Ta có: \(6x^3+6x^2=2x\)
\(\Leftrightarrow6x^3+6x^2-2x=0\)
\(\Leftrightarrow2x\left(3x^2+3x-1\right)=0\)
\(\Leftrightarrow6x\left(x^2+x-\frac{1}{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+x-\frac{1}{6}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+\frac{1}{2}\right)^2=\frac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+\frac{1}{2}=\frac{\sqrt{5}}{2\sqrt{3}}\\x+\frac{1}{2}=-\frac{\sqrt{5}}{2\sqrt{3}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-3+\sqrt{15}}{6}\\x=\frac{-3-\sqrt{15}}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-3+\sqrt{15}}{6};\frac{-3-\sqrt{15}}{6}\right\}\)
