Điều kiện: $x\neq 2;x \neq \pm \sqrt 2;x \neq 0;x^3+x^2+x-4 \neq 0.$
Đặt $\color{red}{a=x-2;b=x^2-2;c=x^3,}$ PT đã cho trở thành:
$\color{green}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Leftrightarrow (a+b)(b+c)(c+a)=0}$
$\color{green}{\Leftrightarrow \left[\ \begin{array}{l} x^2+x-4=0 \\ x^3+x-2=0 \\ x^3+x^2-2=0 \end{array} \right.}$
$\color{green}{\Leftrightarrow \left[\ \begin{array}{l} x^2+x-4=0 \\ (x-1)(x^2+x+2)=0 \\ (x-1)(x^2+2x+2)=0 \end{array} \right.}$
$\color{green}{\Leftrightarrow \left[\ \begin{array}{l} x=\frac{-1 \pm \sqrt{17}}{2}\\ x=1 \end{array} \right.}$ $\color{red}{\bigstar \bigstar \bigstar} $