Ta có: 4x2+x=0
⇔ x(4x+1)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Ta có: \(4x^2+x=0\)
\(\Leftrightarrow x\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{4}\end{matrix}\right.\)