ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
\(\dfrac{2}{x^3-x^2-x+1}=\dfrac{3}{1-x^2}-\dfrac{1}{x+1}\)
=>\(\dfrac{2}{x^2\left(x-1\right)-\left(x-1\right)}=\dfrac{-3}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}\)
=>\(\dfrac{2}{\left(x-1\right)^2\cdot\left(x+1\right)}=\dfrac{-3\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
=>\(2=-3\left(x-1\right)-\left(x-1\right)^2\)
=>\(2+3\left(x-1\right)+\left(x-1\right)^2=0\)
=>\(2+3x-3+x^2-2x+1=0\)
=>\(x^2+x=0\)
=>x(x+1)=0
=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
