CÁC BT MẪU CÓ LỜI GIẢI DÀNH CHO HỌC SINH

TRÖÔØNG THCS PHUÙ THOÏ QUAÄN 11 GV: CAO MINH TAØI TRANG 1

CAÙC BAØI TAÄP MAÃU

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 (voâ 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 (voâ 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/ *Caùch 1: 16 x

2

= 1

⇔ x

2 =

1

16

⇔ x = ±

1

4

*Caùch 2: 16 x

2

= 1

⇔ 16 x

2 –1 = 0

⇔ (4x)2

–12

= 0

⇔ (4x–1)(4x+1) = 0

⇔ 4x– 1 = 0 hay 4x+1 = 0

⇔ x =

1

4

hay x =

−1

4

22/ *Caùch 1: 8 x

2

= 2

⇔ x

2 =

2

8

⇔ x

2 =

1

4

⇔ x = ±

1

2

*Caùch 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 = 9x2

⇔ (2x–1)2 –9x2

= 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