\(\Leftrightarrow\dfrac{-2x-4}{x-4x^2}=\dfrac{x^2+2x}{1-4x}\)
\(\Leftrightarrow\dfrac{-2\left(x+2\right)}{x\left(1-4x\right)}=\dfrac{x\left(x+2\right)}{1-4x}\) (*)
ĐKXĐ\(\left\{{}\begin{matrix}x\ne0\\1-4x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne\dfrac{1}{4}\end{matrix}\right.\)
(*) \(\Leftrightarrow\dfrac{-2x\left(x+2\right)}{x\left(1-4x\right)}=\dfrac{x^2\left(x+2\right)}{x\left(1-4x\right)}\)
\(\Rightarrow-2x\left(x+2\right)=x^2\left(x+2\right)\)
\(\Leftrightarrow-2x=x^2\)
\(\Leftrightarrow-x^2-2x=0\)
\(\Leftrightarrow-x\left(x+2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy S=\(\left\{0;-2\right\}\)
