ĐKXĐ : \(x\ne\pm1\)
PT : \(\Leftrightarrow\dfrac{x-1-x^2-x+2}{x+1}=\dfrac{x+1-\left(x+2\right)\left(x-1\right)}{x-1}\)
\(\Leftrightarrow\dfrac{1-x^2}{x+1}=1-x=\dfrac{3-x^2}{x-1}\)
\(\Leftrightarrow x^2-3=\left(x-1\right)^2=x^2-2x+1\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\left(TM\right)\)
Vậy ...
ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x-1}{x+1}-\dfrac{x^2+x-2}{x+1}=\dfrac{x+1}{x-1}-x-2\)
\(\Leftrightarrow\dfrac{x-1-x^2-x+2}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)
\(\Leftrightarrow\dfrac{-x^2+1}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)
\(\Leftrightarrow\dfrac{-\left(x^2-1\right)}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)
\(\Leftrightarrow\dfrac{-\left(x-1\right)\left(x+1\right)}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)
\(\Leftrightarrow-\left(x-1\right)-\dfrac{x+1}{x-1}+x+2=0\)
\(\Leftrightarrow\dfrac{-\left(x-1\right)^2}{x-1}-\dfrac{x+1}{x-1}+\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
Suy ra: \(-\left(x^2-2x+1\right)-x-1+x^2-x+2x-2=0\)
\(\Leftrightarrow-x^2+2x-1-x-1+x^2+x-2=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(nhận)
Vậy: S={2}
