Lời giải:
$ĐK: x\neq -2$
Đặt $x+2=a(a\neq 0)$ thì pt trở thành:
\((a-1)^2+(\frac{a-1}{a})^2=8\)
\(\Leftrightarrow a^2-2a+1+1+\frac{1}{a^2}-\frac{2}{a}=8\)
\(\Leftrightarrow a^2+\frac{1}{a^2}-2a-\frac{2}{a}-6=0\)
\(\Leftrightarrow (a+\frac{1}{a})^2-2(a+\frac{1}{a})-8=0\)
Đặt $a+\frac{1}{a}=t$ thì $t^2-2t-8=0$
$\Leftrightarrow (t-4)(t+2)=0$
Nếu $t-4=0\Leftrightarrow a+\frac{1}{a}-4=0$
$\Rightarrow a^2+4a-1=0$
$\Rightarrow a=-2\pm \sqrt{5}$
$\Rightarrow x=-4\pm \sqrt{5}$
Nếu $t+2=0\Leftrightarrow a+\frac{1}{a}+2=0$
$\Rightarrow a^2+2a+1=0\Rightarrow (a+1)^2=0\Rightarrow a=-1$
$\Rightarrow x=-3$
Vậy.......
