CÁC BT MẪU CÓ LỜI GIẢI DÀNH CHO HỌC SINH

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

3
–2)(
x
2
+3) = 0
⇔ x = 0 hay
x
3
–2 = 0 hay
x
2
+3 = 0
⇔ x= 0 hay
x
3
= 2 hay
x
2
=–3 (vô lí)
⇔ x = 0 hay x = 6
4/ (
x
2
+2)(
x
2
–2x+1) = 0
⇔ (
x
2

+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

⇔ (x+4)(2x–5–x+3) =0
⇔ (x+4)(x–2) =0
⇔ x = –4 hay x =2
8/ (x–5)
2
–2(x–5) =0
⇔ (x–5)(x–5)–2(x–5)=0
⇔ (x–5)[(x–5)–2] =0
⇔ (x–5)(x–7)=0
⇔ x=5 hay x = 7
9/ 5x(x+1)–7(1+x) = 0
⇔ 5x(x+1)–7(x+1)= 0
⇔ (x+1)(5x–7) = 0
⇔ x = –1 hay x =
7
5
10/ 2x(5x–2)–3(2–5x) = 0
⇔ 2x(5x–2)+3(5x–2)= 0
⇔ (5x–2)(2x+3)=0
⇔ x =
2
5
hay x =
−3
2
25/ 4
x
2
–4x+1 = (x–3)
2

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

x
2
)–(x+2) = 0
⇔
x
2
(x+2)–(x+2) = 0
⇔ (x+2)(
x
2
–1)= 0
⇔ (x+2)(x–1)(x+1)=0
⇔ x =–2 hay x = 1 hay x = –1
14/
x
2
=2x
⇔
x
2
–2x = 0
⇔ x(x–2) = 0
⇔ x = 0 hay x–2 = 0
⇔ x = 0 hay x = 2
15/ (3x+2)(x–4)=2x(x–4)
⇔ (3x+2)(x–4)–2x(x–4) = 0
⇔ (x–4)[(3x+2)–2x] = 0
⇔ (x–4)(3x+2–2x) =0
⇔ (x–4)(x+2)=0
⇔ x= 4 hay x = –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/

x
2
–4 = 0
⇔ (x–2)(x+2) = 0
⇔ x–2 = 0 hay x+2 = 0
⇔ x = 2 hay x = –2
21/ *Cách 1: 16
x
2
= 1
⇔
x
2
=
1
16
⇔ x =
±
1
4
*Cách 2: 16
x
2
= 1
⇔ 16
x
2
–1 = 0
⇔ (4x)
2

–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 = ±

1
2
*Cách 2: 8
x
2
= 2
⇔ 8
x
2
–2 = 0
⇔ 2(4
x
2
–1) = 0
⇔ 2(2x–1)(2x+1) = 0
⇔ 2x–1 = 0 hay 2x+1 = 0
⇔ x =
1
2
hay x =
−1
2
23/ (2x–1)
2
= 9x
2

⇔ (2x–1)
2
–9x

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

TRƯỜNG THCS PHÚ THỌ QUẬN 11 GV: CAO MINH TÀI TRANG 2
⇔ (2x–1)
2
–(x–3)
2
= 0
⇔[(2x–1)–(x–3)][(2x–1)+(x–3)]= 0
⇔ (2x–1–x+3)(2x–1+x–3) = 0
⇔ (x+2)(3x–4) = 0
⇔ x+2 = 0 hay 3x–4 = 0
⇔ x = –2 hay x =
4
3

26/ (2x+7)
2
–9(x+2)
2
= 0
⇔ (2x+7)
2
–3
2
.(x+2)
2
= 0
⇔ (2x+7)
2
– [ 3(x+2) ]
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)

2
= 0
⇔ 4x+1 = 0
⇔ x =
−1
4
29/
x
2
+3x–4 = 0
⇔
x
2
–x+4x – 4 = 0
⇔ (
x
2
–x)+(4x – 4) = 0
⇔ x(x – 1)+4(x–1) = 0
⇔ (x–1)(x+4) = 0
⇔ x = 1 hay x = –4
30/ 2x–7 ≤ 0
⇔ 2x ≤ 7
⇔ x ≤
7
2
⇔ x ≤ 3,5
31/ –3x + 9 ≥ 0
⇔ –3x ≥ –9
⇔ x ≤

−
−
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

⇔ 4x > 4
⇔ x > 1
36/ x– 8 ≥ 2(x +
1
2
) + 7
⇔ x– 8 ≥ 2x + 1 + 7
⇔ x– 2x ≥ 1 +7 + 8
⇔ –x ≥ 16
⇔ x ≤ –16
37/ (x–3)
2
<
x
2
–3
⇔
x
2
– 6x+ 9 <
x
2
–3
⇔
x
2
–
x
2
–6x < –3 – 9

⇔ –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

–3x–3x < –3 –9
⇔ –6x < –12
⇔ x > 2
38/ x(x–3) – (x–2)
2
< 0
⇔
x
2
–3x – [
x
2
–4x +4 ] < 0
⇔
x
2
–3x –
x
2
+ 4x – 4 < 0
⇔ x < 4
39/ (x–2)
2
≥ 3– (4–x)(x+5)
⇔
x
2
–4x+4 ≥ 3– [4x+20–
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

–4x+1) + 6 ≥ 8(
x
2
–9)
⇔ 8
x
2
–8x + 2 + 6 ≥ 8
x
2
– 72
⇔ 8
x
2
– 8
x
2
–8x ≥ –72 –2 – 6
⇔ –8x ≥ –80
⇔ x ≤
−
−
80
8
⇔ x ≤ 10
⇔ 2(2+3x) < 3(x–2)
⇔ 4 + 6x < 3x– 6
⇔ 6x – 3x < –6 –4
⇔ 3x < –10
⇔ x <

−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

2
⇔ x > 1
45/
x
<
−
2
0
3
⇔ 3–x < 0
⇔ –x < –3
⇔ x > 3
46/
x
−
<
−
2
0
3 1
⇔ 3x– 1 > 0 ⇔ 3x > 1 ⇔ x >
1
3
47/
x x x− − − −
− ≥
1 2 13 5 3
3 12 4
(MTC:12)
⇔ 4(1–2x)–1(13–5x) ≥ 3(–x–3)

⇔ 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

3
Giải các phương trình sau:
TRƯỜNG THCS PHÚ THỌ QUẬN 11 GV: CAO MINH TÀI TRANG 3
49/
x− =1 2 5
x x
(dung)
hay
≥

⇔

− = − = −

5 0
1 2 5 1 2 5
x x⇔ − = − − = − −2 5 1 2 5 1 hay
x x⇔ − = − = −2 4 2 6 hay
x x⇔ = − =2 3 hay

Vậy S=
{ }
;−2 3
50/
x − = −3 5 4
Vì –4 < 0 nên phương trình
trên vô nghiệm
51/
x − − = −1 2 9 4
x ⇔ − = − +1 2 4 9

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
≥

⇔


− = =

0
5 5 5 hay
x
x (loai) x (nhân)
≥

⇔

= − =

0
5 1 hay
Vậy S=
{ }
1
53/
x x− + =4 3 2
x x⇔ − = −4 3 2
x
x x x x
− ≥

⇔

− = − − = − +

2 0
4 3 2 4 3 2 hay

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)
≥


⇔

= =



2
3
2
Vậy S= ∅
54/
x + + =2 1 1 9
x⇔ + = −2 1 9 1
x⇔ + =2 1 8
x⇔ + =
8
1
2
x⇔ + =1 4
x x
(dung)
hay
≥

⇔

+ = + = −

4 0
1 4 1 4
x x hay ⇔ = = −3 5
Vậy S=
{ }
;−3 5
55/
x

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 )
(1)
−
⇔ − =
+ − − +
1 1 3 12
2 2 2 2
* MTC: (x+2)(x–2)
*ĐKXĐ: x ≠ –2; x ≠ 2
* QĐ:
(1) ⇔ 1(x–2) – 1(x+2) = 3x–12
⇔ x– 2 –x –2 = 3x– 12
⇔ –3x = –8
⇔ x =
8
3
(nhận)
Vậy S=
 
 
 
8
3
57/
x x x
(x ) x (x )(x )
+ =
− + + −
2
2 3 2 2 1 3

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 )

−
− =
+ − − +
3 1 2
2 1 6 1 1 1
* MTC: 6(x–1)(x+1)
*ĐKXĐ: x ≠ 1; x ≠ –1
* QĐ:
(1) ⇔ 9(x–1) – 1(x+1) = –12x
⇔ 9x – 9 –x –1 = –12x
⇔ 20x = 10
⇔ x =
1
2
(nhận)
Vậy S=
 
 
 
1
2
59/
x x x
x x x x x x
− + +
+ =
+ − −
2 2 3
2 3 1 6
2 3 2 3 4 9

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/

x x x
x x x x x
+ − +
− =
− + −
2 2 2
5 5 5
5 2 10 2 50
x x x
x(x ) x(x )
(x )
+ − +
⇔ − =
− +
−
2
5 5 5
5 2 5
2 25
x x x
x(x ) x(x ) (x )(x )
+ − +
⇔ − =
− + − +
5 5 5
5 2 5 2 5 5
(1)
* MTC : 2x(x–5)(x+5)
* ĐKXĐ : x ≠ 0 ; x ≠ 5 ; x ≠ –5
* QĐ : (1) ⇔ 2(x+5)

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/

x x x x+ + + +
+ = +
1 2 3 4
99 98 97 96
x x x x
( ) ( ) ( ) ( )
+ + + +
⇔ + + + = + + +
1 2 3 4
1 1 1 1
99 98 97 96
x x x x+ + + + + + + +
⇔ + = +
1 99 2 98 3 97 4 96
99 98 97 96
x x x x+ + + +
⇔ + = +
100 100 100 100
99 98 97 96
x x x x+ + + +
⇔ + − − =
100 100 100 100
0
99 98 97 96
(x )
 
⇔ + + − − =
 ÷
 
1 1 1 1

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

x x x x− − − −
⇔ + + + =
2000 2000 2000 2000
0
91 93 95 97
( x)
 
⇔ − + + + =
 ÷
 
1 1 1 1
2000 0
91 93 95 97
x⇔ − =2000 0

x⇔ = 2000
Vậy S=
{ }
2000
62/
x x x x− − − −
+ = +
5 15 1980 1990
1990 1980 15 5
x x x x
( ) ( ) ( ) ( )
− − − −
⇔ − + − = − + −
5 15 1980 1990
1 1 1 1

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