ĐK: $x\ne\pm1$.
Đặt: $y=\dfrac{x+2}{x+1},z=\dfrac{x-2}{x-1}$
Phương trình trở thành:
$y^2+z^2-\dfrac{5}{2}yz=0$
$\Leftrightarrow 2y^2-5yz+2z^2=0$
$\Leftrightarrow (2y-z)(y-2z)=0$
$\Leftrightarrow \left[\begin{array}{l}2y=z\\y=2z\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\dfrac{2(x+2)}{x+1}=\dfrac{x-2}{x-1}\\\dfrac{x+2}{x+1}=\dfrac{2(x-2)}{x-1}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{1}{2}(-3\pm\sqrt{17})\\x=\dfrac{1}{2}(3\pm\sqrt{17})\end{array}\right.$