,sin 0 cot 0,sin 0 cos 0,sin 0

cos 4 1cos

1 2sin 0cos 0,sin 0

t t tt

t tt t t

t t

t

x x xx

xx x

xx x

x x x x x x

xx

txx x

= = = =

= = = + = > > > > > > > >

= =

= >> >

23

x kpi pi = +

VD2. Gii phng trnh 7 3log x log ( x 2)= +

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14

HD. t 7 3log x log ( x 2)= + = t , Ta c: 7

77 7 497 1 22 1x 2 3 7 2 33 3

t

tt tt t

t t t

xxx x

xt

= = = =

= =+ =+ = + =

VD3. Gii phng trnh 3 1 1x x = + HD. t 3 1 1x x = + = t 0, ta c:

33 2

2

12 0 1 0

1x t

t t t xx t

= + = = =

+ =

II. PHNG TRNH GII BNG PHNG PHP I LP. 1. Dng 1. Nu f(x) M, (1) (hay f(x) M, (2)) th: Phng trnh f(x) = M tng ng du ng thc (1) hay (2) xy ra. VD1. Gii phng trnh tanx + cotx + tan2x + cot2x + tan3x + cot3x = 6. HD. Phng trnh cho tanx(1 + tanx + tan2x) + cotx(1 + cotx + cot2x) = 6 (1) 1 + tanx + tan2x > 0, 1 + cotx + cot2x > 0 vi

2x k pi

tanx v cotx cng du. Do vy, t (6) rng v phi dng, suy ra tanx > 0, cotx > 0. Theo Csi: tanx + cotx 2 tan2x + cot2x 2 tanx + cotx + tan2x + cot2x + tan3x + cot3x 6. tan3x + cot3x 2

Phng trnh cho tng ng: 2 2

3 3

tan cot 2tan cot 2tan cot 2tan 0

x x

x x

x x

x

+ =

+ =

+ = >

2 2

3 3

tan cot 1tan cot 1tan cot 1

tan cot 14

x x

x x

x x

x x x kpi pi

= =

= =

= =

= = = +

VD2. Gii phng trnh 2 2 2 21 1 4x yx y

+ + + =

HD. K x 0, y 0. 2 2

2 21 1 4x yx y

+ + + = 2 22 21 1 4x yx y

+ + + =

Ta c: 2 22 21 12, y 2xx y

+ + 2 22 21 1 4x yx y

+ + +

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15

Phng trnh cho tng ng vi: 2 2 2 2

2 2 2 2

2 11 1 1 12 2

x y x y

x y x y

+ = = =

+ = + =

nghim ca

phng trnh cho l (1; 1), (1; - 1), (-1; 1), (- 1; - 1) 2. Dng 2.

Phng trnh : ( ) ( )( ) ( )f x g xf x M g x

=

( )( )

f x Mg x M

=

=

VD1. Gii phng trnh 4(x2- 2)(3 - x2) = 2( 2 5) 1x + HD. (x2- 2)(3 - x2) > 0 2 < x2 < 3 3 - x2 > 0, x2- 2 > 0. Theo Csi:

22 22 2 2 22 3 1( 2)(3 ) 4( 2)(3 ) 1

2 4x x

x x x x +

=

Mt khc 2( 2 5) 1x + 1

Phngtrnh cho tng ng: ( )2 2

2

4( 2)(3 ) 12 5 1 1

x x

x

=

+ =

2 22 35

5 22

x x

xx

=

==

VD2. Gii phng trnh 22 4 6 11x x x x + = + HD. K 2 x 4. Ta c:

2 22 4 2( 2 4 ) 2, 6 11 ( 3) 2 2x x x x x x x + + = + = +

Phngtrnh cho tng ng: 2

2 4 22

( 3) 2 2x x

xx

+ =

= + =

3. Dng 3.

Phng trnh : ( ) ( )( ) , ( )

( : ( ) , ( ) )

f x g x M Nf x M g x Nhay f x M g x N

+ = +

( )( )

f x Mg x N

=

=

VD1. Gii phng trnh 36 4 28 4 2 12 1

x yx y

+ =

HD. Pt cho 36 44 2 1 282 1

x yx y

+ + + =

(1) 36 44 2 24, 1 4

2 1x y

x y+ +

Nh th (1) tng ng:

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16

36 364 2 24 4 2112 2

4 4 51 4 11 1

x xxx x

yy yy y

+ = =

=

= + = =

VD2. Gii phng trnh 1 1

cos3x - 1 cosx - 1 = 1cos3x cosx

+

HD. Pt cho tng ng:

1 - cos3x 1 - cosxcos3x cosx = 1

cos3x cosx+

K: cos3x > 0, cosx > 0. PT cos3x(1 - cos3x) cosx(1 - cosx) = 1 + (1) Ta bit rng a(1 - a) 1 ,

4a . Suy ra: 0 cos3x(1 - cos3x) 1

4

1

cos3x(1 - cos3x) 2

Tng t 1

cosx(1 - cosx) 2

Nh th Ptrnh (1)

31 1 1cos3x(1 - cos3x) = cos3x = 4cos x - 3cosx = 2 2 21 1 1

cos3x(1 - cos3x) = cosx = cosx = 2 2 2

: V nghim

4. Dng 4.

Phng trnh : 1 21 2

( ) ( ) ... ( ) 0( ) 0, ( ) 0,..., ( ) 0

n

n

f x f x f xf x f x f x

+ + + =

1

2

( ) 0( ) 0

.............

( ) 0n

f xf x

f x

=

= =

VD1. Gii phng trnh x2 - 2xsinxy + 1 = 0 HD. Pt cho tng ng: (x - sinxy)2 + 1 - (sinxy)2 = 0

2

1sin 1 sin 1

21 1sin 0 sin 2sin 11 sin 0 sin 1 sin( ) 1 1

1 1 22

x

xy yy k

x xx xy x xyxyxy xy y x

x x y k

pipi

pipi

= = = = + = = = = =

= = = = = = = +

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VD2. Tm tt c cc cp s thc (x, y) tho mn :

2 2x + 2y - 2xy - 2x + 4y + 2 = 0

(Thi HSG L9 Qung Bnh 2007 - 2008) HD. Ta c: 2 2x + 2y - 2xy - 2x + 4y + 2 = 0

( ) ( )22x - 2 y + 1 x + 2 y + 1 0 (1) =

Xt phng trnh bc hai (1) n x v y l tham s Ta c: ' 2 2 2( 1) 2( 1) ( 1) 0, yy y y = + + = + Do , phng trnh (1) c nghim x khi v ch khi

' 20 ( 1) 0 1y y = + = =

Khi phng trnh (1) c nghim kp x = 0. Vy cp s (x, y) cn tm l ( 0, -1). Ghi ch: C th gii bi ton bng cch a v dng 2 2A + B = 0

2 2x + 2y - 2xy - 2x + 4y + 2 = 0 ( ) ( )2 2y - x + 1 1 0y+ + =

III. PHNG TRNH GII BNG PHNG PHP D ON NGHIM V CHNG MINH KHNG CN NGHIM. Phng php gm hai bc:

1. D on nghim, th vo phng trnh. 2. Chng minh khng cn nghim.

VD1. Gii phng trnh 3x + 4x = 5x HD. Bc 1. D on: x = 2 l nghim Chng minh: 32 + 42 = 52 . Bc 2. Chng minh khng cn nghim na.

Tht vy: Pt tng ng vi 3 4 15 5

x x

+ =

i) Nu x > 0 i) Nu x > 0 th 2 23 4 3 4 1

5 5 5 5

x x

+ < + =

: Khng tho pt.

ii) Nu x > 0 i) Nu x < 0 th 2 23 4 3 4 1

5 5 5 5

x x

+ > + =

: Khng tho pt.

VD2. Gii phng trnh 4 4 24 52 2 1956 49x x x+ ++ + = HD. Bc 1. D on: x = 0 l nghim Chng minh: 24 + 25 + 19560 = 49 . Bc 2. Chng minh khng cn nghim na. Tht vy: Nu x 0 th x4 > 0, x4 + 4 > 4, x5 + 5 > 5

4 4 44 4 5 5 02 2 16,2 2 32,1956 1956 1x x x+ +> = > = > =

4 4 44 52 2 1956 16 32 1 49x x x+ ++ + > + + =

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18

VD3. Gii phng trnh 2 2 21 1 120 9 1956 1985x x x + + = HD. x = 0 l nghim x 0 x2 > 0 1 - x2 < 1 2 2 21 1 120 20, 9 9, 1956 1956x x x < < <

2 2 21 1 120 9 + 1956 1985x x x + < VD4. Gii phng trnh 4 4 21 1 119 5 1890 3x x x + + = HD. x = 1 l nghim - 1 < x < 1 1 - x2 > 0 4 4 21 1 1 0 0 019 5 1890 19 5 1890 3x x x + + > + + = x < - 1 hoc x > 1 1 - x2 < 0 4 4 21 1 1 0 0 019 5 1890 19 5 1890 3x x x + + < + + = VD5. Gii phng trnh 5 32 228 2 23 1 2 9x x x x+ + + + + = + HD. x = 1 l nghim VD6. Gii phng trnh 3 2 26 3 3 8x x x+ + + + = HD. x = 1 l nghim VD7. Gii phng trnh 1956 19812007 2008 1x x + = HD. x = 2007, x= 2008 l nghim i) x < 2007 x - 2008 < - 1 19812008 1 2008 1x x > >

1956 1981 2007 2008 1x x + >

ii) x > 2008 x - 2007 > 1 19562007 1 2007 1x x > >

1956 1981 2007 2008 1x x + >

iii) 2007 < x < 2008 0 < x - 2007 < 1 2007x = x - 2007 2007x < 1 19562007 2007 2007x x x < = (1) Tung t: - 1 < x - 2008 < 0 2008x = 2008 - x 2008x < 1 19812008 2008 2008x x x < = (2) T (1)&(2) suy ra: 1956 1981 2007 2008 1x x + <

IV. BIN LUN S NGHIM PHNG TRNH BNG PHNG PHP O HM. VD1. Bin lun theo m s nghim phng trnh 4 444 4 6x x m x x m+ + + + + = HD. t 44 4 0x x m t+ + = Pt cho t2 + t - 6 = 0 t = 2, t = - 3(loi) Ta c 4 44 4 2 4 16x x m x x m+ + = + = (1) Pt cho c nghim khi v ch khi pt(1) c nghim. t f(x) = x4 + 4x, x R. f '(x) = 4x + 4

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19

x - -1 + f '(x) - 0 +

f(x)

+ +

- 3

Ta c kt qu: i) 16 - m < - 3 m > 19: V nghim. ii) 16 - m = - 3 m = 19: x = - 1. iii) 16 - m > - 3 m < 19: Hai nghim phn bit VD2. Bin lun theo m s nghim phng trnh 2 1x m m x+ = + HD. x = 0 l nghim vi mi m. x 0: Pt cho ( )2 1 1x m x= + 2 1 1x mx =+ t

2( )

1 1xf x

x=

+ , x 0

( ) ( )2

22

2 22 2 2

1 11 11

'( )1 1 1 1 1

xx x

xxf xx x x

+ ++

= =

+ + + < 0, x 0

x - 0 + f '(x) - +

f(x)

1 +

- 1

Ta c kt qu: i) m = 1 x = 0. ii) m 1 x = 0 v 1 nghim khc. Bi tp tng t: BT1. Chng minh rng nu n l s t nhin chn v a l mt s ln hn 3 th phng trnh sau v nghim: (n + 1)xn + 2 - 3(n + 2)xn + 1 + an + 2 = 0 BT2. Tm k phng trnh sau c 4 nghim phn bit: x

4 - 4x3 + 8x - k = 0.

Gii phng trnh khi k = 5. BT3. Cho 3 n N . Tm nghim x 0;

2pi

ca phng trnh:

22cos sin 2

n

n nx x

+ =

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20

Ch rng, bi ton ny Trn Phng c mt cch gii khc cch lp bng bin thin ca hm s, mt cch gii y " n tng":

( )( )

(2 ) 22 22 2 2 2

(2 ) 22 22 2 2 2

sin sin 2 ...2 2 sin .2 sin

cos cos 2 ...2 2 cos .2 cos

n n n n n

n n nn

n n n n n

n n nn

x x n x n x

x x n x n x

+ + + =

+

+ + + =

2 2 22 22 2 2

22

2(sin cos ) ( 2).2 .2 (sin cos ) .2

sin cos .2 (1)

n n n

n n

n

n n

x x n n x x n

x x n

+ + + =

+

rng sinx > 0, cosx > 0. Du ng thc (1) xy ra khi ch khi cosx = sinx

4x

pi = .

V. BIN LUN PHNG TRNH V H PHNG TRNH BNG CCH XT CC DU HIU CN V

VD1. Tm tt c cc nghim nguyn ca phng trnh

2 12 1 36x x x+ + + = HD. Pt cho 212 1 36x x x+ =

Du hiu cn: x = 0 l nghim th 21 0 1 1451 6

236 0x

xx x

+ +

> = > 1 x (2) T (1)&(2)suy ra 22 1x x x x+ > + Nh th x = 0 l nghim duy nht. Vy m = 0 tho. VD3. Tm tt c cc gi tr m phng trnh sau c nghim duy nht 4 5x x m + + = HD. Du hiu cn: x l nghim 4 5x x m + = 4 ( 1 ) 5 (1 )x x x m + + =

Trong 2 v trn c n - 2

hng t 22n

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21

- 1 - x l nghim vy, cn Pt cho c nghim duy nht l x = - 1 - x x = - 1

2 m = 3 2

Du hiu : Khi m = 3 2 pt cho tr thnh 4 5 3 2x x + + = Gii Ptrnh ny thy c ng mt nghim x = - 1

2. Suy ra m = 3 2 tho.

VD4. Tm tt c cc gi tr m phng trnh sau c nghim duy nht

4 4 1 1x x x x m+ + + = . HD. Du hiu cn: x l nghim 4 4 1 1x x x x m+ + + = 4 41 1 (1 ) 1 1 (1 )x x x x m + + + = 1 - x l nghim vy, cn Pt cho c nghim duy nht l x = 1 - x x = 1

2

m = 4 41 1 1 12 2 2 2

+ + + = 441 12 2 8 2 2 2 22 2

+ = + = +

Du hiu : Khi m = 2 2 2+ pt cho tr thnh:

4 4 1 1x x x x+ + + = 2 2 2+

Ta c ( )4 4 1 2 1 2 2( 1 ) 2 2x x x x x x+ + + = 1 2( 1 ) 2x x x x+ + =

Nh th Pt tng ng vi 4 4 1 1

21

x xx

x x

= =

= l nghim duy nht.

Suy ra m = 2 2 2+ tho.

VD5. Tm tt c cc gi tr a h phng trnh sau c nghim vi mi b

( ) ( )2 22

1 1 2

1

a yx b

a bxy x y

+ + + = + + =

HD. Du hiu cn: H c nghim vi mi b th c nghim vi b = 0.

Khi h tr thnh ( )22

1 1 (1)1 (2)

a

x

a x y

+ = + =

T (1) suy ra x = 0 th a tu .T (2) suy ra a = 1 Cng t (1) suy ra x 0 th a = 0. Du hiu :

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22

i) a = 0 : h tr thnh ( )22

1 1 (3)1 (4)

yb

bxy x y

+ = + =

Khi b 0 : (3) y = 0 khng tho (4). Suy ra a = 0 khng tho. ii) a = 1 : h tr thnh ( )2 2

2

1 1 (3)0 (4)

yx b

bxy x y

+ + = + =

Khi b = 0 th (4) x = y = 0 tho (3) vi mi b. Suy ra a = 1 tho. Bi tp tng t: BT1. Tm m phng trnh sau c nghim duy nht: 3 6x x m+ + = BT2. Tm a h sau c nghim duy nht:

2

2 2

2

1

xx y x a

x y

+ = + +

+ =

BT3. Tm a h sau c nghim duy nht:

2

2 2

1 sin

tan 1

ax a y x

x y

+ =

+ =

BT4. Tm a h sau c nhiu hn 5 nghim:

2 2

2 2

( )1 0

x y a x y x y a

x y bxy + + = +

+ + =

BT5. Tm x phng trnh sau nghim ng vi mi a: 2

2 3 2 22 2log ( 5 6 ) log (3 1)aa x a x x x+ =

VI. BIN LUN NGHIM CA PHNG TRNH BNG PHNG PHP DNG MIN, MAX. Vi f(x) lin tc trn D, phng trnh f(x) = m c nghim khi v ch khi m thuc tp gi tr ca f(x). Vi f(x) lin tc trn D th t gi tr ln nht v nh nht trn D. Khi phng trnh f(x) = m c nghim khi v ch khi: min ( ) max ( )

x D x Df x m f x

VD1. Tm tt c cc gi tr m phng trnh sau c nghim

2 21 1x x x x m+ + + = HD. t f(x) = 2 21 1x x x x+ + + , x R . Cch 1. f '(x) =

2 2

2 1 2 11 1

x x

x x x x

+

+ + +

-12

x 12

2x + 1 0, 2x - 1 0 f '(x) > 0

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23

x > 12

: f '(x) > 0 2 2

2 1 2 1 01 1

x x

x x x x

+ > >

+ + +

2 2

2 2(2 1) (2 1)

1 1x x

x x x x

+ >

+ + + 2 2 2 2(2 1) ( 1) (2 1) ( 1)x x x x x x+ + > + + x > 0

Vy x > 12

f '(x) > 0

x < - 12

: f '(x) > 0 0 > 2 2

2 1 2 11 1

x x

x x x x

+ >

+ + +

2 2

2 2(2 1) (2 1)

1 1x x

x x x x

+ 0, x [ ]3;6 . f(-3) = - 3, f(6) = 3, f(x) lin tc trn [ ]3;6 [ ] [ ]3;6 3;6min ( ) 3,max ( ) 3f x f x = = Suy ra, Pt cho c nghim khi v ch khi - 3 m 3.

VD4. Tm tt c cc gi tr m phng trnh sau c nghim

sinx + cosx = m

2sinx + cosx + 3

HD. t y = sinx + cosx 2sinx + cosx + 3

Vi mi x: 2sinx 2 , cosx 1 2sin cos 3x x + > (du ng thc khng xy ra v sinx v cosx khng ng thi nhn gi tr - 1) Suy ra 2sin cos 3 0,x x x+ + TX: R. Ta tm tp gi tr ca hm s: y l mt gi tr thuc tp gi tr phng trnh y = sinx + cosx

2sinx + cosx + 3c nghim.

Ptrnh y = sinx + cosx 2sinx + cosx + 3

(2y - 1)sinx + (y - 1)cosx + 3y = 0 Ptrnh ny c nghim khi v ch khi (2y - 1)2 + (y - 1)2 9y2 2y2 + 3y - 1 0

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25

3 17 3 17

4 4y + .

Suy ra, tp gi tr ca y: 3 17 3 17;4 4

+

Bi tp tng t: BT1. Tm m phng trnh sau c nghim 2 21 1x x x x m+ + + + = .

BT2. Tm m phng trnh sau c nghim 22cos sin 2

2sin 2 3cos 2 5x x

mx x

+=

+ .

VII. PHNG PHP TO V HNH HC VD1. Cho h phng trnh

2 2

2

2(1 )( ) 4x y a

x y

+ = +

+ =

1) Gii h khi a = 1. 2) Tm tt c cc gi tr ca a h c ng hai nghim. HD. Cch 1.

2 2 2 2

2

2(1 ) 2(1 )( 2)( 2) 0( ) 4

x y a x y ax y x yx y

+ = + + = +

+ + + =+ =

2 2 2 2 2

2 2 2 2 2

2(1 ) (2 ) 2(1 ) 2 1 0 (1)2 0 2 2 (2)

2(1 ) (2 ) 2(1 ) 2 1 0 (3)2 0 2 2

x y a x x a x x ax y y x y x

x y a x x a x x ax y y x y x

+ = + + = + + =

+ = = = + = + + + = + + + = + + = = = (4)

1) a = 1: H cho tr thnh

2

2

2 0 0 2 2 2

0 2 2 0 2 2

x x x x

y x y x

x xx x

y xy x

= = =

= = = = + = = =

Suy ra 4 nghim (0; 2), (2; 0), (0; - 2), (- 2; 0). 2) H c ng hai nghim. Nhn xt rng (1) v (3) c cng bit s ' = a. Suy ra a 0 a > 0: Mi phng trnh (1) v (3) c 2 nghim phn bit, trong khi t (2) v (4) ta c 2 - x - 2 - x vi x nn h c t nht 4 nghim. Suy ra a > 0 khng tho.

a = 0: H (1)&(2) c nghim (1; 1), h (3)&(4) c nhim (- 1; - 1). Vy a = 0 tho. Cch 2 (PP Hnh hc). Thy ngay a 0 . Trong h to -cc Oxy: Xem Pt 2 2 2(1 )x y a+ = + , a 0 l Pt ng trn (O, R), R = 2(1 )a+ Xem (x + y)2 = 4 (x + y - 2)(x + y + 2) = 0 l phng trnh hai ng thng:

1 : x + y - 2 = 0, 2 : x + y + 2 = 0

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26

Hai ng thng ny i xng nhau qua O. Pt c ng hai nghim 1 tip xc vi (O, R)( do 2 cng tip xc vi (O, R)) d(O, 1 ) = R 0 0 2 2(1 )2 a

+ = + a = 0.

VD2. Cho h phng trnh 2 200

x ay a

x y x

+ =

+ =

1) Tm tt c cc gi tr ca a h c hai nghim phn bit. 2) Gi hai nghim l 1 1 2 2(x ; y ), (x ; y ) l hai nghim. Chng minh rng:

2 21 2 1 2(x - x ) + (y - y ) 1

HD. 1) Trong h to -cc Oxy: Xem phng trnh x + ay - a = 0 l phng trnh ng thng d. Xem phng trnh x2 + y2 - x = 0 l phng trnh ng trn I( 1

2; 0), R = 1

2.

H c hai nghim phn bit khi ch khi ng thng ct ng trn ti hai im phn bit

d(I, d) < R 2 2 22

11 42 1 1 2 1 4 4 1 02 31

a

a a a a a aa

< + > + > + < - 6 1 2' 9 0, ' 6 0a a = + > = + > . suy ra cc nghim ca pt cho: 3 9, 2 6x a x a= + = + VD3. Cho h phng trnh:

2

2

2

ax bx c yay by c zaz bz c x

+ + =

+ + = + + =

, trong 20, ( 1) 4 0.a b ac <

Chng minh h phng trnh trn v nghim. HD. (Chng minh phn chng) Khng mt tnh tng qut, gi s a > 0. Gi s h c nghim (x0, y0, z0). Khi :

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29

20 0 020 0 020 0 0

(1) (2) (3)

ax bx c yay by c zaz bz c x

+ + =

+ + = + + =

Cng tng v (1)(2)(3) ta c:

2 2 20 0 0 0 0 0( 1) ( 1) ( 1) 0ax b x c ay b y c az b z c + + + + + + + + = (4)

t 2( ) ( 1)f t at b t c= + + th (4) f(x0) + f(y0) +f(z0) = 0 (5) Do 20, ( 1) 4 0.a b ac < af(t) > 0, t. V a > 0 nn f(t) > 0, t f(x0) > 0, f(y0) > 0, f(z0) > 0. Tri vi (5). Vy, h cho v nghim.

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