ĐKXĐ: \(x\ne\pm1\)
Đặt \(\left\{{}\begin{matrix}\frac{x+2}{x+1}=a\\\frac{x-2}{x-1}=b\end{matrix}\right.\) pt trở thành:
\(a^2+b^2-\frac{5}{2}ab=0\Leftrightarrow2a^2-5b^2+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+2}{x+1}=\frac{2x-4}{x-1}\\\frac{x-2}{x-1}=\frac{2x+4}{x+1}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)\left(x-1\right)=\left(2x-4\right)\left(x+1\right)\\\left(x-2\right)\left(x+1\right)=\left(2x+4\right)\left(x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
