x3-16x=0
⇔x(x\(^2\)-16)=0
⇔x(x\(^2\)-4\(^2\))=0
⇔x(x-4)(x+4)=0
⇔\(\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy x ∈\(\left\{0;4;-4\right\}\)
\(x^3-16x=0\)
=> x\(\left(x^2-4^2\right)=0\)
=>x(x-4)(x+4)=0
=>\(\left[{}\begin{matrix}x=0\\x-4=0< =>x=4\\x+4=0< =>x=-4\end{matrix}\right.\)
vậy x=4 hoặc x=-4 hoặc x=o