Question

Giải các phương trình tích sau:

1.a)(3x – 2)(4x + 5) = 0 b) (2,3x – 6,9)(0,1x + 2) = 0

c)(4x + 2)(x2 + 1) = 0  d) (2x + 7)(x – 5)(5x + 1) = 0

2. a)(3x + 2)(x2 – 1) = (9x2 – 4)(x + 1) 

    b)x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0

c)2x(x – 3) + 5(x – 3) = 0         d)(3x – 1)(x2 + 2) = (3x – 1)(7x – 10)

3.a)(2x – 5)2 – (x + 2)2 = 0       b)(3x2 + 10x – 8)2 = (5x2 – 2x + 10)2

c)(x2 – 2x + 1) – 4 = 0              d)4x2 + 4x + 1 = x2

4. a) 3x2 + 2x – 1 = 0 b) x2 – 5x + 6 = 0

c) x2 – 3x + 2 = 0 d) 2x2 – 6x + 1 = 0

         e) 4x2 – 12x + 5 = 0 f) 2x2 + 5x + 3 = 0

Answer 1

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

Answer 2

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)

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

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

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

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