x³ - 4x = (2x + 3)(2 - x)
x(x² - 4) = (2x + 3)(2 - x)
x(x - 2)(x + 2) - (2x + 3)(2 - x) = 0
x(x - 2)(x + 2) + (2x + 3)(x - 2) = 0
(x - 2)[x(x + 2) + (2x + 3)] = 0
(x - 2)(x² + 2x + 2x + 3) = 0
(x - 2)(x² + x + 3x + 3) = 0
(x - 2)[(x² + x) + (3x + 3)] = 0
(x - 2)[x(x + 1) + 3(x + 1)] = 0
(x - 2)(x + 1)(x + 3) = 0
x - 2 = 0 hoặc x + 1 = 0 hoặc x + 3 = 0
*) x - 2 = 0
x = 2
*) x + 1 = 0
x = -1
*) x + 3 = 0
x = -3
Vậy x = -3; x = -1; x = 2
\(x\left(x^2-4\right)\) + \(\left(2x+3\right)\left(x-2\right)\) = 0
\(x\left(x-2\right)\left(x+2\right)+\left(2x+3\right)\left(x-2\right)\) = 0
\(\left(x-2\right)\left\lbrack x\left(x+2\right)\right.+2x+3\rbrack\) =0
\(\left(x-2\right)\left\lbrack x^2+2x\right.+2x+3\rbrack\) =0
\(\left(x-2\right)\left(x^2+4x+3\right)\) = 0
\(\left(x-2\right)\left(x+1\right)\left(x+3\right)=0\)
TH1: x -2 = 0
x = 2
TH2: x + 1 = 0
x = -1
TH3: x + 3 = 0
x = -3
Vậy ....
