TRƯỜNG THCS PHÚ THỌ QUẬN 11 GV: CAO MINH TÀI TRANG 1
CÁC BÀI TẬP MẪU
1/ x(x–1)(2–3x) = 0
⇔ x = 0 hay x– 1 = 0 hay 2–3x = 0
⇔ x= 0 hay x = 1 hay –3x = –2
⇔ x = 0 hay x = 1 hay x =
2
3
2/ (
1
2
x +1)(9
x
2
–25 ) = 0
⇔ (
1
2
x +1)(3x–5)(3x+5) = 0
⇔
1
2
x = –1 hay 3x = 5 hay 3x = –5
⇔ x = –2 hay x =
5
3
hay x =
−5
3
3/ x(
x

+2)(x–1)
2
= 0
⇔
x
2
+2 = 0 hay (x–1)
2
= 0
⇔
x
2
= –2 (vô lí) hay x– 1 = 0
⇔ x = 1
5/ 2x(x +3) – 4(x +3) = 0
⇔ (x+3)(2x–4) = 0
⇔ x+3=0 hay 2x–4=0
⇔ x = –3 hay x = 2
6/
x
2
(x+1) – 4(x+1)=0
⇔ (x+1)(
x
2
–4) = 0
⇔ (x+1)(x–2)(x+2) = 0
⇔ x= –1 hay x = 2 hay x = –2
7/ (2x–5)(x+4) –(x+4)(x–3) = 0
⇔ (x+4).[(2x–5)–(x–3)]=0


11/ x(x–2)–3x+6 = 0
⇔ x(x–2)–(3x–6) = 0
⇔ x(x–2)–3(x–2) = 0
⇔ (x–2)(x–3) = 0
⇔ x = 2 hay x = 3
12/ 6(x+2) –
x
2
+4 = 0
⇔ 6(x+2) –(
x
2
–4) = 0
⇔ 6(x+2) –(x+2)(x–2) = 0
⇔ (x+2)[6–(x–2)] = 0
⇔ (x+2)(6–x+2) = 0
⇔ (x+2)(8–x)=0
⇔ x = –2 hay x = 8
13/
x
3
+2
x
2
–x–2 = 0
⇔ (
x
3
+2

16/ 5(x+3)(x–2)= 3(x+5)(x–2)
⇔ 5(x+3)(x–2)– 3(x+5)(x–2) = 0
⇔ (x–2)[5(x+3)–3(x+5)] =0
⇔ (x–2)(5x+15–3x–15) = 0
⇔ (x–2)(2x) = 0
⇔ x–2 = 0 hay 2x = 0
⇔ x = 2 hay x = 0
17/ (x–1)(2x–1)=x(1–x)
⇔ (x–1)(2x–1)–x(1–x) = 0
⇔ (x–1)(2x–1) +x(x–1) = 0
⇔ (x–1)(2x–1+x) = 0
⇔ (x–1)(x–1)=0
⇔ x–1 = 0
⇔ x = 1
18/ (2x+2)(x–3)= 3(x+1)
⇔ 2(x+1)(x–3)= 3(x+1)
⇔ 2(x+1)(x–3)– 3(x+1) = 0
⇔ (x+1)[2(x–3)–3] = 0
⇔ (x+1)(2x–6–3) = 0
⇔ (x+1)(2x–9) = 0
⇔ x = –1 hay x = 4,5
34/ 2x – 5 < 5x –3
19/ 3x–12 = 5x(x–4)
⇔ 3(x–4) = 5x(x–4)
⇔ 3(x–4)– 5x(x–4) = 0
⇔ (x–4)(3–5x)=0
⇔ x = 4 hay x =
3
5
20/

–1
2
= 0
⇔ (4x–1)(4x+1) = 0
⇔ 4x– 1 = 0 hay 4x+1 = 0
⇔ x =
1
4
hay x =
−1
4
22/ *Cách 1: 8
x
2
= 2
⇔
x
2
=
2
8

⇔
x
2
=
1
4

⇔ x = ±

2
= 0
⇔ (2x–1)
2
–(3x)
2
= 0
⇔ [(2x–1)–3x][(2x–1)+3x] = 0
⇔ (–x–1)(5x–1) = 0
⇔ –x–1 = 0 hay 5x–1 = 0
⇔ x = –1 hay x =
1
5
24/ (5x–3)
2
–(4x–7)
2
= 0
⇔[(5x–3)–(4x–7)][(5x–3)+(4x–7)]=0
⇔ (5x–3–4x+7)(5x–3+4x–7) = 0
⇔ (x+4)(9x–10) = 0
⇔ x+4 = 0 hay 9x– 10 = 0
⇔ x = –4 hay x =
10
9
41/
x x+ −
<
2 3 2
3 2

= 0
⇔ (2x+7)
2
–(3x+6)
2
= 0
⇔(–x+1)(5x+13) =0
⇔ x = 1 hay x =
−13
5
27/ 4
x
2
+4x+1 = 0
⇔ (2x+1)
2
= 0
⇔ 2x+1 = 0
⇔ x =
−1
2
28/ 16
x
2
+8x = –1
⇔ 16
x
2
+8x +1 = 0
⇔ (4x+1)

−
−
9
3
⇔ x ≤ 3
32/ 15–3x > 9
⇔ –3x > 9–15
⇔ –3x > –6
⇔ x <
−
−
6
3
⇔ x < 2
33/ –2 > 1– 3x
⇔ 3x > 1+2
⇔ 3x > 3
⇔ x > 1
⇔ 2x – 5x < –3 +5
⇔ –3x < 2
⇔ x >
−
2
3
⇔ x >
−2
3
35/ 1 + 2(x–1) > 3– 2x
⇔ 1+2x – 2 > 3– 2x
⇔ 2x + 2x > 3–1 +2

⇔ –6x < –12
⇔ x >
−
−
12
6
⇔ x > 2
*Cách 2 : (x–3)
2
<
x
2
–3
⇔ (x–3)(x–3) <
x
2
–3
⇔
x
2
–3x–3x + 9 <
x
2
–3
⇔
x
2
–
x
2

–5x]
⇔
x
2
–4x+4 ≥ 3– 4x–20 +
x
2
+5x
⇔
x
2
–
x
2
–4x +4x –5x ≥ 3–20–4
⇔ –5x ≥ –21
⇔ x ≤
−
−
21
5
⇔ x ≤
21
5
40/ 2(2x–1)
2
+ 6 ≥ 8(x+3)(x–3)
⇔ 2(4
x
2

−10
3
42/
x −
>
3 1
2
4
⇔ 3x– 1 > 4 . 2
⇔ 3x > 8 + 1 ⇔ 3x > 9 ⇔ x > 3
43/
x− <
1
5
3
⇔ –x < 3 . 5
⇔ –x < 15 ⇔ x > –15
44/
x−
<
2 2
0
3
⇔ 2–2x < 0 . 3
⇔ 2–2x < 0
⇔ –2x < –2
⇔ x >
−
−
2

⇔ 4–8x –13 +5x ≥ –3x –9
⇔ –8x + 5x +3x ≥ –9 –4 +13
⇔ 0x ≥ 0
⇔ 0 ≥ 0 (đúng)
Vậy bất phương trình trên vô số
nghiệm
48/
x
x
−
− > −
2 2
1
2 3
⇔
x x− −
− >
2 2 1
2 3 1
(MTC: 6)
⇔ 3(x–2) –4 > 6(x–1)
⇔ 3x – 6 –4 > 6x – 6
⇔ 3x – 6x > –6 + 6 +4
⇔ –3x > 4
⇔ x <
−
4
3
⇔ x <
−4

x ⇔ − =1 2 5
(câu 49)
52/
x x− =2 5 3
x
x x x x hay
≥

⇔

− = − = −

3 0
2 5 3 2 5 3
x
x x x x
≥

⇔

− = + =

0
2 3 5 2 3 5 hay
x
x x
≥

⇔


x
x x x x hay
≥

⇔

− − = − − − + = −

2
3 2 4 3 2 4
x
x x
≥

⇔

− = − − = −

2
4 6 2 2 hay
x
x x (loai) hay 1 (loai)
≥


⇔

= =



x
+
=
−
2 1
1
2
(1)
* MTC: (x–2)
*ĐKXĐ: x ≠ 2
* QĐ:
(1) ⇔ 2x+1 = x–2
⇔ 2x– x = –2 –1
⇔ x = –3 (nhận)
Vậy S=
{ }
−3
56/
x
x x
x
−
+ =
+ −
−
2
1 1 3 12
2 2
4
x

x x x
(x ) (x ) (x )(x )
⇔ + =
− + + −
2
2 3 2 1 1 3
(1)
* MTC: 2(x–3)(x+1)
*ĐKXĐ: x ≠ 3; x ≠ –1
* QĐ: (1)⇔ x(x+1) + x(x–3) = 4x
⇔
x
2
+ x +
x
2
–3x –4x = 0
⇔ 2
x
2
–6x = 0
⇔ 2x(x – 3) = 0
⇔ 2x = 0 hay x– 3 = 0
⇔ x = 0 (nhận) hay x = 3 (loại)
Vậy S=
{ }
0
58/
x
(x ) (x ) (x )(x )

x x x
x( x ) x( x )
x( x )
− + +
⇔ + =
+ −
−
2
2 3 1 6
2 3 2 3
4 9
x x x
x( x ) x( x ) x( x )( x )
− + +
⇔ + =
+ − − +
2 3 1 6
2 3 2 3 2 3 2 3
* MTC: x(2x+3)(2x–3)
*ĐKXĐ: x ≠ 0; x ≠
−3
2
; x ≠
3
2
* QĐ: (1) ⇔ (2x–3)
2
+(x+1)(2x+3) = x+6
…………
59/

2
– (x–5)
2
= x(x+5)
⇔ 2(
x
2
+10x+25)–(
x
2
–10x+25) =
x
2
+5x
⇔ 2
x
2
+20x+50–
x
2
+10x –25 –
x
2
–5x = 0
⇔ 25x = –25
⇔ x = –1 (nhận)
Vậy S=
{ }
−1
60/

100 0
99 98 97 96
x⇔ + =100 0

x⇔ = −100
Vậy S=
{ }
−100
61/
x x x x− − − −
+ + + + =
1909 1907 1905 1903
4 0
91 93 95 97
x x x
( ) ( ) ( )
− − −
⇔ + + + + +
1909 1907 1905
1 1 1
91 93 95

x
( )
−
+ + + =−
1903
1 4 0
97
4

1990 1980 15 5
x x x x
− − − − − − − −
⇔ + = +
5 1990 15 1980 1980 15 1990 5
1990 1980 15 5
x x x x− − − −
⇔ + = +
1995 1995 1995 1995
1990 1980 15 5
(x )
 
⇔ − + − − =
 ÷
 
1 1 1 1
1995 0
1990 1980 15 5
x⇔ − =1995 0

x⇔ = 1995

Vậy S=
{ }
1995