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