Question

TÌM X : (x^2-9)×(x^2+x+1)=0

Answer 1

Giải : 

\(\left(x^2-9\right)\left(x^2+x+1\right)=0\\ =>\left(x-3\right)\left(x+3\right)\left(x^2+x+1\right)=0\\ =>\left[{}\begin{matrix}x-3=0\\x+3=0\\x^2+x+1=0\left(\ge0\forall x\right)\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{3;-3\right\}\)

Answer 2

(x² - 9)(x² + x + 1) = 0

x² - 9 = 0

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

x - 3 = 0 hoặc x + 3 = 0

*) x - 3 = 0

x = 3

*) x + 3 = 0

x = -3

Vậy x = -3; x = 3