Đáp án:
$3x^3-6x=0$
$⇔3x(x^2-2)=0$
$⇔3x(x-\sqrt[]{2})(x+\sqrt[]{2})=0$
⇔\(\left[ \begin{array}{l}3x=0\\x-\sqrt[]{2}=0\\x+\sqrt[]{2}=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=\sqrt[]{2}\\x=-\sqrt[]{2}\end{array} \right.\)
$\text{Vậy x ∈ {$0; -\sqrt[]{2} ; \sqrt[]{2}$}}$
