ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\frac{x-2}{x+1}=a\\\frac{x+2}{x-1}=b\end{matrix}\right.\) pt trở thành:
\(5a^2-44b^2+12ab=0\) \(\Leftrightarrow\left(a-2b\right)\left(5a+22b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\5a=-22b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{x-2}{x+1}=\frac{2x+4}{x-1}\\\frac{5x+10}{x-1}=\frac{-22x-44}{x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-2\right)-\left(2x-4\right)\left(x+1\right)=0\\\left(5x+10\right)\left(x-1\right)+\left(22x+44\right)\left(x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)