Question

[Nguyễn Hồ Quỳnh Trang]

Tìm x,biết:

\(6\left(x+1\right)^2-2\left(x+1\right)^3+2\left(x-1\right)\left(x^2+x+1\right)=0\)

Answer 1

[KAl(SO4)2·12H2O]

6(x + 1)² - 2(x + 1)³ + 2(x - 1)(x² + x + 1) = 0

<=> 6(x² + 2x + 1) - 2(x³ + 3x² + 3x + 1) + 2(x - 1)(x² + x + 1) = 0

<=> 6.x² + 6.2x + 6.1 + (-2).x³ + (-2).3x² + (-2).3x + (-2).1 + 2.x³ + 2(-1) = 0

<=> 6x² + 12x + 6 - 2x³ - 6x² - 6x - 2 + 2x³ - 2 = 0

<=> (6x² - 6x²) + (12x - 6x) + (6 - 2 - 2) + (-2x³ + 2x²) = 0

<=> 6x + 2 = 0

<=> 6x = 0 - 2

<=> 6x = -2

<=> x = -2/6 = -1/3

=> x = -1/3