Bat phuong trinh tuyen tap

Part 1 : Cc bi ton

Bi 1 : Gii bt phng trnh (x 1)x2 2x+ 5 4xx2 + 1 2 (x+ 1)Li gii tham kho :

(x 1)x2 2x+ 5 4xx2 + 1 2 (x+ 1)

(x+ 1) (2 +x2 2x+ 5)+ 2x (2x2 + 1x2 2x+ 5) 0 (x+ 1) (2 +x2 2x+ 5)+ 2x (4x2 + 4 x2 + 2x 5)

2x2 + 1 +

x2 2x+ 5 0

(x+ 1) (2 +x2 2x+ 5)+ 2x (x+ 1) (3x 1)2x2 + 1 +

x2 2x+ 5 0

(x+ 1)[(2 +x2 2x+ 5)+ 2x (3x 1)

2x2 + 1 +

x2 2x+ 5

] 0

(x+ 1)[4x2 + 1 + 2

x2 2x+ 5 + 2(x2 + 1) (x2 2x+ 5) + (7x2 4x+ 5)

2x2 + 1 +

x2 2x+ 5

] 0

C 7x2 4x+ 5 = 7(x2 4

7x+

4

49

)+31

7 31

7nn biu thc trong ngoc lun > 0.

Do bt phng trnh x+ 1 0 x 1Vy tp nghim ca bt phng trnh l T = (;1]

Bi 2 : Gii bt phng trnhx+ 2 + x2 x+ 2 3x 2

Li gii tham kho :

iu kin : x 23

bpt x+ 23x 2 + x2 x 2 0

2 (x 2)x+ 2 +

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

(x 2)[ 2

x+ 2 +3x 2 + x+ 1

] 0

BT PHNG TRNH V T

Maths287 BT PHNG TRNH V T

Xt f (x) =2

x+ 2 +3x 2 + x+ 1 f

(x) =

1x+ 2

+3

3x 2(x+ 2 +

3x 2) + 1 > 0

f (x) f (23

)> 0

Do bt phng trnh x 2 0 x 2

Vy tp nghim ca bt phng trnh l T =

[2

3; 2

]Bi 3 : Gii bt phng trnh 4

x+ 1 + 2

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

Li gii tham kho :

iu kin : x 1Nhn thy x = - 1 l mt nghim ca bt phng trnh

Xt x > - 1 ta c bt phng trnh tng ng vi

4(

x+ 1 2)+ 2 (2x+ 3 3) x3 x2 2x 12 4 (x 3)

x+ 1 + 2+

4 (x 3)2x+ 3 + 3

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

(x 3)(

4x+ 1 + 2

+4

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

) 0

V x > - 1 nnx+ 1 > 0 v

2x+ 3 > 1 4

x+ 1 + 2+

42x+ 3 + 3

< 3

Do 4

x+ 1 + 2+

42x+ 3 + 3

(x+ 1)2 3 < 0

Suy ra bt phng trnh x 3 0 x 3Vy tp nghim ca bt phng trnh l T = {1} [3; +)

Bi 4 : Gii bt phng trnh

x (x+ 2)

(x+ 1)3 x 1

Li gii tham kho :

iu kin : x 0 . Khi x 0 ta c(x+ 1)3 x > 0


x (x+ 2)

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

x2 + 2x x3 + 3x2 + 4x+ 1 2 (x+ 1)x (x+ 1) x3 + 2x2 + 2x+ 1 2 (x+ 1)x2 + x 0 (x+ 1) (x2 + x+ 1 2x2 + x) 0 x2 + x+ 1 2x2 + x 0 (x2 + x 1)2 0 x2 + x = 1 x = 1

5

2

Kt hp vi iu kin ta c nghim ca bt phng trnh l x =

5 12

Bi 5 : Gii bt phng trnh1x+ 2

1x 1 2

3x 1

Li gii tham kho :

iu kin : 2 < x < 1 ()

bpt 3(

1x+ 2

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

3 x+ 2x 1 (x+ 2x 1)t a =

x+ 2x 1 x+ 2.x 1 = 1 a

2

2

Ta c bt phng trnha a32 3 a3a+6 0 (a+ 2) (a2 2a+ 3) 0

a 2 x+ 2x 1 2 x+ 2 + 2 x 1 x+ 6 + 4x+ 2 x 1 4x+ 2 (2x+ 7) (1)(1) lun ng vi iu kin (*). Vy tp nghim ca bt phng trnh l T = (2;1)


x+ 1

x+ 13 x > x1

2

Li gii tham kho :

iu kin : x [1; 3] \ {1}


bptx+ 1

(x+ 1 +

3 x)

2 (x 1) > x1

2 x+ 1 +

x2 + 2x+ 32 (x 1) > x

1

2()

Trng hp 1 : 1 < x 3 (1)() x+ 1 +x2 + 2x+ 3 > 2x2 3x+ 1 2 (x2 + 2x+ 3) +x2 + 2x+ 3 6 > 0

x2 + 2x+ 3 > 32 x

(27

2;2 +7

2

)

Kt hp vi (1) ta c x (1;2 +7

2

)Trng hp 2 : 1 < x < 1 (2)() x+ 1 +x2 + 2x+ 3 < 2x2 3x+ 1 2 (x2 + 2x+ 3) +x2 + 2x+ 3 6 < 0

0 x2 + 2x+ 3 < 32 x

[1; 2

7

2

)(2 +7

2; 3

]

Kt hp vi (2) ta c x [1; 2

7

2

)


[1; 2

7

2

)(1;2 +7

2

)

Bi 7 : Gii bt phng trnh6x2 2 (3x+ 1)x2 1 + 3x 6

x+ 1x 12 x2 (x2 + 2) 0Li gii tham kho :

iu kin : 1 x 2Ta c

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


bpt 6x2 2 (3x+ 1)x2 1 + 3x 6 0 4 (x2 1) 2 (3x+ 1)x2 1 + 2x2 + 3x 2 0

(

x2 1 x+ 12

)(x2 1 x

2 1) 0 (1)

Xt 1 x 2 ta c x2 1 x2 1 3 2 < 0

Do bt phng trnh x2 1 x+ 12 0 1 x 5

4


[1;5

4

]

Bi 8 : Gii bt phng trnh 2x3 +

5 4xxx+

10

x 2

Li gii tham kho :

iu kin : x > 0

bpt 2x2 4x+ 5 x2 2x+ 10 2 (x2 2x+ 10)x2 2x+ 10 15 0 x2 2x+ 10 3 x2 2x+ 10 9bt phng trnh cui lun ng. Vy tp nghim ca bt phng trnh l T = (0;+)

Bi 9 : Gii bt phng trnh 3(2x2 xx2 + 3) < 2 (1 x4)

Li gii tham kho :

bpt 2 (x4 + 3x2) 3xx2 (x2 + 3) 2 < 0t x

x3 + 3 = t x4 + 3x2 = t2

Khi bpt 2t2 3t 2 < 0 12< t < 2 1

2< xx2 + 3 < 2

* Vi x 0 ta c

bpt{

x 0xx2 + 3 < 2

{

x 0x4 + 3x2 4 < 0

{x 0x2 < 1

0 x < 1

* Vi x < 0 ta c


bpt{

x < 0

12< xx2 + 3

{

x < 012> xx2 + 3

{x < 0

x4 + 3x2 14< 0

x < 0x2 < 3 +10

2

3 +10

2< x < 0


(3 +10

2; 1

)


x+ 24 +

x

x+ 24x 0

bptx+ 24 +

x

x+ 24x 35

12

Li gii tham kho

iu kin : |x| > 1Nu x < - 1 th x+

xx2 1 < 0 nn bt phng trnh v nghim

Do bpt x > 1x2 + x2

x2 1 +2x2x2 1

1225

144> 0

x > 1x4

x2 1 + 2.x2x2 1

1225

144> 0

t t =x2x2 1 > 0

Khi ta c bpt t2 + 2t 1225144

> 0 t > 2512

Ta c

x > 1x2x2 1 >

25

12

x > 1x4

x2 1 >625

144

x (1;5

4

)(5

3;+

)


Vy tp nghim ca bt phng trnh l

(1;5

4

)(5

3;+

)

Bi 16 : Gii bt phng trnhx2 8x+ 15 +x2 + 2x 15 4x2 18x+ 18

Li gii tham kho

iu kin : x (;5] [5; +) {3}D thy x = 3 l mt nghim ca bt phng trnh

Vi x 5 ta cbpt(x 5) (x 3) +(x+ 5) (x 3) (x 3) (4x 6) x 3 (x 5 +x+ 5) x 3.4x 6 x 5 +x+ 5 4x 6 2x+ 2x2 25 4x 6 x2 25 x 6 x2 25 x2 6x+ 9

x 173

Kt hp ta c 5 x 173

Vi x 5 ta c(5 x) (3 x) +(x 5) (3 x) (3 x) (6 4x) 5 x+x 5 6 4x 5 x x 5 + 2x2 25 6 4x x2 25 3 x x2 25 9 6x+ x2

x 173

Kt hp ta c x 5

Vy tp nghim ca bt phng trnh l T = (;5] [5;17

3

] {3}


Bi 17 : Gii bt phng trnh2x+ 4 22 x > 12x 8

9x2 + 16

Li gii tham kho

iu kin : 2 x 2

bpt 2x+ 4 22 x > 2.(2x+ 4) 4 (2 x)9x2 + 16

2x+ 4 22 x > 2.(

2x+ 4 22 x) (2x+ 4 + 22 x)9x2 + 16

(2x+ 4 22 x)(1 2 (2x+ 4 + 22 x)9x2 + 16

)> 0

(2x+ 4 22 x) (2x+ 4 + 22 x)(1 2 (2x+ 4 + 22 x)9x2 + 16

)> 0

(6x 4) (9x2 + 16 2 (2x+ 4 + 22 x)) > 0 (3x 2) (9x2 + 16 2 (2x+ 4 + 22 x)) (9x2 + 16 + 2 (2x+ 4 + 22 x)) > 0 (3x 2)

(9x2 + 16 4(2x+ 4 + 22 x)2) > 0

(3x 2) (9x2 + 8x 32 168 2x2) > 0 (3x 2) (8x 168 2x2 + x2 4 (8 2x2)) > 0 (3x 2) (8 (x 28 2x2)+ (x 28 2x2) (x+ 28 2x2)) > 0 (3x 2) (x 28 2x2) (8 + x+ 28 2x2) > 0 (3x 2) (x 28 2x2) > 0 [ 2 x < 23

43

3< x 2

Bi 18 : Gii bt phng trnh 32x+ 1 + 3

6x+ 1 > 3

2x 1

Li gii tham kho

bpt 32x 1 32x+ 1 < 36x+ 1 2 3 3(2x 1) (2x+ 1) ( 32x 1 32x+ 1) < 6x+ 1 3(2x 1) (2x+ 1) ( 32x 1 32x+ 1)+ 2x+ 1 > 0


32x+ 1[

3

(2x 1)2 + 3(2x 1) (2x+ 1) + 3(2x+ 1)2] > 0

32x+ 1 > 0

x > 12

( do biu thc trong ngoc lun dng)


(12;+

)

Bi 19 : Gii bt phng trnh (4x2 x 7)x+ 2 > 10 + 4x 8x2

Li gii tham kho

iu kin : x 2bpt (4x2 x 7)x+ 2 + 2 (4x2 x 7) > 2 [(x+ 2) 4] (4x2 x 7) (x+ 2 + 2) > 2 (x+ 2 2) (x+ 2 + 2) 4x2 x 7 > 2x+ 2 4 4x2 > x+ 2 + 2x+ 2 + 1

4x2 > (x+ 2 + 1)2

{

x+ 2 > 2x 1 (1)x+ 2 < 2x 1 (2) (I){ x+ 2 < 2x 1 (3)x+ 2 > 2x 1 (4) (II)

Xt (I) t (1) v (2) suy ra

{x 22x 1 < 2x 1 2 x < 0

Khi h (I) {2 x < 0x+ 2 < 2x 1

{2 x 1/2x+ 2 < (2x 1)2 x [2;1)

Xt (II) t (3) v (4)

{x 22x 1 < 2x 1 x > 0

Khi h (II) {

x > 0x+ 2 < 2x 1

{x > 1/2

x+ 2 < (2x 1)2 x (

5+41

8; +

)Vy tp nghim ca bt phng trnh l T = [2;1)

(5+41

8; +

)


Bi 20 : Gii bt phng trnh 4x+ 1 +

4x+ 42x+ 3 + 1

(x+ 1) (x2 2x) 0

Li gii tham kho

iu kin : x 1

bpt x+ 1 = 04 +

4x+ 1

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

Xt (*)

Nu 0 x 2 suy ra VT > 0 v VP < 0 bt phng trnh v nghimNu 1 x < 0 suy ra VT > 4 v VP < 3 bt phng trnh v nghim

Nu x > 2 ta c bpt 4x+ 1

+4

2x+ 3 + 1 x2 2x

f (x) =4x+ 1

+4

2x+ 3 + 1nghch bin trn (2;+)

g (x) = x2 2x ng bin trn (2;+)Vi x < 3 ta c f (x) > f (3) = 6 = g (3) > g (x) bt phng trnh v nghim

Vi x 3 ta c f (x) f (3) = 6 = g (3) g (x)Vy tp nghim ca bt phng trnh l T = [3;+) {1}

Bi 21 : Gii bt phng trnh 32x 1 4x 1 4

2x2 3x+ 1

36

Li gii tham kho

iu kin : x 1Ta thy x = 1 l nghim ca bt phng trnh.

Xt x 6= 1 chia hai v ca bt phng trnh cho 42x2 3x+ 1 ta c

3. 42x 1x 1 4.

4

x 12x 1

16

t t = 42x 1x 1

4

x 12x 1 =

1

ta ( iu kin t > 0)


Khi ta c bpt 3t 4t 1

6 36t2 t 46 0

t 1666(l)

t 3

2(n)

Vi t

32ta c 4

2x 1x 1

3

2 2x 1

x 1 9

4 x+ 5

4 (x 1) 0 1 < x 5

Vy tp nghim ca bt phng trnh l T = [1; 5]

Bi 22 : Gii bt phng trnh x+ 1 +x2 4x+ 1 3x

Li gii tham kho

iu kin :

[0 x 23x 2 +3

Vi x = 0 bt phng trnh lun ng

Vi x > 0 chia hai v bt phng trnh chox ta c

bpt x+ 1x+

x+

1

x 4 3 (1)

t t =x+

1x 2 t2 = x+ 1

x+ 2

Ta c bt phng trnht2 6 3 t

3 t < 0{ 3 t 0t2 6 (3 t)2

t 52

Do x+

1x 5

2 x 2 x 1

2 x

(0;1

4

] [4; +)

chnh l tp nghim ca bt phng trnh

Bi 23 : Gii bt phng trnh 8

2x 3x+ 1

+ 3 62x 3 + 4x+ 1

Li gii tham kho

iu kin : x 32


8

2x 3x+ 1

+ 3 62x 3 + 4x+ 1

82x 3 + 3x+ 1 6(2x 3) (x+ 1) + 4 64 (2x 3) + 9 (x+ 1) + 48(2x 3) (x+ 1) 36 (2x 3) (x+ 1)+16 + 48

(2x 3) (x+ 1)

72x2 173x 91 0

79 x 13

8

Kt hp vi iu kin ta c tp nghim ca bt phng trnh l T =

[3

2;13

8

]

Bi 24 : Gii bt phng trnh5

2

x3 + x+ 2 x2 + 3

Li gii tham kho

iu kin : x 1Nhn thy x = - 1 l mt nghim ca bt phng trnh

bpt 52

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

t

{a =x2 x+ 2 0

b =x+ 1 0

C a2b2 = x2x+2x1 = x22x+1 = (x 1)2 0 (a b) (a+ b) 0 a bKhi bt phng trnh tr thnh

5

2ab a2 + b2 2a2 5ab+ b2 0 (a 2b) (2a b) 0 a 2b 0 a 2b x2 x+ 2 2x+ 1 x2 x+ 2 4x+ 4 x2 5x 2 0

x (; 5

33

2

][5 +33

2;+

)


[5 +33

2;+

)

{1}


Bi 25 : Gii bt phng trnh 3x3 1 2x2 + 3x+ 1

Li gii tham kho

iu kin : x 1Nhn thy x = 1 l mt nghim ca bt phng trnh

bpt 2x (x3 + x)x+ 1

+ 2 (x+ 2)x+ 1 > x3 + x+ 2x (x+ 2)

(x3 + x)(

2xx+ 1

1) (x+ 2)x+ 1

(2xx+ 1

1)> 0

(x3 + x (x+ 2)x+ 1) (2xx+ 1) > 0

{

x3 + x (x+ 2)x+ 1 > 02xx+ 1 > 0{x3 + x (x+ 2)x+ 1 < 02xx+ 1 < 0

Xt hm s f (t) = t3 + t f (t) = 3t2 + 1 > 0 tNn hm f(t) ng bin trn R.

Trng hp 1 :

{f (x) > f

(x+ 1

)2xx+ 1 > 0

{x >x+ 1

2x >x+ 1

x > 1 +5

2

Trng hp 2 :

{f (x) < f

(x+ 1

)2xx+ 1 < 0

{x 3 x+x2 6x+ 11

(x 1)2 + 2 +x 1 >

(3 x)2 + 2 +3 x

Xt hm s f (t) =t2 + 2 +

t

Ta c f (t) =t

t2 + 2+

1

2t> 0 t [1; 3]


Nn f(t) ng bin nn f (x 1) > f (3 x) x 1 > 3 x x > 2Kt hp vi iu kin ta c tp nghim ca bt phng trnh l T = (2; 3]

Bi 27 : Gii bt phng trnhx3 3x2 + 2x

x4 x2 12

Li gii tham kho

iu kin : x (;1) (1;+)x (x 1) (x 2)|x|x2 1

12

Nu x < - 1 ta c

bpt (1 x) (x 2)x2 1

12

x (;1){

1 x > 0x 2 < 0

(1 x) (x 2)x2 1 < 0 0x 2 0

(1 x) (x 2)x2 1 0 0 th x > - 1 ta c bt phng trnh tr thnh ( chia cho y3)

bpt(x

y

)3+ 3

(x

y

)2 4 0

(x

y 1)(

x

y+ 2

)2 0

[x/y 1x/y = 2

Trng hp 1 :x

y= 2 x = 2x+ 1 x = 2 22

Trng hp 2: xy 1 x x+ 1 1 x 1 +

5

2



[1; 1 +

5

2

]

Bi 30 : Gii bt phng trnh 2

x2 + x+ 1

x+ 4+ x2 4 2

x2 + 1

Li gii tham kho

iu kin : x > 4

bpt 2(

x2 + x+ 1

x+ 4 1)+ x2 3 2

x2 + 1

x2 + 1

2.x2 + x+ 1

x+ 4 1

x2 + x+ 1

x+ 4+ 1

+ x2 3 4 (x2 + 1)(

2 +x2 + 1

)x2 + 1

2 (x2 3)

(x+ 4) (x2 + x+ 1) + x+ 4+ x2 3 + d x

2 3(2 +x2 + 1

)x2 + 1

0

(x2 3)[

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

+ 1 +1(

2 +x2 + 1

)x2 + 1

] 0

x2 3 0 3 x 3Kt hp iu kin ta c tp nghim ca bt phng trnh l T =

[3;3]

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