đk : x khác 1 ; -1
<=> \(-x\left(x+1\right)+x^2+2=2\left(x-1\right)\)
\(\Leftrightarrow-x+2=2x-2\Leftrightarrow x=\dfrac{4}{3}\)(tm)
\(\Leftrightarrow-x\left(x+1\right)+x^2+2=2x-2\)
\(\Leftrightarrow-x^2-x+x^2+2-2x+2=0\)
=>-3x+4=0
hay x=4/3(nhận)
\(\Leftrightarrow\dfrac{-x\left(x+1\right)+\left(x^2+2\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow-x^2-x+x^2+2-2x+2=0\left(quy\cdotđồng\cdot và\cdot khử\cdot mẫu\right)\)
\(\Leftrightarrow-3x+4=0\)
\(\Leftrightarrow-3x=-4\)
\(\Leftrightarrow x=\dfrac{-4}{3}\)
Vậy \(S=\left\{-\dfrac{4}{3}\right\}\)
