Đặt \(\dfrac{x+2}{x+1}=a;\dfrac{x-2}{x-1}=b\), pt trở thành:
\(a^2+b^2-\dfrac{5}{2}ab=0\)
\(\Leftrightarrow\left(a-\dfrac{1}{2}b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{1}{2}b\\a=2b\end{matrix}\right.\)
To be continued. . .