Chuyen de 02 PT BPT HPT N.M.hieu Mathvn.com

Chuyn 2

Phng Trnh - Bt Phng Trnh & HPhng Trnh i S

1. Phng Trnh - Bt Phng Trnh Khng Cha CnBi tp 2.1. Gii cc bt phng trnh sau

a) x2 6x + 6 > 0. b) 4x2 + x 2 0.c) x4 4x3 + 3x2 + 8x 10 0. d) x4 + x2 + 4x 3 0.

Li gii.

a) Ta c x2 6x + 6 > 0[x > 3 +

3

x < 33 . Vy tp nghim S =(; 33) (3 +3; +).

b) Ta c = 31 < 0 4x2 + x 2 < 0,x R. Vy bt phng trnh v nghim.c) Bt phng trnh tng ng vi

x4 + 3x2 10 4x3 + 8x 0 (x2 2) (x2 + 5) 4x (x2 2) 0 (x2 2) (x2 4x + 5) 0 x2 2 0 2 x 2

Vy bt phng trnh c tp nghim S =[2;2].

d) Bt phng trnh tng ng vi

x4 + 2x2 + 1 x2 4x + 4 (x2 + 1)2 (x 2)2 (x2 + x 1) (x2 x + 3) 0 x2 + x 1 0 [ x 1+52

x 15

2

Vy bt phng trnh c tp nghim S =(; 1

5

2

][1+5

2 ; +).

Bi tp 2.2. Gii cc bt phng trnh sau

a)x 2

x2 9x + 8 0. b)x2 3x 2

x 1 2x + 2.

c)x + 5

2x 1 +2x 1x + 5

> 2. d)1

x2 5x + 4 0

x2 12x + 362x2 + 9x 5 > 0.

Ta c bng xt du

x 5 12 6 +x2 12x + 36 + | + | + 0 +2x2 + 9x 5 + 0 0 + | +

VT + || || + 0 +Vy bt phng trnh c tp nghim S = (;5) ( 12 ; 6) (6; +).d) Bt phng trnh tng ng vi

x2 7x + 10 x2 + 5x 4(x2 5x + 4) (x2 7x + 10) < 0

2x + 6(x2 5x + 4) (x2 7x + 10) < 0.

Ta c bng xt du

x 1 2 3 4 5 +2x + 6 + | + | + 0 | |

x2 5x + 4 + 0 | | 0 + | +x2 7x + 10 + | + 0 | | 0 +

VT + || || + 0 || + || Vy bt phng trnh c tp nghim S = (1; 2) (3; 4) (5; +).

Bi tp 2.3. Gii cc phng trnh saua) x3 5x2 + 5x 1 = 0. b) x3 33x2 + 7x3 = 0.c) x4 4x3 x2 + 16x 12 = 0. d) (x 3)3 + (2x + 3)3 = 18x3.e)(x2 + 1

)3+ (1 3x)3 = (x2 3x + 2)3. f) (4 + x)2 (x 1)3 = (1 x) (x2 2x + 17).

Li gii.

a) Ta c x3 5x2 + 5x 1 = 0 (x 1) (x2 4x + 1) = 0 [ x = 1x = 23 .

Vy phng trnh c ba nghim x = 1, x = 23.b) Ta c x3 33x2 + 7x3 = 0 (x3) (x2 23x + 1) = 0 [ x = 3

x =

32 .Vy phng trnh c ba nghim x =

3, x =

32.

c) Ta c x4 4x3 x2 + 16x 12 = 0 (x 1) (x3 3x2 4x + 12) = 0 x = 1x = 3x = 2

.

Vy phng trnh c ba nghim x = 1, x = 3, x = 2.d) Phng trnh tng ng vi

(x 3 + 2x + 3)3 3(x 3)(2x + 3)(x 3 + 2x + 3) = 18x3 9x3 9x (2x2 3x 9) = 0 9x (7x2 + 3x + 9) = 0 x = 0

Vy phng trnh c nghim duy nht x = 0.e) Phng trnh tng ng vi(

x2 + 1 + 1 3x)3 3(x2 + 1)(1 3x)(x2 + 1 + 1 3x) = (x2 3x + 2)3 3(x2 + 1)(1 3x)(x2 3x + 2) = 0

x = 13x = 1x = 2

Vy phng trnh c ba nghim x = 13 , x = 1, x = 2.f) Phng trnh tng ng vi

(4 + x)2

= (x 1)3 (x 1) (x2 2x + 17) (4 + x)2 = (x 1) (x2 2x + 1 x2 + 2x 17) = 0 x2 + 8x + 16 = 16x + 16 x2 + 24x = 0

[x = 0x = 24

Vy phng trnh c hai nghim x = 0, x = 24.

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Chuyn 2. Phng Trnh - Bt Phng Trnh & H Phng Trnh i S

Bi tp 2.4. Gii cc phng trnh saua)(x2 4x + 3)2 (x2 6x + 5)2 = 0. b) x4 = (2x 5)2.

c) x4 + 3x2 + 3 = 2x. d) x4 4x 1 = 0.e) x4 = 6x2 12x + 8. f) x4 = 2x3 + 3x2 4x + 1.

Li gii.

a) Ta c(x2 4x + 3)2 (x2 6x + 5)2 = 0 (2x2 10x + 8) (2x 2) = 0 [ x = 1

x = 4.

Vy phng trnh c hai nghim x = 1, x = 4.b) Ta c x4 = (2x 5)2 (x2 + 2x 5) (x2 2x + 5) = 0 x = 16.Vy phng trnh c hai nghim x = 16.c) Ta c x4 + 3x2 + 3 = 2x (x2 + 2)2 = (x + 1)2 (x2 + x + 3) (x2 x + 1) = 0.Vy phng trnh v nghim.d) Ta c x4 4x 1 = 0 (x2 + 1)2 = 2(x + 1)2 (x2 +2x + 1 +2) (x2 2x + 12) = 0.Vy phng trnh c hai nghim x =

2

4

2 22

.

e) Ta c x4 = 6x2 12x + 8 (x2 1)2 = (2x 3)2 (x2 + 2x 4) (x2 2x + 2) = 0 x = 15.Vy phng trnh c hai nghim x = 15.f) Ta c x4 = 2x3 + 3x2 4x + 1 (x2 x)2 = (2x 1)2 (x2 + x 1) (x2 3x + 1) = 0 [ x = 152

x = 35

2

.

Vy phng trnh c bn nghim x =15

2, x =

352

.

Bi tp 2.5. Gii cc phng trnh saua) (x + 3)4 + (x + 5)4 = 2. b) (x + 1)4 + (x + 3)4 = 16.c) (x + 3)4 + (x 1)4 = 82. d) x4 + (x 1)4 = 418 .

Li gii.a) t x + 4 = t. Phng trnh tr thnh

(t 1)4 + (t + 1)4 = 2 2t4 + 12t2 = 0[t2 = 0t2 = 6 (loi) t = 0

Vi t = 0 x = 4. Vy phng trnh c nghim duy nht x = 4.b) t x + 2 = t. Phng trnh tr thnh

(t 1)4 + (t + 1)4 = 16 2t4 + 12t2 14 = 0[t2 = 1t2 = 7 (loi) t = 1

Vi t = 1 x = 1; t = 1 x = 3. Vy phng trnh c hai nghim x = 1, x = 3.c) t x + 1 = t. Phng trnh tr thnh

(t + 2)4

+ (t 2)4 = 16 2t4 + 48t2 50 = 0[t2 = 1t2 = 25 (loi) t = 1

Vi t = 1 x = 0; t = 1 x = 2. Vy phng trnh c hai nghim x = 0, x = 2.d) t x 12 = t. Phng trnh tr thnh(

t +1

2

)4+

(t 1

2

)4=

41

8 2t4 + 3t2 5 = 0

[t2 = 1t2 = 52 (loi)

t = 1

Vi t = 1 x = 32 ; t = 1 x = 12 . Vy phng trnh c hai nghim x = 32 , x = 12 .Bi tp 2.6. Gii cc phng trnh sau

a) (x + 1) (x + 2) (x + 3) (x + 4) = 3. b)(x2 1) (x + 3) (x + 5) + 16 = 0.

c) (x 1) (x 2) (x 3) (x 6) = 3x2. d) (x2 2x + 4) (x2 + 3x + 4) = 14x2.Li gii.

a) Phng trnh tng ng vi

(x + 1) (x + 4) (x + 2) (x + 3) = 3 (x2 + 5x + 4) (x2 + 5x + 6) = 3t x2 + 5x + 4 = t. Phng trnh tr thnh t (t + 2) = 3

[t = 1t = 3 .

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Nguyn Minh Hiu

Vi t = 1 x2 + 5x + 4 = 1 x = 5

13

2; t = 3 x2 + 5x + 4 = 3 (v nghim).

Vy phng trnh c hai nghim x =513

2.

b) Phng trnh tng ng vi

(x 1) (x + 5) (x + 1) (x + 3) + 16 = 0 (x2 + 4x 5) (x2 + 4x + 3)+ 16 = 0t x2 + 4x 5 = t. Phng trnh tr thnh t (t + 8) + 16 = 0 t = 4.Vi t = 4 x2 + 4x 5 = 4 x = 25. Vy phng trnh c hai nghim x = 25.c) Nhn thy x = 0 khng phi l nghim phng trnh. Vi x 6= 0, phng trnh tng ng vi

(x 1) (x 6) (x 2) (x 3) = 3x2 (x2 7x + 6) (x2 5x + 6) = 3x2(x 7 + 6

x

)(x 5 + 6

x

)= 3

t x 7 + 6x = t. Phng trnh tr thnh t (t + 2) = 3[t = 1t = 3 .

Vi t = 1 x 7 + 6x = 1 x2 8x + 6 = 0 x = 4

10;t = 3 x 7 + 6x = 3 x2 4x + 6 = 0 (v nghim).

Vy phng trnh c hai nghim x = 410.d) Nhn thy x = 0 khng phi l nghim phng trnh. Vi x 6= 0, phng trnh tng ng vi(

x 2 + 4x

)(x + 3 +

4

x

)= 14

t x 2 + 4x = t. Phng trnh tr thnh t (t + 5) = 14[t = 2t = 7 .

Vi t = 2 x 2 + 4x = 2 x2 4x+ 4 = 0 x = 2; t = 7 x 2 + 4x = 7 x2 + 5x+ 4[x = 1x = 4 .

Vy phng trnh c ba nghim x = 2, x = 1, x = 4.Bi tp 2.7. Gii cc phng trnh sau

a) x4 4x3 + 6x2 4x + 1 = 0. b) 2x4 + 3x3 9x2 3x + 2 = 0.c) 2x4 + 3x3 27x2 + 6x + 8 = 0. d) x4 5x3 + 8x2 10x + 4 = 0.

Li gii.a) Nhn thy x = 0 khng phi l nghim phng trnh. Vi x 6= 0, phng trnh tng ng vi

x2 4x + 6 4x

+1

x2= 0 x2 + 1

x2 4

(x +

1

x

)+ 6 = 0

t x +1

x= t x2 + 1

x2= t2 2. Phng trnh tr thnh t2 2 4t + 6 = 0 t = 2.

Vi t = 2 x + 1x

= 2 x2 2x + 1 = 0 x = 1.Vy phng trnh c nghim duy nht x = 1.b) Nhn thy x = 0 khng phi l nghim phng trnh. Vi x 6= 0, phng trnh tng ng vi

2x2 + 3x 9 3x

+2

x2= 0 2

(x2 +

1

x2

)+ 3

(x 1

x

) 9 = 0

t x 1x

= t x2 + 1x2

= t2 + 2. Phng trnh tr thnh 2(t2 + 2

)+ 3t 9 = 0

[t = 1t = 52

.

Vi t = 1 x 1x

= 1 x2 x 1 = 0 x = 1

5

2.

Vi t = 52 x 1

x= 5

2 2x2 + 5x 2 = 0 x = 5

41

4


2, x =

5414

.

c) Nhn thy x = 0 khng phi l nghim phng trnh. Vi x 6= 0, phng trnh tng ng vi

2x2 + 3x 27 + 6x

+8

x2= 0 2

(x2 +

4

x2

)+ 3

(x +

2

x

) 27 = 0

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t x +2

x= t x2 + 4

x2= t2 4. Phng trnh tr thnh 2 (t2 4)+ 3t 27 = 0 [ t = 5

t = 72.

Vi t = 5 x + 2x

= 5 x2 + 5x + 2 = 0 x = 5

17

2.

Vi t =7

2 x + 2

x=

7

2 2x2 7x + 4 = 0 x = 7

17

4.


2, x =

7174

.

d) Nhn thy x = 0 khng phi l nghim phng trnh. Vi x 6= 0, phng trnh tng ng vi

x2 5x + 8 10x

+4

x2= 0 x2 + 4

x2 5

(x +

2

x

)+ 8 = 0

t x +2

x= t x2 + 4

x2= t2 4. Phng trnh tr thnh t2 4 5t + 8 = 0

[t = 4t = 1

.

Vi t = 4 x + 2x

= 4 x2 4x + 2 = 0 x = 2

2.

Vi t = 1 x + 2x

= 1 x2 x + 2 = 0 (v nghim).Vy phng trnh c hai nghim x = 22.

Bi tp 2.8. Gii cc phng trnh saua)(x2 + 5x

)2 2 (x2 + 5x) 24 = 0. b) (x2 + x + 1) (x2 + x + 2) = 12.c)(x2 2x 2)2 2x2 + 3x + 2 = 0. d) (4x + 3)2 (x + 1) (2x + 1) = 810.

Li gii.

a) t x2 + 5x = t. Phng trnh tr thnh t2 2t 24 = 0[t = 6t = 4 .

Vi t = 6 x2 + 5x = 6[x = 1x = 6 . Vi t = 4 x

2 + 5x = 4[x = 1x = 4 .

Vy phng trnh c bn nghim x = 1, x = 4, x = 6.b) t x2 + x + 1 = t. Phng trnh tr thnh t(t + 1) = 12

[t = 3t = 4 .

Vi t = 3 x2 + x + 1 = 3[x = 1x = 2 . Vi t = 4 x

2 + x + 1 = 4 (v nghim).Vy phng trnh c hai nghim x = 1, x = 2.c) Phng trnh tng ng vi (x2 2x 2)2 (x2 2x 2) x2 + x = 0.t x2 2x 2 = t. Phng trnh tr thnh

t2 t x2 + x = 0 (t x)(t + x) (t x) = 0 (t x)(t + x 1) = 0[t = xt = 1 x

Vi t = x x2 2x 2 = x x = 3

17

2; t = 1 x x2 2x 2 = 1 x x = 1

13

2.


2, x =

1132

.

d) Phng trnh tng ng vi(16x2 + 24x + 9

) (2x2 + 3x + 1

)= 810 [8(2x2 + 3x + 1) + 1] (2x2 + 3x + 1) = 810

t 2x2 + 3x + 1 = t. Phng trnh tr thnh (8t + 1) t = 810[t = 10t = 818

.

Vi t = 10 2x2 + 3x + 1 = 10[x = 3x = 32

. Vi t = 818 2x2 + 3x + 1 = 818 (v nghim).Vy phng trnh c hai nghim x = 3, x = 32 .

Bi tp 2.9. Gii cc phng trnh sau

a)1

2x2 x + 1 +1

2x2 x + 3 =6

2x2 x + 7 . b)4x

4x2 8x + 7 +3x

4x2 10x + 7 = 1.

c)x2 + 1

x+

x

x2 + 1= 5

2. d)

(x 1x + 2

)2+x 3x + 2

2(x 3x 1

)2= 0.

e) x2 +(

x

x + 1

)2= 3. f)

(1

x2 + x + 1

)2+

(1

x2 + x + 2

)2=

13

36.

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Nguyn Minh Hiu

Li gii.a) t 2x2 x + 1 = t (t > 0). Phng trnh tr thnh

1

t+

1

t + 2=

6

t + 6 (t + 2) (t + 6) + t (t + 6) = 6t (t + 2) 4t2 2t 12 = 0

[t = 2t = 32 (loi)

Vi t = 2 2x2 x + 1 = 2[x = 1x = 12

.

Vy phng trnh c hai nghim x = 1, x = 12 .b) Nhn thy x = 0 khng phi l nghim phng trnh. Vi x 6= 0, phng trnh tng ng vi

4

4x 8 + 7x+

3

4x 10 + 7x= 1

t 4x 8 + 7x = t. Phng trnh tr thnh

4

t+

3

t 2 = 1 4 (t 2) + 3t = t (t 2) t2 9t + 8 = 0

[t = 1t = 8

Vi t = 1 4x 8 + 7x = 1 4x2 9x + 7 = 0 (v nghim).Vi t = 8 4x 8 + 7x = 8 4x2 16x + 7 = 0

[x = 12x = 72

.

Vy phng trnh c hai nghim x = 12 , x =72 .

c) iu kin: x 6= 0.t

x2 + 1

x= t. Phng trnh tr thnh t + 1t = 52

[t = 2t = 12

.

Vi t = 2 x2 + 1

x= 2 x2 + 2x + 1 = 0 x = 1.

Vi t = 12 x

2 + 1

x= 1

2 2x2 + x + 2 = 0 (v nghim).

Vy phng trnh c nghim x = 1.d) iu kin: x 6= 1, x 6= 2.t

x 1x + 2

= u,x 3x 1 = v. Phng trnh tr thnh u

2 + uv 2v2 = 0[u = vu = 2v .

Vi u = v x 1x + 2

=x 3x 1 x

2 2x + 1 = x2 x 6 x = 7.

Vi u = 2v x 1x + 2

= 2.x 3x 1 x

2 2x + 1 = 2x2 + 2x + 12 3x2 4x 11 = 0 x = 2

37

3.

Vy phng trnh c ba nghim x = 7, x =237

3.

e) iu kin: x 6= 1. Phng trnh tng ng vi(x x

x + 1

)2+ 2x.

x

x + 1= 3

(x2

x + 1

)2+ 2

x2

x + 1 3 = 0

tx2

x + 1= t. Phng trnh tr thnh t2 + 2t 3 = 0

[t = 1t = 3 .

Vi t = 1 x2

x + 1= 1 x2 x 1 = 0 x = 1

5

2.

Vi t = 3 x2

x + 1= 3 x2 + 3x + 3 = 0 (v nghim).


2.

f) Phng trnh tng ng vi(1

x2 + x + 1 1x2 + x + 2

)2+ 2.

1

x2 + x + 1.

1

x2 + x + 2=

13

36

(

1

(x2 + x + 1) (x2 + x + 2)

)2+

2

(x2 + x + 1) (x2 + x + 2) 13

36= 0

t1

(x2 + x + 1) (x2 + x + 2)= t (t > 0). Phng trnh tr thnh t2 + 2t 13

36= 0

[t = 16t = 136 (loi)

.

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Vi t =1

6 1

(x2 + x + 1) (x2 + x + 2)=

1

6 (x2 + x + 1) (x2 + x + 2) = 6.

t x2 + x + 1 = u (u > 0). Phng trnh tr thnh u (u + 1) = 6[u = 2u = 3 (loi) .

Vi u = 2 x2 + x + 1 = 2 x = 1

5

2.


2.

Bi tp 2.10. Gii cc phng trnh saua) |x 1| = x2 3x + 1. b) x2 + 4x 5 = x2 + 5.c)x2 5x + 4 x = 4. d) x2 + 4x + 4 = 5 x2.

e)x2 5x + 4 = x2 + 6x + 5. f) x2 5x + 5 = 2x2 + 10x 11.

Li gii.

a) Ta c |x 1| = x2 3x + 1 [ x 1 = x2 3x + 1x 1 = x2 + 3x 1

x = 22x = 0x = 2

.

Vy phng trnh c bn nghim x = 22, x = 0, x = 2.

b) Ta cx2 + 4x 5 = x2 + 5 [ x2 + 4x 5 = x2 + 5

x2 + 4x 5 = x2 5 x = 52x = 0x = 2

.

Vy phng trnh c ba nghim x = 52 , x = 0, x = 2.c) Vi x2 5x + 4 0

[x 4x 1 , phng trnh tr thnh x

2 5x + 4 x = 4[x = 0x = 6

(tha mn).

Vi x2 5x+ 4 < 0 1 < x < 4, phng trnh tr thnh x2 + 5x 4 x = 4 x2 4x+ 8 = 0 (v nghim).Vy phng trnh c hai nghim x = 0, x = 6.d) Phng trnh tng ng vi |x + 2| = 5 x2.Vi x + 2 0 x 2, phng trnh tr thnh x + 2 = 5 x2 x2 + x 3 = 0

[x = 1+

13

2

x = 113

2 (loi)

Vi x + 2 < 0 x < 2, phng trnh tr thnh x 2 = 5 x2 x2 x 7 = 0[x = 1+

29

2 (loi)x = 1

29

2

Vy phng trnh c hai nghim x =1 +13

2, x =

1292

.

e) Vi x2 5x + 4 0[x 4x 1 , phng trnh tr thnh x

2 5x + 4 = x2 + 6x + 5 x = 111 (tha mn).Vi x2 5x + 4 < 0 1 < x < 4, PT tr thnh x2 + 5x 4 = x2 + 6x + 5 2x2 + x + 9 = 0 (v nghim).Vy phng trnh c hai nghim x = 111 .

f) Vi x2 5x+ 5 0[x 5+

5

2

x 55

2

, PT tr thnh x2 5x+ 5 = 2x2 + 10x 11 x = 1533

2 (tha mn).

Vi x2 5x + 5 < 0 55

2 < x 0). Phng trnh tr thnh3

t2 t 2 = 0 t3 + 2t 3 = 0 t = 1.

Vi t = 1 x + 12x 1

= 1 |x + 1| = |2x 1| [ x + 1 = 2x 1x + 1 = 2x + 1 [x = 2x = 0

.

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Nguyn Minh Hiu


c) Ta cx2 + 3x 10+ x2 4 = 0 { x2 + 3x 10 = 0

x2 4 = 0 [x = 2x 5

x = 2 x = 2.

Vy phng trnh c nghim duy nht x = 2.

d) Ta cx2 + 3x 4+x2011 + 2011x 2012 = 0 { x2 + 3x 4 = 0

x2011 + 2011x 2012 = 0 [x = 1 (tha mn)x = 4 (loi)

x2011 + 2011x 2012 = 0.



a) |x 2| < |2x + 1|. b)2x 3x 3

1.c)x2 5x + 4 x2 + 6x + 5. d) x2 2x+ x2 4 > 0.

Li gii.

a) Ta c |x 2| < |2x + 1| (x 2)2 < (2x + 1)2 3x2 + 8x 3 > 0[x > 13x < 3 .

Vy bt phng trnh c tp nghim S = (;3) ( 13 ; +).b) iu kin: x 6= 3. Bt phng trnh tng ng vi

|2x 3| |x 3| (2x 3)2 (x 3)2 3x2 6x 0 0 x 2 (tha mn)Vy bt phng trnh c tp nghim S = [0; 2].

c) Vi x2 5x + 4 0[x 4x 1 , bt phng trnh tr thnh

x2 5x + 4 x2 + 6x + 5 x 111 S1 =

[ 1

11; 1

] [4; +)

Vi x2 5x + 4 < 0 1 < x < 4, bt phng trnh tr thnhx2 + 5x 4 x2 + 6x + 5 2x2 + x + 9 0 (ng x (1; 4)) S2 = (1; 4)

Vy bt phng trnh c tp nghim S = S1 S2 =[ 111 ; +).

d) Vi x2 2x 0[x 2x 0 , bt phng trnh tr thnh

x2 2x + x2 4 > 0[x > 2x < 1 (tha mn) S1 = (;1) (2; +)

Vi x2 2x < 0 0 < x < 2, bt phng trnh tr thnhx2 + 2x + x2 4 > 0 x > 2 (loi) S2 =

Vy bt phng trnh c tp nghim S = S1 S2 = (;1) (2; +).Bi tp 2.13. Gii cc phng trnh sau

a) |9 x| = |6 5x|+ |4x + 3|. b) x2 5x + 4+ x2 5x = 4.c) |7 2x| = |5 3x|+ |x + 2|. d) |x 1| 2 |x 2|+ 3 |x 3| = 4.e)x2 2x + 1 +x2 + 4x + 4 = 5. f)

x + 2

x 1 +

x 2x 1 = 2.

Li gii.a) Ta c bng xt du

x 34 65 9 +9 x + | + | + 0 6 5x + | + 0 | 4x + 3 0 + | + | +

Vi x (; 34], phng trnh tr thnh 9 x = 6 5x 4x 3 x = 34 (tha mn).Vi x ( 34 ; 65], phng trnh tr thnh 9 x = 6 5x + 4x + 3 9 = 9 (ng ,x ( 34 ; 65]).Vi x ( 65 ; 9], phng trnh tr thnh 9 x = 6 + 5x + 4x + 3 x = 65 (loi).Vi x (9; +), phng trnh tr thnh 9 + x = 6 + 5x + 4x + 3 x = 34 (loi).Vy phng trnh c tp nghim S =

[ 34 ; 65].b) Ta c bng xt du

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x 0 1 4 5 +x2 5x + 4 + | + 0 0 + | +x2 5x + 0 | | 0 +

Vi x (; 0], phng trnh tr thnh x2 5x + 4 + x2 5x = 4[x = 0 (tha mn)x = 5 (loi)

.

Vi x (0; 1], phng trnh tr thnh x2 5x + 4 x2 + 5x = 4 4 = 4 (ng ,x (0; 1]).Vi x (1; 4], phng trnh tr thnh x2 + 5x 4 x2 + 5x = 4

[x = 4 (tha mn)x = 1 (loi)

.

Vi x (4; 5], phng trnh tr thnh x2 5x + 4 x2 + 5x = 4 4 = 4 (ng ,x (4; 5]).Vi x (5; +), phng trnh tr thnh x2 5x + 4 + x2 5x = 4

[x = 0 (loi)x = 5 (loi) .

Vy phng trnh c tp nghim S = [0; 1] [4; 5].c) Ta c bng xt du

x 2 53 72 +7 2x + | + | + 0 5 3x + | + 0 | x + 2 0 + | + | +

Vi x (;2], phng trnh tr thnh 7 2x = 5 3x x 2 x = 2 (tha mn).Vi x (2; 53], phng trnh tr thnh 7 2x = 5 3x + x + 2 7 = 7 (ng ,x (2; 53]).Vi x ( 53 ; 72], phng trnh tr thnh 7 2x = 5 + 3x + x + 2 x = 53 (loi).Vi x ( 72 ; +), phng trnh tr thnh 7 + 2x = 5 + 3x + x + 2 x = 2 (loi).Vy phng trnh c tp nghim S =

[2; 53].d) Ta c bng xt du

x 1 2 3 +x 1 0 + | + | +x 2 | 0 + | +x 3 | | 0 +

Vi x (; 1], phng trnh tr thnh x + 1 2 (x + 2) + 3 (x + 3) = 4 x = 1 (tha mn).Vi x (1; 2], phng trnh tr thnh x 1 2 (x + 2) + 3 (x + 3) = 4 4 = 4 (ng ,x (1; 2]).Vi x (2; 3], phng trnh tr thnh x 1 2 (x 2) + 3 (x + 3) = 4 x = 2 (loi).Vi x (3; +), phng trnh tr thnh x 1 2 (x 2) + 3 (x 3) = 4 x = 5 (tha mn).Vy phng trnh c tp nghim S = [1; 2] {5}.e) Phng trnh tng ng vi |x 1|+ |x + 2| = 5.Ta c bng xt du

x 2 1 +x 1 | 0 +x + 2 0 + | +

Vi x (;2], phng trnh tr thnh x + 1 x 2 = 5 x = 3 (loi).Vi x (2; 1], phng trnh tr thnh x + 1 + x + 2 = 5 3 = 5 (v l).Vi x (1; +), phng trnh tr thnh x 1 + x + 2 = 5 x = 2 (tha mn).Vy phng trnh c nghim x = 2.f) Phng trnh tng ng vi

x 1 + 1 + x 1 1 = 2.

Vix 1 1 0 x 2, PT tr thnh x 1 + 1 +x 1 1 = 2 x 1 = 1 x = 2 (tha mn).

Vix 1 1 < 0 1 x < 2, PT tr thnh x 1 + 1x 1 + 1 = 2 2 = 2 (ng x [1; 2)).

Vy phng trnh c tp nghim S = [1; 2].

2. Phng Trnh - Bt Phng Trnh Cha CnBi tp 2.14. Gii cc phng trnh sau

a) xx 1 7 = 0. b) 2x + 9 = 4 x +3x + 1.c)

3x 35 x = 2x 4. d)

2x +

6x2 + 1 = x + 1.e) 3

2x 1 + 3x 1 = 33x + 1. f) 3x + 1 + 3x + 2 + 3x + 3 = 0.

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Li gii.a) Phng trnh tng ng vi

x 1 = x 7

{x 7x 1 = x2 14x + 49

x 7[ x = 5 (loi)x = 10

x = 10

Vy phng trnh c nghim duy nht x = 5.b) iu kin: 13 x 4. Phng trnh tng ng vi

2x + 9 = 4 x + 3x + 1 + 2

(4 x) (3x + 1) 4 = 23x2 + 11x + 4

3x2 + 11x + 4 = 4[x = 0x = 113

(tha mn).

Vy phng trnh c hai nghim x = 0, x = 113 .c) iu kin: 2 x 5. Phng trnh tng ng vi

3x 3 = 5 x +2x 4 3x 3 = 5 x + 2x 4 + 2

(5 x) (2x 4)

2x 4 = 2

(5 x) (2x 4) (2x 4)2 = 4 (5 x) (2x 4)

(2x 4) (2x 4 20 + 4x) = 0[x = 2x = 4

(tha mn).

Vy phng trnh c hai nghim x = 2, x = 4.d) Phng trnh tng ng vi

{x + 1 02x +

6x2 + 1 = x2 + 2x + 1

{

x 16x2 + 1 = x4 + 2x2 + 1

x 1 x = 0x = 2x = 2 (loi)

[x = 0x = 2

Vy phng trnh c hai nghim x = 0, x = 2.e) Phng trnh tng ng vi

2x 1 + x 1 + 3 3

(2x 1) (x 1) ( 32x 1 + 3x 1) = 3x + 1 3

(2x 1) (x 1) (3x + 1) = 1 6x3 7x2 = 0[x = 0x = 76

Th li ta thy x = 0 khng phi l nghim phng trnh. Vy phng trnh c nghim duy nht x = 76 .f) Phng trnh tng ng vi

3x + 1 + 3

x + 2 = 3x + 3 x + 1 + x + 2 + 3 3

(x + 1) (x + 2)

(3x + 1 + 3

x + 2

)= x 3

3

(x + 1) (x + 2) (x + 3) = x + 2 (x + 1) (x + 2) (x + 3) = x + 2 x = 2Th li ta thy x = 2 l nghim phng trnh. Vy phng trnh c nghim duy nht x = 2.

Bi tp 2.15. Gii cc bt phng trnh saua)x2 4x 12 > 2x + 3. b) x2 4x 12 x 4.

c) 3

6x 9x2 < 3x. d) x3 + 1 x + 1.Li gii.

a) Bt phng trnh tng ng vi{

2x + 3 < 0x2 4x 12 0{2x + 3 0x2 4x 12 > 4x2 + 12x + 9

x <

32[

x 6x 2{

x 323 < x < 73

x 2

Vy bt phng trnh c tp nghim S = (;2].b) Bt phng trnh tng ng vi x 4 0x2 4x 12 0

x2 4x 12 x2 8x + 16

x 4[x 6x 2

x 7 6 x 7

Vy bt phng trnh c tp nghim S = [6; 7].c) Bt phng trnh tng ng vi 6x 9x2 < 27x3 27x3 + 9x2 6x > 0. Ta c bng xt du

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x 23 0 13 +VT 0 + 0 0 +

Vy bt phng trnh c tp nghim S =( 23 ; 0) ( 13 ; +).

d) Bt phng trnh tng ng vi

{

x + 1 < 0x3 + 1 0{x + 1 0x3 + 1 x2 + 2x + 1

{

x < 1x3 1 (v nghim) x 1[ 1 x 0x 2

[ 1 x 0x 2

Vy bt phng trnh c tp nghim S = [1; 0] [2; +).Bi tp 2.16. Gii cc bt phng trnh sau

a) (C-09)x + 1 + 2

x 2 5x + 1. b) (A-05) 5x 1x 1 > 2x 4.

c)

2x +

6x2 + 1 > x + 1. d) (A-04)

2 (x2 16)x 3 +

x 3 > 7 x

x 3 .

Li gii.a) iu kin: x 2. Bt phng trnh tng ng vi

x + 1 + 4 (x 2) + 4

(x + 1) (x 2) 5x + 1 x2 x 2 4 2 x 3Kt hp iu kin bt phng trnh c tp nghim S = [2; 3].b) iu kin: x 2. Bt phng trnh tng ng vi

5x 1 > x 1 +2x 4 5x 1 > x 1 + 2x 4 + 2

(x 1) (2x 4)

x + 2 >

(x 1) (2x 4) x2 + 4x + 4 > 2x2 6x + 4 0 < x < 10

Kt hp iu kin bt phng trnh c tp nghim S = [2; 10).c) Bt phng trnh tng ng vi

{x + 1 < 0

2x +

6x2 + 1 0{x + 1 02x +

6x2 + 1 > x2 + 2x + 1

{

x < 16x2 + 1 2x{

x 16x2 + 1 > x4 + 2x2 + 1

{

x < 16x2 + 1 4x2 (ng,x R) x 1[ x < 2

0 < x < 2

[x < 10 < x < 2

Vy bt phng trnh c tp nghim S = (;1) (0; 2).d) iu kin: x 4. Bt phng trnh tng ng vi

2 (x2 16) + x 3 > 7 x

2 (x2 16) > 10 2x

10 2x < 0{ 10 2x 02x2 32 > 100 40x + 4x2

x > 5{ x 5

1034 < x < 10 +34[x > 5

1034 < x 5 x > 10

34

Kt hp iu kin bt phng trnh c tp nghim S =(1034; +).

Bi tp 2.17. Gii cc phng trnh saua) (D-05) 2

x + 2 + 2

x + 1x + 1 = 4. b)

x 1 + 2x 2

x 1 2x 2 = 1.

c) x +x + 12 +

x + 14 = 9. d)

x + 2

x 1 +

x 2x 1 = x + 3

3.

Li gii.a) Phng trnh tng ng vi 2

(x + 1 + 1

)x + 1 = 4 x + 1 = 2 x = 3.Vy phng trnh c nghim x = 3.

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b) Phng trnh tng ng vix 2 + 1 x 2 1 = 1 x 2 x 2 1 = 0

Vix 2 1 0 x 3, PT tr thnh x 2x 2 + 1 = 0 1 = 0 (v l).

Vix 2 1 < 0 2 x < 3, PT tr thnh x 2 +x 2 1 = 0 4(x 2) = 1 x = 94 (tha mn).

Vy phng trnh c nghim x = 94 .c) Phng trnh tng ng vi

x +

x +

1

4+

1

2= 9

x +

1

4=

17

2 x

{

172 x 0x + 14 =

2894 17x + x2

x

172[

x = 12 (loi)x = 6

x = 6

d) Phng trnh tng ng vix 1 + 1 + x 1 1 = x+33 .

Vix 1 1 0 x 2, phng trnh tr thnh

x 1 + 1 +x 1 1 = x + 3

3 6x 1 = x + 3

{

x + 3 036(x 1) = x2 + 6x + 9

{x 3x = 15 65 x = 15 6

5

Vix 1 1 < 0 1 x < 2, phng trnh tr thnh

x 1 + 1x 1 + 1 = x + 3

3 6 = x + 3 x = 3 (loi)

Vy phng trnh c nghim x = 15 65.Bi tp 2.18. Gii cc bt phng trnh sau

a)

x4 +x 4 8 x. b) (D-02) (x2 3x)2x2 3x 2 0.

c) (x 2)x2 + 4 < x2 4. d) (x + 2)9 x2 x2 2x 8.e)x2 3x + 2 +x2 4x + 3 2x2 5x + 4. f) x2 + x 2 +x2 + 2x 3 x2 + 4x 5.

Li gii.a) Bt phng trnh tng ng vi

x + 4x 4 16 2x x 4 + 2 16 2x x 4 14 2x

{

14 2x < 0x 4 0{14 2x 0x 4 196 56x + 4x2

{

x > 7x 4{x 7254 x 8

[x > 7254 x 7

x 254

Vy bt phng trnh c tp nghim S =[254 ; +

).

b) Bt phng trnh tng ng vi

2x2 3x 2 = 0{ 2x2 3x 2 > 0x2 3x 0

x = 2x = 12[x > 2x < 12[x 3x 0

x = 2x = 12x 3x < 12

x = 2x 3x 12

Vy bt phng trnh c tp nghim S =(; 12] [3; +) {2}.

c) Bt phng trnh tng ng vi

(x 2)x2 + 4 < (x 2) (x + 2) (x 2)

(x2 + 4 x 2

)< 0

{

x 2 > 0x2 + 4 < x + 2{

x 2 < 0x2 + 4 > x + 2

{

x > 2x2 + 4 < x2 + 4x + 4{x < 2x2 + 4 > x2 + 4x + 4

[x > 2x < 0

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Vy bt phng trnh c tp nghim S = (; 0) (2; +).d) Bt phng trnh tng ng vi

(x + 2)

9 x2 (x + 2) (x 4) (x + 2)(

9 x2 x + 4) 0

{

x + 2 09 x2 x 4{

x + 2 09 x2 x 4

x 2x 4 09 x2 09 x2 x2 8x + 16

(v nghim)

{x 29 x2 0

3 x 2

Vy bt phng trnh c tp nghim S = [3;2].e) iu kin:

[x 4x 1 . BPT tng ng vi

(x 1) (x 2) +(x 1) (x 3) 2(x 1) (x 4).

Nhn thy x = 1 l nghim ca bt phng trnh.Vi x 4, bt phng trnh tr thnh x 1 (x 2 +x 3 2x 4) 0 x 2+x 3 2x 4.Vx 2 > x 4 v x 3 > x 4 nn bt phng trnh nghim ng x [4; +).

Vi x < 1, bt phng trnh tr thnh

1 x (2 x +3 x 24 x) 0 2 x+3 x 24 x.V

2 x < 4 x v 3 x > 4 x nn bt phng trnh v nghim.Vy bt phng trnh cho c tp nghim S = [4; +) {1}.f) iu kin:

[x 1x 5 . BPT tng ng vi

(x 1) (x + 2) +(x 1) (x + 3) 2(x 1) (x + 5).

Nhn thy x = 1 l nghim ca bt phng trnh.Vi x > 1, bt phng trnh tr thnh

x 1 (x + 2 +x + 3x + 5) 0 x + 2 +x + 3 x + 5

x + 2 + x + 3 + 2

(x + 2) (x + 3) x + 5 2

(x + 2) (x + 3) x (v nghim)

Vi x 5, bt phng trnh tr thnh

1 x (x 2 +x 3x 5) 0 x 2 +x 3 x 5 x 2 x 3 + 2

(x + 2) (x + 3) x 5 2

(x + 2) (x + 3) x

Vy bt phng trnh cho c nghim x = 1.

Bi tp 2.19. Gii cc phng trnh saua) (D-06)

2x 1 + x2 3x + 1 = 0. b)

7 x2 + xx + 5 = 3 2x x2.

c)

2x2 + 8x + 6 +x2 1 = 2x + 2. d) 3

(2 +x 2) = 2x +x + 6.

e) x2 + 3x + 1 = (x + 3)x2 + 1. f)

x2 7

x2+

x 7

x2= x.

Li gii.a) Phng trnh tng ng vi

2x 1 = x2 + 3x 1

{ x2 + 3x 1 02x 1 = x4 + 9x2 + 1 6x3 + 2x2 6x

{ x2 + 3x 1 0

x4 6x3 + 11x2 8x + 2 = 0 { x2 + 3x 1 0

(x 1)2 (x2 4x + 2) = 0

x2 + 3x 1 0 x = 1x = 2 +2 (loi)x = 22

[x = 0

x = 22

Vy phng trnh c hai nghim x = 0, x = 22.b) Ta c

7 x2 + xx + 5 =

3 2x x2 7 x2 + xx + 5 = 3 2x x2

xx + 5 = 2x 4 x2 (x + 5) = 4x2 + 16x + 16[x = 1x = 4

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Th li ta thy x = 4 khng phi l nghim phng trnh. Vy phng trnh c nghim x = 1.

c) iu kin:

x 1x = 1x 3

. Phng trnh tng ng vi

2 (x + 1) (x + 3) +

(x 1) (x + 1) = 2(x + 1)

Nhn thy x = 1 l nghim ca phng trnh.Vi x 1, phng trnh tr thnh

x + 1

(2x + 6 +

x 1 2x + 1) = 0 2x + 6 +x 1 = 2x + 1

2x + 6 + x 1 + 2

(2x + 6) (x 1) = 4 (x + 1) 2

2x2 + 4x 6 = x 1

4 (2x2 + 4x 6) = x2 2x + 1 7x2 + 18x 25 = 0 [ x = 1x = 257 (loi)

Vi x 3, phng trnh tr thnhx 1 (2x 6 +1 x 2x 1) = 0 2x 6 +1 x = 2x 1

2x 6 + 1 x + 2

(2x + 6) (x 1) = 4 (x 1) 2

2x2 + 4x 6 = 1 x

4 (2x2 + 4x 6) = x2 2x + 1 7x2 + 18x 25 = 0 [ x = 1 (loi)x = 257

d) iu kin: x 2. Phng trnh tng ng vi

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

3x 2 +x + 6 = 2x 6

8 (x 3) = 2 (x 3) (3x 2 +x + 6) 2 (x 3) (3x 2 +x + 6 4) = 0[x = 33x 2 +x + 6 = 4

[x = 3

9 (x 2) + x + 6 + 6(x 2) (x + 6) = 16[x = 3

3x2 + 4x 12 = 14 5x

x = 314 5x 09(x2 + 4x 12) = 196 160x + 25x2

x = 3x 145

16x2 196x + 304 = 0[x = 3

x = 4920972 (loi)

Vy phng trnh c nghim duy nht x = 3.e) Ta c phng trnh h qu

x4 + 9x2 + 1 + 6x3 + 2x2 + 6x =(x2 + 6x + 9

) (x2 + 1

) x4 + 6x3 + 11x2 + 6x + 1 = x4 + 6x3 + 10x2 + 6x + 9 x = 2

2

Th li ta thy x = 22 l nghim phng trnh. Vy phng trnh c hai nghim x = 22.f) Ta c phng trnh h qu

x2 7x2

= xx 7

x2 x2 7

x2= x2 + x 7

x2 2x

x 7

x2

x(

1 2x 7

x2

)= 0 2

x 7

x2= 1 4

(x 7

x2

)= 1

4x3 x2 28 = 0 x = 2

Th li ta thy x = 2 l nghim phng trnh. Vy phng trnh c nghim duy nht x = 2.


a)11 4x2

x< 3. b)

121 4x + x2x + 1

0.

c)2x

2x + 1 1 > 2x + 2. d)x2(

1 +

1 + x)2 > x 4.

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Li gii.a) iu kin x [ 12 ; 12] \ {0}. Phng trnh tng ng vi

1 (1 4x2)x(1 +

1 4x2) < 3 4x < 3 + 31 4x2 31 4x2 > 4x 3V 4x 3 < 0,x [ 12 ; 12] \ {0} nn bt phng trnh nghim ng x [ 12 ; 12] \ {0}.Vy bt phng trnh c tp nghim S =

[ 12 ; 12] \ {0}.b) iu kin: x 6= 1. Bt phng trnh tng ng vi

1 (21 4x + x2)(x + 1)

(1 +

21 4x + x2) 0 x2 + 4x 20x + 1 0 x + 1 < 0 x < 1Kt hp iu kin bt phng trnh c tp nghim S = (;1).c) iu kin: x 12 , x 6= 0. Bt phng trnh tng ng vi

2x + 1 + 1 > 2x + 2 2x + 1 > 2x + 1 2x + 1 > 4x2 + 4x + 1 1

2< x < 0

Kt hp iu kin bt phng trnh c tp nghim S =( 12 ; 0).

d) iu kin: x 1. Nhn thy x = 0 l nghim ca bt phng trnh.Vi x 6= 0, bt phng trnh tng ng vi(

1x + 1)2 > x 4 1 + x + 1 2x + 1 > x 4 x + 1 < 3 x < 8Kt hp iu kin bt phng trnh c tp nghim S = [1; 8).

Bi tp 2.21. Gii cc phng trnh saua) (x + 5) (2 x) = 3x2 + 3x. b)

(x + 1) (2 x) = 1 + 2x 2x2.

c)x + 1 +

4 x +(x + 1) (4 x) = 5. d) 3x 2 +x 1 = 4x 9 + 23x2 5x + 2.

Li gii.a) Phng trnh tng ng vi x2 3x + 10 = 3x2 + 3x x2 + 3x + 3x2 + 3x 10 = 0.tx2 + 3x = t (t 0). Phng trnh tr thnh t2 + 3t 10 = 0

[t = 2t = 5 (loi) .

Vi t = 2 x2 + 3x = 2 x2 + 3x 4 = 0[x = 1x = 4 .

Vy phng trnh c hai nghim x = 1, x = 4.b) Phng trnh tng ng vi

2 + x x2 = 1 + 2 (x x2).

t

2 + x x2 = t (t 0). Phng trnh tr thnh t = 1 + 2 (t2 2) 2t2 t 3 = 0 [ t = 1 (loi)x = 32

Vi t = 32

2 + x x2 = 32 4(2 + x x2) = 9 4x2 4x + 1 = 0 x = 12 .

Vy phng trnh c nghim duy nht x = 12 .c) iu kin: 1 x 4. t x + 1 +4 x = t (t 0)(x + 1) (4 x) = t252 . Phng trnh tr thnh

t +t2 5

2= 5 t2 + 2t 15 = 0

[t = 3t = 5 (loi)

Vi t = 3 x2 + 3x + 4 = 2 x2 + 3x + 4 = 4[x = 0x = 3

(tha mn).

Vy phng trnh c nghim x = 0, x = 3.d) iu kin: t 1. t 3x 2 +x 1 = t (t 0) 4x+ 23x2 5x + 2 = t2 + 3. Phng trnh tr thnh

t = t2 + 3 9 t2 t 6 = 0[t = 3t = 2 (loi)

Vi t = 3

3x2 5x + 2 = 3 2x{

x 323x2 5x + 2 = 9 12x + 4x2 x =

7212

(tha mn).

Vy phng trnh c nghim x =721

2.

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Bi tp 2.22. Gii cc phng trnh sau

a) x +

4 x2 = 2 + 3x4 x2. b) (x 3) (x + 1) + 4 (x 3)

x+1x3 = 3.

c)4

x2+

x2

4 x2 +5

2

(4 x2x

+x

4 x2

)+ 2 = 0. d) (B-2011) 3

2 + x62 x+44 x2 = 103x.

Li gii.a) iu kin: 2 x 2. t x +4 x2 = t x4 x2 = t242 . Phng trnh tr thnh

t = 2 +3(t2 4)

2 3t2 2t 8 = 0

[t = 2t = 43

Vi t = 2 4 x2 = 2 x 4 x2 = 4 4x + x2 [x = 0x = 2

(tha mn).

Vi t = 43

4 x2 = 43 x{

x 439(4 x2) = (4 + 3x)2

{x 43x = 2

14

3

x = 214

3 (tha mn).

b) iu kin:[x > 3x 1 . t (x 3)

x+1x3 = t (x 3) (x + 1) = t2.

Phng trnh tr thnh t2 + 4t + 3 = 0[t = 1t = 3 .

Vi t = 1 (x 3)

x+1x3 = 1

{x < 3(x 3) (x + 1) = 1

{x < 3

x = 15 x = 1

5 (tha mn).

Vi t = 3 (x 3)

x+1x3 = 3

{x < 3(x 3) (x + 1) = 9

{x < 3

x = 113 x = 1

13 (tha mn).

Vy phng trnh c hai nghim x = 15, x = 113.c) iu kin: 2 < x < 2, x 6= 0. t

4 x2x

+x

4 x2 = t4 x2x2

+x2

4 x2 = t22 4

x2+

x2

4 x2 = t21.

Phng trnh tr thnh t2 1 + 52t + 2 = 0

[t = 2t = 12

.

Vi t = 2

4 x2x

+x

4 x2 = 2 4 = 2x

4 x2 {

x < 0

x = 2 x =

2 (tha mn).

Vi t = 12

4 x2x

+x

4 x2 = 1

2 4 = 1

2x

4 x2 {

x < 0x4 4x2 + 64 = 0 (v nghim).

d) iu kin: 2 x 2. Phng trnh tng ng vi 3 (2 + x 22 x)+ 44 x2 = 10 3x.t

2 + x 22 x = t 44 x2 = 10 3x t2. Phng trnh tr thnh 3t t2 = 0[t = 0t = 3

.

Vi t = 0 2 + x = 22 x 2 + x = 4 (2 x) x = 65 (tha mn).Vi t = 3 2 + x = 22 x + 3 122 x = 5x 15 (v nghim v 5x 15 < 0,x [2; 2]).Vy phng trnh c nghim duy nht x = 65 .

Bi tp 2.23. Gii cc phng trnh saua) x2 + 3x + 2 2x2 + 3x + 5. b) x2 +2x2 + 4x + 3 6 2x.c) x (x + 1)x2 + x + 4 + 2 0. d) x2 2x + 8 6

(4 x) (2 + x) 0.

e)x

x + 1 2x + 1

x> 3. f)

x + 2 +

x 1 + 2x2 + x 2 11 2x.

Li gii.

a) tx2 + 3x + 5 = t (t 0). Bt phng trnh tr thnh t2 3 2t

[t 3t 1 (loi) .

Vi t 3 x2 + 3x + 5 9[x 1x 4 . Vy bt phng trnh c tp nghim S = (;4] [1; +).

b) t

2x2 + 4x + 3 = t (t 0). Bt phng trnh tr thnh t232 + t 6[t 3t 5 (loi) .

Vi t 3 2x2 + 4x + 3 9[x 1x 3 . Vy bt phng trnh c tp nghim S = (;3] [1; +).

c) tx2 + x + 4 = t (t 0). Bt phng trnh tr thnh t2 4 t + 2 0

[t 2t 1 (loi) .

Vi t 2 x2 + x + 4 4[x 0x 1 . Vy bt phng trnh c tp nghim S = (;1] [0; +).

d) Bt phng trnh tng ng vi x2 2x + 8 68 + 2x x2 0.

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t

8 + 2x x2 = t (t 0). Bt phng trnh tr thnh 8 t2 + 8 6t 0[t 2t 8 (loi) .

Vi t 2 8 + 2x x2 4 15 < x < 1 +5.Vy bt phng trnh c tp nghim S =

(15; 1 +5).

e) iu kin:[x > 0x < 1 . t

x + 1

x= t (t > 0). Bt phng trnh tr thnh

1

t2 2t > 3 2t3 + 3t2 1 < 0 (t + 1)2 (2x 1) < 0 t < 1

2

Vi t 0) v2 u2 = x2 3x + 2. Phng trnh tr thnh

2(v2 u2) = 3uv 2u2 + 3uv 2v2 = 0 [ u = 2v (loi)

v = 2u

Vi v = 2u x2 2x + 4 = 2x + 2 x2 2x + 4 = 4 (x + 2) x = 313.Vy phng trnh c hai nghim x = 313.

Bi tp 2.26. Gii cc phng trnh saua) x2 +

x + 5 = 5. b) x3 + 2 = 3 3

3x 2.

c) x3 + 1 = 2 3

2x 1. d) x 335 x3 (x + 335 x3) = 30.Li gii.

a) tx + 5 = t (t 0). Phng trnh tr thnh

{x2 = t + 5 (1)t2 = x + 5 (2) .

Tr theo v (2) v (1) ta c t2 x2 = x + t (x + t) (t x 1) = 0[t = xt = x + 1

.

Vi t = x x + 5 = x{ x 0

x + 5 = x2{

x 0x = 1

21

2

x = 1

21

2.

Vi t = x + 1 x + 5 = x + 1{

x + 1 0x + 5 = x2 + 2x + 1

{

x 1x = 1

17

2

x = 1 +

17

2.


2, x =

1 +172

.

b) t 3

3x 2 = t. Phng trnh tr thnh{

x3 + 2 = 3t (1)t3 + 2 = 3x (2) .

Tr theo v (1) v (2) ta c

x3 t3 = 3t 3x (x t) (x2 + xt + t2) = 3 (t x) (x t) (x2 + xt + t2 + 3) = 0 [ x = t

x2 + xt + t2 + 3 = 0 (v nghim)

Vi t = x 33x 2 = x 3x 2 = x3 [x = 1x = 2 . Vy phng trnh c hai nghim x = 1, x = 2.

c) t 3

2x 1 = t. Phng trnh tr thnh{

x3 + 1 = 2t (1)t3 + 1 = 2x (2) .

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x3 t3 = 2t 2x (x t) (x2 + xt + t2) = 2 (t x) (x t) (x2 + xt + t2 + 2) = 0 [ x = t

x2 + xt + t2 + 2 = 0 (v nghim)

Vi t = x 32x 1 = x 2x 1 = x3 [x = 1

x = 15

2

.

Vy phng trnh c ba nghim x = 1, x =15

2.

d) t 3

35 x3 = t. Phng trnh tr thnh{xt(x + t) = 30t3 + x3 = 35

{

xt(x + t) = 30

(t + x)3 3xt (x + t) = 35

{xt(x + t) = 30

(t + x)3

= 125{

xt = 6t + x = 5

[x = 2x = 3


Bi tp 2.27. Gii cc phng trnh, bt phng trnh sau

a) (B-2012) x + 1 +x2 4x + 1 3x. b) (A-2010) x

x

12 (x2 x + 1) 1.c) 3x2 2 = 2 x3. d) x +

3 (1 x2) = 2 (1 2x2).

Li gii.

a) iu kin:[

0 x 23x 2 +3 . Nhn thy x = 0 l mt nghim ca bt phng trnh.

Vi x > 0, bt phng trnh tng ng vix +

1x

+

x +

1

x 4 3.

tx +

1x

= t (t > 0) x + 1x = t2 2, bt phng trnh tr thnh

t2 6 3 t

3 t < 0{ 3 t 0t2 6 9 6t + t2

[t > 352 t 3

t 52

Vi t 52 x + 1

x 5

2 2x 5x + 2 0

[ x 2x 12

[x 40 < x 14

.

Kt hp ta c tp nghim ca bt phng trnh l S =[0; 14] [4; +).

b) iu kin: x 0. Nhn thy x2 x + 1 34

2 (x2 x + 1) > 1. Do PT tng ng vi

xx 1

2 (x2 x + 1)

2x2 2x + 2 1 +x x

{

1 +x x 0

2x2 2x + 2 1 + x + x2 + 2x 2x 2xx {

x x 11 + x + x2 2x 2x + 2xx 0

{

x x 1(1x x)2 0

{ x x 1

1x x = 0 {

x x 1x = 1 x

x x 1

1 x 0x = 1 2x + x2

{

x 1x = 3

5

2

x = 3

5

2

c) iu kin: x 32. Nhn thy 2 x3 0 3x2 2 0[x 2x 2 .

T iu kin ta c x 2. Khi phng trnh tng ng vi (x2 2)2 = (2 x3)3 x4 4x2 + 4 = 8 12x3 + 6x6 x9 = 0 x9 6x6 + x4 + 12x3 4x2 4 = 0

x9 5x6 (x3 1

2x

)2+ 12x3 15

4x2 4 = 0 (v nghim).

Vy phng trnh v nghim.

Bi tp 2.28. Gii cc phng trnh saua)

4x 1 +4x2 1 = 1. b) x 1 = x3 4x + 5.c)

2x 1 +x2 + 3 = 4 x. d) x5 + x3 1 3x + 4 = 0.e) x3 + 4x (2x + 7)2x + 3 = 0. f) (C-2012) 4x3 + x (x + 1)2x + 1 = 0.

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Nguyn Minh Hiu

Li gii.a) iu kin: x 12 . Nhn thy x = 12 l mt nghim ca phng trnh.Xt hm s y =

4x 1 +4x2 1 trn [ 12 ; +) c y = 24x1 + 4x4x21 > 0,x ( 12 ; +).

Do hm s ng bin trn[12 ; +

)suy ra x = 12 l nghim duy nht ca phng trnh.

Vy phng trnh c nghim duy nht x = 12 .b) iu kin: x 1. Phng trnh tng ng vi x 1 + x3 + 4x = 5.Nhn thy x = 1 l mt nghim ca phng trnh.Xt hm s y =

x 1 + x3 + 4x trn [1; +) c y = 1

2x1 + 3x

2 + 4 > 0,x (1; +).Do hm s ng bin trn [1; +) suy ra x = 1 l nghim duy nht ca phng trnh.Vy phng trnh c nghim duy nht x = 1.c) iu kin: x 12 . Phng trnh tng ng vi

2x 1 +x2 + 3 + x = 4.

Nhn thy x = 1 l mt nghim ca phng trnh.Xt hm s y =

2x 1 +x2 + 3 + x trn [ 12 ; +) c y = 12x1 + xx2+3 + 1 > 0,x ( 12 ; +).

Do hm s ng bin trn[12 ; +

)suy ra x = 1 l nghim duy nht ca phng trnh.

Vy phng trnh c nghim duy nht x = 1.d) iu kin: x 13 . Nhn thy x = 1 l mt nghim ca phng trnh.Xt hm s y = x5 + x3 1 3x + 4 trn (; 13] c y = 5x4 + 3x2 + 3213x > 0,x (; 13).Do hm s ng bin trn

(; 13] suy ra x = 1 l nghim duy nht ca phng trnh.Vy phng trnh c nghim duy nht x = 1.e) t

2x + 3 = u (u 0). Phng trnh tr thnh x3 + 4x (u2 + 4)u = 0 x3 + 4x = u3 + 4u.

Xt hm s f(t) = t3 + 4t trn [0; +) c f (t) = 3t2 + 4 > 0,t [0; +).Do phng trnh tng ng vi

u = x 2x + 3 = x{

x 02x + 3 = x2

x = 3

Vy phng trnh c nghim duy nht x = 3.f) iu kin: x 12 . Phng trnh tng ng vi 8x3 + 2x = (2x + 2)

2x + 1.

t

2x + 1 = u (u 0). Phng trnh tr thnh 8x3 + 2x = (u2 + 1)u (2x)3 + 2x = u3 + u.Xt hm s f(t) = t3 + t trn [0; +) c f (t) = 3t2 + 1 > 0,t [0; +).Do phng trnh tng ng vi

u = 2x 2x + 1 = 2x{

x 02x + 1 = 4x2

x = 1 +

5

4

Vy phng trnh c nghim duy nht x =1 +

5

4.

Bi tp 2.29. Gii cc phng trnh saua)x2 2x + 5 +x 1 = 2. b) x 2 +4 x = x2 6x + 11.

c) 2(

x 2 1)2 +x + 6 +x 2 2 = 0. d) 5x3 + 3x2 + 3x 2 = 12x2 + 3x 12 .Li gii.

a) Phng trnh tng ng vi

(x 1)2 + 4 +x 1 = 2.

Ta c

{ (x 1)2 + 4 2x 1 0

(x 1)2 + 4 +x 1 2.

Du bng xy ra khi v ch khi

{ (x 1)2 + 4 = 2x 1 = 0

x = 1.

Vy phng trnh c nghim duy nht x = 1.b) Ta c x2 6x + 11 = (x 3)2 + 2 2 (1).Xt hm s y =

x 2 +4 x trn [2; 4] c y = 1

2x 2

1

2

4 x ; y = 0 x = 3.

Ta c y(2) =

2, y(4) =

2, y(3) = 2 max[2;4]

y = y(3) = 2 x 2 +4 x 2 (2).

T (1) v (2) ta c phng trnh tng ng vi{

x2 6x + 11 = 2x 2 +4 x = 2 x = 3.


c) iu kin: x 2. Khi 2

(x 2 1)2 0

x + 6 > 2x 2 0

2(x 2 1)2 +x + 6 +x 2 > 2.20

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Do phng trnh cho v nghim.d) Phng trnh tng ng vi

(5x 2) (x2 + x + 1) = 12

(x2 + 6x 1).

Theo bt ng thc Cauchy ta c

(5x 2) (x2 + x + 1) 12(x2 + 6x 1).

Du bng xy ra khi v ch khi

5x 2 = x2 + x + 1 x2 4x + 3 = 0[x = 1x = 3

.


3. H Phng Trnh i SBi tp 2.30. Gii cc h phng trnh sau

a){

x2 + y2 + xy = 7x + y + xy = 5

. b){

x + y + xy = 1

x3 + y3 + 3(x y)2 4 = 0 .

c) (DB-05){

x2 + y2 + x + y = 4x (x + y + 1) + y (y + 1) = 2

. d){

x2 xy + y2 = 3 (x y)x2 + xy + y2 = 7(x y)2 .

Li gii.

a) H cho tng ng vi{

(x + y)2 xy = 7

x + y + xy = 5.

t x + y = S, xy = P (S2 4P ). H tr thnh{

S2 P = 7 (1)S + P = 5 (2) .

T (2) P = 5 S thay vo (1) ta c S2 + S 12 = 0[S = 3S = 4 .

Vi S = 3 P = 2{

x + y = 3xy = 2

{

x = 2y = 1

hoc{

x = 1y = 2

. Vi S = 4 P = 9 (loi).Vy h c hai nghim (x; y) = (2; 1) v (x; y) = (1; 2).

b) H cho tng ng vi{

x + y + xy = 1

(x + y)3 3xy (x + y) + 3(x + y)2 12xy 4 = 0 .

t x + y = S, xy = P (S2 4P ). H tr thnh{

S + P = 1 (1)S3 3PS + 3S2 12P 4 = 0 (2) .

T (1) P = 1 S thay vo (2) ta c S3 3S (1 S) + 3S2 12 (1 S) 4 = 0 S = 1.Vi S = 1 P = 0

{x + y = 1xy = 0

{

x = 0y = 1

hoc{

x = 1y = 0

.

Vy h c hai nghim (x; y) = (0; 1) v (x; y) = (1; 0).

c) H cho tng ng vi{

(x + y)2 2xy + x + y = 4

(x + y)2 xy + x + y = 2 . t x + y = S, xy = P (S

2 4P ).

H tr thnh{

S2 2P + S = 4S2 P + S = 2

{P = 2S2 + S = 0

{

S = 0P = 2 hoc

{S = 1P = 2 .

Vi{

S = 0P = 2

{x + y = 0xy = 2

{x =

2

y = 2 hoc{

x = 2y =

2.

Vi{

S = 1P = 2

{x + y = 1xy = 2

{x = 1y = 2 hoc

{x = 2y = 1

.

Vy h c bn nghim (x; y) =(

2;2) , (x; y) = (2;2) , (x; y) = (1;2) v (x; y) = (2; 1).d) H cho tng ng vi

{(x y)2 + xy = 3 (x y)(x y)2 + 3xy = 7(x y)2

{(x y)2 + xy = 3 (x y)xy = 2(x y)2 .

t x y = S, xy = P . H tr thnh{S2 + P = 3SP = 2S2

{

3S2 3S = 0P = 2S2

{

S = 0P = 0

hoc{

S = 1P = 2

Vi{

S = 0P = 0

{

x y = 0xy = 0

{

x = 0y = 0

; vi{

S = 1P = 2

{

x y = 1xy = 2

{

x = 2y = 1

hoc{

x = 1y = 2 .

Vy h c ba nghim (x; y) = (0; 0) , (x; y) = (2; 1) v (x; y) = (1;2).

Bi tp 2.31. Gii cc h phng trnh sau

a){

x2 2y2 = 2x + yy2 2x2 = 2y + x . b)

x 3y = 4y

x

y 3x = 4xy

.

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Nguyn Minh Hiu

c)

2x + y =

3

x2

2y + x =3

y2

. d) (B-03)

3y =

y2 + 2

x2

3x =x2 + 2

y2

.

Li gii.

a) Xt h{

x2 2y2 = 2x + y (1)y2 2x2 = 2y + x (2) .

Tr theo v (1) v (2) ta c 3x2 3y2 = x y (x y) (3x + 3y 1) = 0[x = yy = 13x3

.

Vi x = y thay vo (1) ta c x2 = 3x[x = 0x = 3 h c nghim (x; y) = (0; 0) hoc (x; y) = (3;3).

Vi y = 13x3 thay vo (1) ta c x2 2(1 3x)

2

9= 2x +

1 3x3

9x2 3x + 5 = 0 (v nghim).Vy h c hai nghim (x; y) = (0; 0) v (x; y) = (3;3).b) H cho tng ng vi

{x2 3xy = 4y (1)y2 3xy = 4x (2) .

Tr theo v (1) v (2) ta c x2 y2 = 4y 4x (x y) (x + y + 4) = 0[x = yy = x 4 .

Vi x = y thay vo (1) ta c 2x2 = 4x[x = 0x = 2 h c nghim (x; y) = (0; 0) hoc (x; y) = (2;2).

Vi y = x 4 thay vo (1) ta c x2 3x (x 4) = 4 (x 4) x = 2 h c nghim (x; y) = (2;2).Vy h c hai nghim (x; y) = (0; 0) v (x; y) = (2;2).c) H cho tng ng vi

{2x3 + x2y = 3 (1)2y3 + xy2 = 3 (2) .

Tr theo v (1) v (2) ta c 2x3 2y3 + x2y xy2 = 0 (x y) (2x2 + 3xy + 2y2) = 0 x = y.Vi x = y thay vo (1) ta c 3x3 = 3 x = 1. Vy h c nghim duy nht (x; y) = (1; 1).d) T v phi ca cc phng trnh ta c x, y > 0. H cho tng ng vi

{3x2y = y2 + 2 (1)3xy2 = x2 + 2 (2) .

Tr theo v (1) v (2) ta c 3x2y 3xy2 = y2 x2 (x y) (3xy + x + y) = 0 x = y.Vi x = y thay vo (1) ta c 3x3 = x2 + 2 x = 1. Vy h c nghim duy nht (x; y) = (1; 1).


a){

x2 xy = 22x2 + 4xy 2y2 = 14 . b)

{x2 2xy + 3y2 = 9x2 4xy + 5y2 = 5 .

c){

x3 + y3 = 1x2y + 2xy2 + y3 = 2

. d) (DB-06){

(x y) (x2 + y2) = 13(x + y)

(x2 y2) = 25 .

Li gii.

a) H cho tng ng vi{

7x2 7xy = 14 (1)2x2 + 4xy 2y2 = 14 (2) .

Tr theo v (1) v (2) ta c 5x2 11xy + 2y2 = 0[x = 2yy = 5x

.

Vi x = 2y thay vo (1) ta c 14y2 = 14 y = 1 h c nghim (x; y) = (2; 1) hoc (x; y) = (2;1).Vi y = 5x thay vo (1) ta c 28x2 = 14 (v nghim).Vy h c hai nghim (x; y) = (2; 1) v (x; y) = (2;1).b) H cho tng ng vi

{5x2 10xy + 15y2 = 45 (1)9x2 36xy + 45y2 = 45 (2) .

Tr theo v (1) v (2) ta c 4x2 + 26xy 30y2 = 0[x = 5yx = 32y

.

Vi x = 5y thay vo (1) ta c 90y2 = 45 y = 12 h c nghim (x; y) =

( 5

2; 1

2

).

Vi y = 32x thay vo (1) ta c954 x

2 = 45 x = 619 h c nghim (x; y) =

( 6

19; 9

19

).

Vy h c bn nghim (x; y) =(

52; 1

2

), (x; y) =

( 5

2; 1

2

), (x; y) =

(619

; 919

)v (x; y) =

( 6

19; 9

19

).

c) H cho tng ng vi{

2x3 + 2y3 = 2 (1)x2y + 2xy2 + y3 = 2 (2) .

Tr theo v (1) v (2) ta c 2x3 x2y 2xy2 + y3 = 0 x = yx = yy = 2x

.

Vi x = y thay vo (1) ta c 4x3 = 2 x = 132 h c nghim (x; y) =(

132 ;

132

).

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Vi x = y thay vo (1) ta c 0 = 2 (v nghim).Vi y = 2x thay vo (1) ta c 18x3 = 2 x = 139 h c nghim (x; y) =

(139 ;

239

).

Vy h c hai nghim (x; y) =(

132 ;

132

)v (x; y) =

(139 ;

239

).

d) H cho tng ng vi{

x3 x2y + xy2 y3 = 13x3 + x2y xy2 y3 = 25

{25x3 25x2y + 25xy2 25y3 = 325 (1)13x3 + 13x2y 13xy2 13y3 = 325 (2) .

Tr theo v (1) v (2) ta c 12x3 38x2y + 38xy2 12y3 = 0 x = yx = 32yx = 23y

.

Vi x = y thay vo (1) ta c 0 = 325 (v nghim).Vi x = 32y thay vo (1) ta c

3258 y

3 = 325 y = 2 h c nghim (x; y) = (3; 2).Vi x = 23y thay vo (1) ta c 32527 y3 = 325 y = 3 h c nghim (x; y) = (2;3).Vy h c hai nghim (x; y) = (3; 2) v (x; y) = (2;3).


a){

x + y = 1x3 3x = y3 3y . b) (DB-06)

{x2 + 1 + y (y + x) = 4y(x2 + 1

)(y + x 2) = y .

c) (B-08){

x4 + 2x3y + x2y2 = 2x + 9x2 + 2xy = 6x + 6

. d) (D-09){

x (x + y + 1) 3 = 0(x + y)

2 5x2 + 1 = 0.

Li gii.

a) Xt h{

x + y = 1 (1)x3 3x = y3 3y (2) . T (1) y = x 1 thay vo (2) ta c

x3 3x = (x 1)3 3 (x 1) 2x3 + 3x2 3x 2 = 0 x = 2x = 1x = 12

Vy h c ba nghim (x; y) = (2; 1) , (x; y) = (1;2) v (x; y) = ( 12 ; 12).b) Xt h

{x2 + 1 + y(y + x) = 4y (1)(x2 + 1)(y + x 2) = y (2) . T (1) x

2 + 1 = y(4 y x) thay vo (2) ta c

y (4 y x) (x + y 2) = y y(

(x + y)2 6(x + y) + 9

)= 0

[y = 0y = 3 x

Vi y = 0 thay vo (1) ta c x2 + 1 = 0 (v nghim).

Vi y = 3 x thay vo (1) ta c x2 + x 2 = 0[x = 1x = 2 .

Vy h c hai nghim (x; y) = (1; 2) v (x; y) = (2; 5).c) H cho tng ng vi

{(x2 + xy)2 = 2x + 9 (1)x2 + 2xy = 6x + 6 (2) . T (2) xy =

6x + 6 x22

thay vo (1) ta c

(x2 +

6x + 6 x22

)2= 2x + 9 x4 + 12x3 + 48x2 + 64x = 0

[x = 0x = 4

Vi x = 0 thay vo (2) ta c 0 = 6 (v nghim).Vi x = 4 thay vo (2) ta c y = 174 .Vy h c nghim duy nht (x; y) =

(4; 174 ).d) Xt h

{x(x + y + 1) 3 = 0 (1)(x + y)2 5x2 + 1 = 0 (2)

. T (1) x + y = 3x 1 thay vo (2) ta c

(3

x 1)2 5x2

+ 1 = 0 4x2 6x

+ 2 = 0{

x 6= 02x2 6x + 4 = 0

[x = 1x = 2

Vi x = 1 thay vo (1) ta c y = 1; x = 2 thay vo (1) ta c y = 32 .Vy h c hai nghim (x; y) = (1; 1) v (x; y) =

(2; 32

).


a) (B-02){

3x y = x y

x + y =x + y + 2

. b) (A-03){

x 1x = y 1y2y = x3 + 1

.

c){

x2 + y2 + 2xyx+y = 1x + y = x2 y . d)

{6x2 3xy + x + y = 1x2 + y2 = 1

.

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Nguyn Minh Hiu

Li gii.

a) Xt h{

3x y = x y (1)

x + y =x + y + 2 (2) . iu kin: x y 0, x + y + 2 0.

Ta c (1) (x y)2 = (x y)3 (x y)2 (x y 1) = 0[x = yx = y + 1

.

Vi x = y thay vo (2) ta c 2y =

2y + 2{

y 04y2 = 2y + 2

y = 1 x = 1 (tha mn).

Vi x = y + 1 thay vo (2) ta c 2y + 1 =

2y + 3{

2y + 1 04y2 + 4y + 1 = 2y + 3

y = 12 x = 32 (tha mn).Vy h c hai nghim (x; y) = (1; 1) v (x; y) =

(32 ;

12

).

b) Xt h{

x 1x = y 1y (1)2y = x3 + 1 (2)

. iu kin: x 6= 0, y 6= 0.

Ta c (1) x2y y = xy2 x xy (x y) + x y = 0 (x y) (xy + 1) = 0[y = xy = 1x

.

Vi y = x thay vo (2) ta c 2x = x3 + 1[x = 1

x = 15

2

.

Suy ra h c nghim (x; y) = (1; 1) hoc (x; y) =(15

2 ;15

2

).

Vi y = 1x thay vo (2) ta c 2x = x3 + 1 x4 + x + 2 = 0(x2 12

)2+(x + 12

)2+ 32 = 0 (v nghim).

Vy h c ba nghim (x; y) = (1; 1) v (x; y) =(15

2 ;15

2

).

c) Xt h{

x2 + y2 + 2xyx+y = 1 (1)x + y = x2 y (2) . iu kin: x + y > 0. Ta c

(1)[(x + y)

2 2xy]

(x + y) + 2xy = x + y

(x + y)[(x + y)

2 1] 2xy (x + y 1) = 0

(x + y 1) [(x + y) (x + y + 1) 2xy] = 0

(x + y 1) (x2 + y2 + x + y) = 0 [ y = 1 xx2 + y2 + x + y = 0 (v nghim)

Vi y = 1 x thay vo (2) ta c x2 + x 2 = 0[x = 1x = 2 .

Vy h c hai nghim (1; 0) v (2; 3).d) H cho tng ng vi

{6x2 (3y 1)x + y 1 = 0 (1)x2 + y2 = 1 (2) .

Xt phng trnh (1) c = (3y 1)2 24 (y 1) = 9y2 30y + 25 = (3y 5)2.Do (1)

[x = 3y13y+512x = 3y1+3y512

[x = 13x = 12 (y 1)

.

Vi x = 13 thay vo (2) ta c19 + y

2 = 1 y = 22

3 .

Vi x = 12 (y 1) thay vo (2) ta c 14(y2 2y + 1)+ y2 = 1 [ y = 1

y = 35.

Vy h c bn nghim (x; y) =(13 ;

22

3

), (x; y) =

(13 ; 2

2

3

), (x; y) = (0; 1) v (x; y) =

( 45 ; 35).Bi tp 2.35. Gii cc h phng trnh sau

a) (DB-07){

x4 x3y x2y2 = 1x3y x2 xy = 1 . b) (D-08)

{xy + x + y = x2 2y2x

2y yx 1 = 2x 2y .

c) (D-2012){

xy + x 2 = 02x3 x2y + x2 + y2 2xy y = 0 . d)

{x3 + 2y2 = x2y + 2xy

2x2 2y 1 + 3

y3 14 = x 2 .

Li gii.

a) Xt h{

x4 x3y x2y2 = 1 (1)x3y x2 xy = 1 (2) .

Ta c (2) x2 (xy 1) = xy 1 (xy 1) (x2 1) = 0 [ x = 1y = 1x

.

Vi x = 1 thay vo (1) ta c y2 + y = 0[y = 0y = 1 . Vi x = 1 thay vo (1) ta c y

2 y = 0[y = 0y = 1

.

Vi y = 1x thay vo (1) ta c x4 x2 2 = 0 x2 = 2 x = 2 y = 1

2.

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Vy h c su nghim{

x = 1y = 0

,

{x = 1y = 1 ,

{x = 1y = 0

,

{x = 1y = 1

,

{x =

2

y =

2v

{x = 1

2

y = 12

.

b) Xt h{

xy + x + y = x2 2y2 (1)x

2y yx 1 = 2x 2y (2) . iu kin: x 1, y 0.Ta c (1) y (x + y) + x + y = (x y) (x + y) (x + y) (y + 1 x + y) = 0 x = 2y + 1.Vi x = 2y + 1 thay vo (2) ta c

(2y + 1)

2y y

2y = 2y + 2 (y + 1)

2y = 2 (y + 1)

2y = 2 y = 2 x = 5Vy h c nghim duy nht (x; y) = (5; 2).

c) Xt h{

xy + x 2 = 0 (1)2x3 x2y + x2 + y2 2xy y = 0 (2) .

Ta c (2) 2x (x2 y) y (x2 y)+ x2 y = 0 (x2 y) (2x y + 1) = 0 [ y = x2y = 2x + 1

.

Vi y = x2 thay vo (1) ta c x3 + x 2 = 0 x = 1 y = 1.Vi y = 2x + 1 thay vo (1) ta c 2x2 + 2x 2 = 0 x = 1

5

2 y =

5.

Vy h c ba nghim (x; y) = (1; 1) , (x; y) =(1+5

2 ;

5)

v (x; y) =(15

2 ;

5).

d) Xt h{

x3 + 2y2 = x2y + 2xy (1)2x2 2y 1 + 3

y3 14 = x 2 (2) . iu kin: x

2 2y + 1.

Ta c (1) x2 (x y) = 2y (x y) (x y) (x2 2y) = 0 [ x = yx2 = 2y (loi) .

Vi x = y thay vo (2) ta c 2x2 2x 1 + 3x3 14 = x 2 (*).

tx2 2x 1 = u 0, 3x3 14 = v v3 6u2 = (x 2)3.

Phng trnh (*) tr thnh v3 6u2 = (2u + v)3 2u(u2 + 3(u + v)

2+ 3u

)= 0 u = 0 x = 12.

Vy h c hai nghim (x; y) =(1 +

2; 1 +

2)v (x; y) =

(12; 12).


a){

x2 + y2 + xy = 1x3 + y3 = x + 3y

. b){

x3 + 2xy2 + 12y = 08y2 + x2 = 12

.

c) (DB-06){

x3 8x = y3 + 2yx2 3 = 3 (y2 + 1) . d) (A-2011)

{5x2y 4xy2 + 3y3 2 (x + y) = 0xy(x2 + y2

)+ 2 = (x + y)

2 .

Li gii.

a) Xt h{

x2 + y2 + xy = 1 (1)x3 + y3 = x + 3y (2) .

Thay (1) vo (2) ta c x3 + y3 =(x2 + y2 + xy

)(x + 3y) 4x2y + 4xy2 + 2y3 = 0 y = 0.

Vi y = 0 thay vo h ta c{

x2 = 1x3 = x

x = 1. Vy h c hai nghim (x; y) = (1; 0) v (x; y) = (1; 0).

b) Xt h{

x3 + 2xy2 + 12y = 0 (1)8y2 + x2 = 12 (2) .

Thay (2) vo (1) ta c x3 + 2xy2 +(8y2 + x2

)y = 0 x3 + x2y + 2xy2 + 8y3 = 0 (*)

Nhn thy y = 0 khng phi nghim ca h. Vi y 6= 0, chia hai v phng trnh (*) cho y3 ta c(x

y

)3+

(x

y

)2+ 2

x

y+ 8 = 0 x

y= 2 x = 2y

Vi x = 2y thay vo (2) ta c 12y2 = 12 y = 1 x = 2.Vy h c hai nghim (x; y) = (2;1) v (x; y) = (2; 1).c) H cho tng ng vi

{x3 y3 = 2 (4x + y)x2 3y2 = 6

{3x3 3y3 = 6 (4x + y) (1)x2 3y2 = 6 (2) .

Thay (2) vo (1) ta c 3x3 3y3 = (x2 3y2) (4x + y) x3 + x2y 12xy2 = 0 (*)Nhn thy x = 0 khng phi nghim ca h. Vi x 6= 0, chia hai v phng trnh (*) cho x3 ta c

1 +y

x 12

(yx

)2= 0

[yx =

13

yx = 14

[x = 3yx = 4y

Vi x = 3y thay vo (2) ta c 6y2 = 6 y = 1 x = 3.Vi x = 4y thay vo (2) ta c 13y2 = 6 y =

613 x = 4

614 .

Vy h c bn nghim (x; y) = (1; 3) , (x; y) = (1;3) , (x; y) =(

613 ;4

613

)v (x; y) =

(

613 ; 4

613

).

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Nguyn Minh Hiu

d) Xt h{

5x2y 4xy2 + 3y3 2 (x + y) = 0 (1)xy(x2 + y2

)+ 2 = (x + y)

2 (2).

Ta c (2) xy (x2 + y2)+ 2 = x2 + y2 + 2xy (x2 + y2) (xy 1) = 2 (xy 1) [ x = 1yx2 + y2 = 2

.

Vi x = 1y thay vo (1) ta c3y 6y + 3y3 = 0 3y4 6y2 + 3 = 0 y2 = 1 y = 1 x = 1.

Vi x2 + y2 = 2 (3) thay vo (1) ta c

5x2y 4xy2 + 3y3 (x2 + y2) (x + y) = 0 x3 4x2y + 5xy2 2y3 = 0 (*)Nhn thy y = 0 khng phi nghim ca h. Vi y 6= 0, chia hai v phng trnh (*) cho y3 ta c(

x

y

)3 4(x

y

)2+ 5

x

y 2 = 0

[ xy = 1xy = 2

[x = yx = 2y

Vi x = y thay vo (3) ta c 2y2 = 2 y = 1 x = 1.Vi x = 2y thay vo (3) ta c 5y2 = 2 y =

25 x = 2

25 .

Vy h c bn nghim (x; y) = (1; 1), (x; y) = (1;1), (x; y) =(

2

25 ;

25

)v (x; y) =

(2

25 ;

25

).


a) (B-09){

xy + x + 1 = 7yx2y2 + xy + 1 = 13y2

. b){

2x2 + x 1y = 2y y2x 2y2 = 2 .

c){

8x3y3 + 27 = 9y3

4x2y + 6x + y2 = 0. d)

{x3 y3 = 9x2 + 2y2 = x 4y .

Li gii.a) Nhn thy y = 0 khng phi nghim ca h. Vi y 6= 0, h cho tng ng vi

{x + xy +

1y = 7

x2 + xy +1y2 = 13

{

x + 1y +xy = 7(

x + 1y

)2 xy = 13

t x + 1y = S,xy = P (S

2 4P ). H tr thnh{

S + P = 7 (1)S2 P = 13 (2) .

T (1) P = 7 S thay vo (2) ta c S2 (7 S) = 13[S = 4S = 5 .

Vi S = 4 P = 3{

x + 1y = 4xy = 3

{

x = 1y = 13

hoc{

x = 3y = 1

.

Vi S = 5 P = 12 (khng tha mn). Vy h c hai nghim (x; y) = (1; 13) v (x; y) = (3; 1).b) iu kin: y 6= 0. H cho tng ng vi

{2x2 + x 1y = 21y x 2 = 2y2

{

2x2 + x 1y = 2 (1)2y2 +

1y x = 2 (2)

.

Tr theo v (1) v (2) ta c 2x2 2y2 + 2x 2y = 0(x 1y

)(x + 1y + 1

)= 0

[ 1y = x1y = x 1

.

Vi 1y = x thay vo (1) ta c 2x2 = 2 x = 1 y = 1.

Vi 1y = x 1 thay vo (1) ta c 2x2 + 2x 1 = 0 x = 13

2 y = 13

2 .

Vy h c bn nghim (x; y) = (1; 1) , (x; y) = (1;1) , (x; y) =(1+3

2 ;13

2

)v (x; y) =

(13

2 ;1+3

2

).

c) Nhn thy y = 0 khng phi nghim ca h. Vi y 6= 0, h cho tng ng vi{

8x3y3 + 27 = 9y3 (1)36x2y2 + 54xy = 9y3 (2) .

Cng theo v (1) v (2) ta c 8x3y3 + 36x2y2 + 54xy + 27 = 0 xy = 32 .Vi xy = 32 thay vo (1) ta c 0 = 9y3 y = 0 (khng tha mn). Vy h cho v nghim.d) H cho tng ng vi

{x3 y3 = 9 (1)3x2 + 6y2 = 3x 12y (2) .


x3 3x2 + 3x 1 = y3 + 6y2 + 12y + 8 (x 1)3 = (y + 2)3 y = x 3

Vi y = x 3 thay vo (2) ta c 3x2 + 6(x 3)2 = 3x 12 (x 3) 9x2 27x + 18 = 0[x = 1x = 2

.

Vy h c hai nghim (x; y) = (1;2) v (x; y) = (2;1).

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a){

x (3x + 2y) (x + 1) = 12x2 + 2y + 4x 8 = 0 . b)

{x + y xy = 3x + 1 +

y + 1 = 4

.

c) (C-2010){

2

2x + y = 3 2x yx2 2xy y2 = 2 . d) (DB-05)

{ 2x + y + 1x + y = 1

3x + 2y = 4.

e){

x2 + y2 = 5y 1 (x + y 1) = (y 2)x + y . f) (A-08)

{x2 + y + x3y + xy2 + xy = 54x4 + y2 + xy (1 + 2x) = 54

.

Li gii.

a) H cho tng ng vi{

(3x + 2y)(x2 + x

)= 12

x2 + x + 3x + 2y = 8.

t 3x + 2y = S, x2 + x = P , h tr thnh{

SP = 12S + P = 8

{

S = 2P = 6

hoc{

S = 6P = 2

.

Vi{

S = 2P = 6

{

3x + 2y = 2x2 + x = 6

3x + 2y = 2[ x = 2

x = 3{

x = 2y = 2 hoc

{x = 3y = 112

.

Vi{

S = 6P = 2

{

3x + 2y = 6x2 + x = 2

3x + 2y = 2[ x = 1

x = 2{

x = 1y = 12

hoc{

x = 2y = 4

.

Vy h c bn nghim (x; y) = (2;2) , (x; y) = (3; 112 ) , (x; y) = (1; 12) v (x; y) = (2; 4).b) iu kin: x 1, y 1, xy 0. H cho tng ng vi

{x + y xy = 3x + y + 2

x + y + xy + 1 = 14

.

t x + y = S,xy = P (P 0), h tr thnh

{S P = 3 (1)S + 2

S + P 2 + 1 = 14 (2)

.

T (1) S = P + 3 thay vo (2) ta c

P + 3 + 2P + 3 + P 2 + 1 = 14 2

P 2 + P + 4 = 11 P

{

P 114(P 2 + P + 4

)= 121 22P + P 2

[P = 3P = 353 (loi)

Vi P = 3 S = 6{

x + y = 6xy = 3

{

x = 3y = 3

. Vy h c nghim duy nht (x; y) = (3; 3).

c) H cho tng ng vi{

2

2x + y = 3 (2x + y) (1)x2 2xy y2 = 2 (2) .

t

2x + y = t (t 0). Phng trnh (1) tr thnh 2t = 3 t2 [t = 1t = 3 (loi) .

Vi t = 1 y = 1 2x thay vo (2) ta c x2 2x (1 2x) (1 2x)2 = 2[x = 1x = 3 .

Vy h c hai nghim (x; y) = (1;1) v (x; y) = (3; 7).d) t

2x + y + 1 = u,

x + y = v (u, v 0) 3x+2y = u2+v21. H cho tr thnh

{u v = 1 (1)u2 + v2 = 5 (2) .

T (1) u = v + 1 thay vo (2) ta c (v + 1)2 + v2 = 5[v = 1v = 2 (loi) .

Vi v = 1 u = 2{

2x + y + 1 = 2x + y = 1

{

x = 2y = 1 . Vy h c nghim duy nht (x; y) = (2;1).

e) iu kin: y 1, x + y 0. Xt h{

x2 + y2 = 5 (1)y 1 (x + y 1) = (y 2)x + y (2) .

ty 1 = u,x + y = v (u, v 0). Phng trnh (2) tr thnh

u(v2 1) = (u2 1) v uv (u v) + u v = 0 (u v) (uv + 1) = 0 u = v

Vi u = v y 1 = x + y x = 1. Vi x = 1 thay vo (1) ta c 1 + y2 = 5[y = 2y = 2 (loi) .

Vy h c nghim duy nht (x; y) = (1; 2).f) H cho tng ng vi

{x2 + y + xy

(x2 + y

)+ xy = 54(

x2 + y)2

+ xy = 54.

t x2 + y = S, xy = P , h tr thnh{

S + PS + P = 54 (1)S2 + P = 54 (2)

.

T (2) P = S2 54 thay vo (1) ta c S +(S2 54)S S2 54 = 54 [ S = 0S = 12 .

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Nguyn Minh Hiu

Vi S = 0 P = 54 {

x2 + y = 0xy = 54

{

x =3102

y = 31004

.

Vi S = 12 P = 32 {

x2 + y = 12xy = 32

{

x = 1y = 32

.

Vy h c hai nghim (x; y) =(

3102 ;

31004

)v (x; y) =

(1; 32

).


a){

x + 10 +y 1 = 11

x 1 +y + 10 = 11 . b){

x 1y = 8 x3(x 1)4 = y .

c) (A-2012){

x3 3x2 9x + 22 = y3 + 3y2 9yx2 + y2 x + y = 12

. d) (A-2010){ (

4x2 + 1)x + (y 3)5 2y = 0

4x2 + y2 + 2

3 4x = 7 .

Li gii.

a) iu kin: x, y 1. Xt h{

x + 10 +y 1 = 11 (1)

x 1 +y + 10 = 11 (2) .Tr theo v (1) v (2) ta c

x + 10x 1 = y + 10y 1 (*).

Xt hm s f(t) =t + 10t 1 trn [1; +) c f (t) = 1

2t+10

12t1 < 0,t (1; +).

Suy ra f(t) lun nghch bin trn [1; +). Do () f(x) = f(y) x = y.Vi x = y thay vo (1) ta c

x + 10 +

x 1 = 11 x + 10 + x 1 + 2

(x + 10) (x 1) = 121

x2 + 9x 10 = 56 x

{x 56x2 + 9x 10 = (56 x)2 x = 26 (tha mn)

Vy h c nghim duy nht (x; y) = (26; 26).

b) iu kin: x 1, y 0. Xt h{

x 1y = 8 x3 (1)(x 1)4 = y (2) .

Thay (2) vo (1) ta cx 1 (x 1)2 = 8 x3 x 1 + x3 x2 + 2x 9 = 0 (*).

Nhn thy x = 2 l mt nghim ca phng trnh (*).Xt hm s f(x) =

x 1 + x3 x2 + 2x 9 trn [1; +) c f (x) = 1

2x1 + 3x

2 2x+ 2 > 0,x (1; +).Suy ra f(x) lun ng bin trn [1; +). Do (*) c nghim duy nht x = 2 y = 1.Vy h c nghim duy nht (x; y) = (2; 1).

c) H cho tng ng vi

{(x 1)3 12 (x 1) = (y + 1)3 12 (y + 1) (1)(x 12

)2+(y + 12

)2= 1 (2)

.

T (2) suy ra{ 1 x 12 11 y + 12 1

{ 32 x 1 12 12 y + 1 32 .Xt hm s f(t) = t3 12t trn [ 32 ; 32] c f (t) = 3t2 12 < 0,t [ 32 ; 32].Suy ra f(t) lun nghch bin trn

[ 32 ; 32]. Do (1) f(x 1) = f(y + 1) x 1 = y + 1 y = x 2.Vi y = x 2 thay vo (2) ta c (x 12)2 + (x 32)2 = 1 4x2 8x + 3 = 0 [ x = 12x = 32 .Vy h c hai nghim (x; y) =

(12 ; 32

)v (x; y) =

(32 ; 12

).

d) iu kin: x 34 , y 52 . H cho tng ng vi{ (

4x2 + 1)

2x = (6 2y)5 2y (1)4x2 + y2 + 2

3 4x = 7 (2) .

t

5 2y = u (u 0), phng trnh (1) tr thnh (4x2 + 1) 2x = (u2 + 1)u (2x)3 + 2x = u3 + u (*).Xt hm s f(t) = t3 + t trn [0; +) c f (t) = 3t2 + 1 > 0,t [0; +).Suy ra f(t) lun ng bin trn [0; +). Do () f(2x) = f(u) 2x = u 2x = 5 2y

{x 0y = 54x

2

2

.

Vi y = 54x2

2 thay vo (2) ta c 4x2 +

(54x2

2

)2+ 2

3 4x 7 = 0 4x4 6x2 + 23 4x 34 = 0 (**).Nhn thy x = 12 l mt nghim ca phng trnh (**).Xt hm s f(x) = 4x4 6x2 + 23 4x 34 trn

[0; 34].

Ta c f (x) = 16x3 12x 434x = 4x

(4x2 3) 4

34x < 0,x [0; 34] f(x) ng bin trn [0; 34].

Do phng trnh (**) c nghim duy nht x = 12 y = 2.Vy h cho c nghim duy nht (x; y) =

(12 ; 2).

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4. Phng Trnh - Bt Phng Trnh & H Cha Tham SBi tp 2.40. Tm m phng trnh

(m5)x2 3mx + m + 1 = 0.

a) C nghim. b) V nghim c) C hai nghim tri du.

Li gii. Vi m =

5, phng trnh tr thnh 35x +5 + 1 = 0.Vi m 6= 0, ta c = 9m2 4 (m5) (m + 1) = 5m2 + (45 4)m + 45 > 0,m 6= 5.a) Phng trnh c nghim vi mi m R.b) Khng c gi tr no ca m phng trnh v nghim.c) Phng trnh c hai nghim tri du

(m5) (m + 1) < 0 1 < m < 5.

Bi tp 2.41. Tm m phng trnh x2 + 2 (m + 1)x + 9m 5 = 0 c hai nghim m phn bit.

Li gii. Ta c = (m + 1)2 (9m 5) = m2 7m + 6.Phng trnh c hai nghim m phn bit khi v ch khi

> 0

S < 0P > 0

m

2 7m + 6 > 02 (m + 1) < 09m 5 > 0

[m > 6m < 1

m > 1m > 59

[m > 659 < m < 1

Vy vi m ( 59 ; 1) (6; +) th phng trnh cho c hai nghim m phn bit.Bi tp 2.42. Tm m phng trnh (m 2)x2 2mx + m + 3 = 0 c hai nghim dng phn bit.Li gii. Nhn thy m = 2 khng tha mn yu cu bi ton. Vi m 6= 2 ta c = m2 (m 2) (m + 3) = 6m.

Khi phng trnh c hai nghim dng phn bit khi v ch khi

> 0

S > 0P > 0

6m > 02mm2 > 0m+3m2 > 0

m < 6[m > 2m < 0[m > 2m < 3

[

2 < m < 6m < 3

Vy vi m (;3) (2; 6) th phng trnh cho c hai nghim dng phn bit.Bi tp 2.43. Tm m phng trnh (m 2)x4 2 (m + 1)x2 + 2m 1 = 0.

a) C mt nghim. b) C hai nghim phn bit. c) C bn nghim phn bit.

Li gii. Vi m = 2 phng trnh c hai nghim phn bit.Vi m 6= 0, t x2 = t 0 phng trnh tr thnh (m 2) t2 2 (m + 1) t + 2m 1 = 0.t f(t) = (m 2) t2 2 (m + 1) t + 2m 1 c = (m + 1)2 (m 2) (2m 1) = m2 + 7m 1.a) Phng trnh cho c mt nghim

[f(t) c nghim kp bng 0f(t) c mt nghim 0 v mt nghim m

{

= 0f(0) = 0 > 0

f(0) = 0S < 0

{ m2 + 7m 1 = 0

2m 1 = 0m2 + 7m 1 > 02m 1 = 02(m+1)m2 < 0

m = 12

Vy vi m = 12 th phng trnh cho c mt nghim.

b) Phng trnh cho c hai nghim phn bit m = 2f(t) c nghim kp dngf(t) c hai nghim tri du

m = 2{

= 0S > 0

P < 0

m = 2{ m2 + 7m 1 = 0

2(m+1)m2 > 0

2m1m2 < 0

m = 2m = 7+352

12 < m < 2

Vy vi m ( 12 ; 2] { 7+352 } th phng trnh cho c hai nghim phn bit.29

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Nguyn Minh Hiu

c) Phng trnh cho c bn nghim phn bit f(t) c hai nghim dng phn bit

> 0S > 0P > 0

m2 + 7m 1 > 02(m+1)m2 > 0

2m1m2 > 0

735

2 < m 2m < 1[m > 2m < 12

2 < m < 7 + 3

5

2

Vy vi m (

2; 7+35

2

)th phng trnh cho c bn nghim phn bit.

Bi tp 2.44. (D-04) Tm m h{

x +y = 1

xx + y

y = 1 3m c nghim.

Li gii. H cho tng ng vi{

x +y = 1(

x +y)3 3xy (x +y) = 1 3m

{ x +y = 1

xy = m.

Suy rax,y l hai nghim khng m ca phng trnh t2 t + m = 0 (*).

Do h cho c nghim phng trnh (*) c nghim khng m

0S 0

P 0 1 4m 01 0

m 0 0 m 1

4

Vy vi m [0; 14] th h cho c nghim.Bi tp 2.45. Tm m bt phng trnh

4x 2 +16 4x m c nghim.

Li gii. Xt hm s f(x) =

4x 2 +16 4x trn [ 12 ; 4].o hm f (x) =

24x 2

216 4x ; f

(x) = 0 4x 2 = 16 4x x = 94. Bng bin thin:

x 12

94 4

f (x) + 0

f(t)

14

2

7

14

T bng bin thin suy ra bt phng trnh cho c nghim m min[ 12 ;4]

f(x) m 14.

Bi tp 2.46. Tm m phng trnh (x 3) (x + 1) + 4 (x 3)

x+1x3 = m c nghim.

Li gii. t (x 3)

x+1x3 = t (t R). Phng trnh cho tr thnh t2 + 4tm = 0 (*).

Phng trnh cho c nghim phng trnh (*) c nghim 0 m 4.Vy vi m 4 th phng trnh cho c nghim.

Bi tp 2.47. (DB-07) Tm m BPT m(

x2 2x + 2 + 1)+ x (2 x) 0 c nghim thuc on [0; 1 +3].Li gii. t

x2 2x + 2 = t. Vi x [0; 1 +3] t [1; 2]. Bt phng trnh cho tr thnh

m (t + 1) + 2 t2 0 m t2 2t + 1

(*)

Xt hm s f(t) = t22t+1 trn [1; 2] c f

(t) = t2+2t+2(t+1)2

> 0,t [1; 2] lim[1;2]

f(t) = f(2) = 23 .

Vy bt phng trnh cho c nghim trn[0; 1 +

3] bt phng trnh (*) c nghim trn [1; 2] m 23 .

Bi tp 2.48. (A-07) Tm m phng trnh 3x 1 + mx + 1 = 2 4x2 1 c nghim thc.

Li gii. iu kin: x 1. Phng trnh cho tng ng vi 3

x1x+1 + 2

4

x1x+1 = m.

t 4

x1x+1 = t. Vi x 1 t [0; 1). Phng trnh tr thnh 3t2 + 2t = m (*)

Xt hm s f(t) = 3t2 + 2t trn [0; 1) c f (t) = 6t + 2; f (t) = 0 t = 13 . Bng bin thin:

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t 0 13 1

f (t) + 0

f(t)

0

13

1

Vy phng trnh cho c nghim phng trnh (*) c nghim trn [0; 1) 1 m 13 .Bi tp 2.49. (B-06) Tm m phng trnh

x2 + mx + 2 = 2x + 1 c hai nghim thc phn bit.

Li gii. Phng trnh cho tng ng vi{

2x + 1 0x2 + mx + 2 = 4x2 + 4x + 1

{

x 12m = 3x

2+4x1x

.

Xt hm s f(x) =3x2 + 4x 1

xtrn

[ 12 ; +) \ {0} c f (x) = 3x2 + 1x2 > 0,x [1

2; +

)\ {0}.

Bng bin thin:

x 12 0 + f (x) + +

f(x)

92

+

+

T bng bin thin suy ra phng trnh cho c hai nghim phn bit m 92 .Bi tp 2.50. (B-04) Tm m PT m

(1 + x2 1 x2 + 2) = 21 x4 +1 + x2 1 x2 c nghim.

Li gii. iu kin: 1 x 1.t

1 + x2 1 x2 = t c t = x1+x2

x1x2 ; t

= 0 x = 0; t(0) = 0, t(1) = 2 t [0;2].Phng trnh cho tr thnh m (t + 2) = t2 + t + 2 m = t2+t+2t+2 (*).Xt hm s f(t) = t

2+t+2t+2 trn [0;

2] c f (t) = t

24t(t+2)2

0,t [0;2].Suy ra min

[0;2]f(t) = f

(2)

=

2 1; max[0;2]f(t) = f(0) = 1.

Khi phng trnh cho c nghim phng trnh (*) c nghim trn [0;2] 2 1 m 1.Bi tp 2.51. (A-08) Tm m phng trnh 4

2x +

2x + 2 4

6 x + 26 x = m c hai nghim phn bit.

Li gii. iu kin: 0 x 6. Xt hm s f(x) = 42x +2x + 2 46 x + 26 x trn [0; 6].Ta c f (x) = 1

2 4(2x)3

+ 12x 1

2 4(6x)3

16x =

12

(1

4(2x)3

14(6x)3

)+ 1

2x 1

6x .

t 14(2x)3

14(6x)3 = u(x),

12x 1

6x = v(x).

Nhn thy f (2) = 0 v u(x), v(x) cng dng trn (0; 2), cng m trn (2; 6) nn ta c bng bin thin

x 0 2 6

f (x) + 0

f(x)

2

6 + 2 4

6

3

2 + 6

4

12 + 2

3

Do phng trnh c hai nghim phn bit 26 + 2 46 m < 32 + 6.Bi tp 2.52. (DB-07) Tm m phng trnh 4

x4 13x + m + x 1 = 0 c ng mt nghim.

Li gii. Phng trnh cho tng ng vi

4x4 13x + m = 1 x

{1 x 0x4 13x + m = x4 4x3 + 6x2 4x + 1

{x 1m = 4x3 + 6x2 + 9x + 1

Xt hm s f(x) = 4x3 + 6x2 + 9x + 1 trn [1; +) c f (x) = 12x2 + 12x + 9; f (x) = 0 x = 12 .Bng bin thin

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Nguyn Minh Hiu

x 12 1f (x) 0 +

f(x)

+

32

12

Do phng trnh cho c ng mt nghim khi v ch khi m > 12 hoc m = 32 .Bi tp 2.53. (B-07) Chng minh vi mi m > 0, PT x2 + 2x 8 = m (x 2) c hai nghim phn bit.Li gii. iu kin: x 2. Nhn thy x = 2 l mt nghim ca phng trnh.

Vi x > 2, phng trnh tng ng vi(x2 + 2x 8)2 = m (x 2) x3 + 6x2 32 = m.

Xt hm s f(x) = x3 + 6x2 32 trn (2; +) c f (x) = 3x2 + 12x > 0,x > 2. Bng bin thin

x 2 + f (x) +

f(x)

0

+

T bng bin thin ta thy vi mi m > 0 th phng trnh lun c ng mt nghim trn (2; +).Vy vi mi m > 0 th phng trnh cho c ng hai nghim.

Bi tp 2.54. Chng minh rng vi mi m, phng trnh x4 + x3 2x2 + 3mxm2 = 0 lun c nghim.Li gii. Phng trnh cho tng ng vi m2 3xm x4 x3 + 2x2 = 0 (*).

Ta c = 9x2 4 (x4 x3 + 2x2) = 4x4 + 4x3 + x2 = (2x2 + x)2.Do ()

[m = 3x+2x

2+x2

m = 3x2x2x

2

[m = x2 + 2xm = x x2

[x2 + 2xm = 0 (1)x2 x + m = 0 (2) .

Phng trnh cho c nghim [

(1) c nghim(2) c nghim

[

1 = 4 + 4m 02 = 1 4m 0

[m 1m 14

m R.Vy phng trnh cho c nghim vi mi m.

Bi tp 2.55. (DB-04) Tm m h{

x2 5x + 4 03x2 mxx + 16 = 0 c nghim.

Li gii. H cho tng ng vi

{1 x 4m = 3x

2+16xx

.

Xt hm s f(x) =3x2 + 16

xx

trn [1; 4] c f (x) =3x(x2 16)2x5

0,x [1; 4].Suy ra max

[1;4]f(x) = f(1) = 19; min

x f(x) = f(4) = 8. Do h cho c nghim 8 m 19.

Bi tp 2.56. (D-2011) Tm m h{

2x3 (y + 2)x2 + xy = mx2 + x y = 1 2m c nghim.

Li gii. H cho tng ng vi{ (

x2 x) (2x y) = mx2 x + 2x y = 1 2m .

t x2 x = u, 2x y = v (u 14 ). H tr thnh{

uv = m (1)u + v = 1 2m (2) .

T (2) v = 1 2m u thay vo (1) ta c u (1 2m u) = m m = u2+u2u+1 .Xt hm s f(u) =

u2 + u2u + 1

c f (u) =2u2 + 2u 1

(2u + 1)2 ; f

(u) = 0 u = 1 +

3

2. Bng bin thin:

x 14 1+3

2+

f (x) + 0

f(x) 58

232

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Vy h cho c nghim khi v ch khi m 23

2 .

Bi tp 2.57. Tm m h

1 x2 + 2 31 x2 = m c nghim duy nht.Li gii. Nhn thy nu x0 l nghim phng trnh th x0 cng l nghim phng trnh.

Do gi s phng trnh c nghim duy nht th nghim bng 0 m = 3.Vi m = 3 phng trnh tr thnh

1 x2 + 2 31 x2 = 3 (*).

t 6

1 x2 = t (t 0). Phng trnh (*) tr thnh t3 + 2t2 3 = 0 t = 1 61 x2 = 1 x = 0.Vy vi m = 3 th phng trnh cho c nghim duy nht x = 0.

Bi tp 2.58. Tm m h{

x = y2 y + my = x2 x + m c nghim duy nht.

Li gii. Xt h{

x = y2 y + m (1)y = x2 x + m (2) .

Nhn thy nu h c nghim (x; y) th cng c nghim (y;x).Do h c nghim duy nht khi x = y, thay vo (1) ta c x2 2x + m = 0 (*).Vy h c nghim duy nht khi v ch khi (*) c nghim kp 1m = 0 m = 1.

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Chuyen de 02 PT BPT HPT N.M.hieu Mathvn.com

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