Question

Tìm x

a ) ( 2x-2)^2 - (3x-2)^2 = 0

b) ( 5x - 4)^2 - 1 = 0

c) x(3-x)(x+3)-(3-x)(9+3x+^2)= -27

d) 3x^2 +5x - 8 = 0

Answer 1

b) (5x - 4 )² - 1 = 0

(5x - 4 - 1).(5x - 4+1) = 0

(5x - 5 ).(5x - 3) =0

*) 5x -5 = 0 --> 5x = 5 --> x = 1

*) 5x - 3 = 0 --> 5x = 3 --> x = \(\dfrac{3}{5}\)

Vậy , x = 1 hoặc x =\(\dfrac{3}{5}\)

Answer 2

3x² + 5x - 8 =0

3x² + 5x - 3 - 5 = 0

3(x² -1 ) + 5(x-1) = 0

3(x-1)(x+1) + 5(x -1 ) =0

(x -1).[3(x+1) + 5] = 0

(x -1 ) . (3x +3 +5) = 0

(x-1) .(3x + 8) =0

*) x-1 = 0 --> x =1

*) 3x + 8 = 0 --> 3x = -8 --> x = \(\dfrac{-8}{3}\)

Vậy , x = 1 hoặc x = \(\dfrac{-8}{3}\)