Ta có: \(4x^3-x=0\)
\(\Leftrightarrow x\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(2x\right)^2-1^2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(2x-1\right)\left(2x+1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2};x=\dfrac{-1}{2}\end{matrix}\right.\)
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