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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
Maths287 BT PHNG TRNH V T
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)
Bi 6 : Gii bt phng trnh
x+ 1
x+ 13 x > x1
2
Li gii tham kho :
iu kin : x [1; 3] \ {1}
Maths287 BT PHNG TRNH V T
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
)
Vy tp nghim ca bt phng trnh l T =
[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
Maths287 BT PHNG TRNH V T
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
Vy tp nghim ca bt phng trnh l T =
[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
Maths287 BT PHNG TRNH V T
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
Vy tp nghim ca bt phng trnh l T =
(3 +10
2; 1
)
Bi 10 : Gii bt phng trnh
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;+
)
Maths287 BT PHNG TRNH V T
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}
Maths287 BT PHNG TRNH V T
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
Maths287 BT PHNG TRNH V T
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)
Vy tp nghim ca bt phng trnh l T =
(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; +
)
Maths287 BT PHNG TRNH V T
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)
Maths287 BT PHNG TRNH V T
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
Maths287 BT PHNG TRNH V T
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;+
)
Kt hp vi iu kin ta c tp nghim ca bt phng trnh l T =
[5 +33
2;+
)
{1}
Maths287 BT PHNG TRNH V T
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]
Maths287 BT PHNG TRNH V T
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
Maths287 BT PHNG TRNH V T
Kt hp vi iu kin ta c tp nghim ca bt phng trnh l T =
[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]