đk: x # - 1; x #0
\(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)=\left(\dfrac{1-x\left(2-x\right)}{x^2+x}\right):\left(\dfrac{1+x^2-2x}{x}\right)=\dfrac{1+x^2-2x}{x^2+x}:\dfrac{1+x^2-2x}{x}=\dfrac{1+x^2-2x}{x\left(x+1\right)}\cdot\dfrac{x}{1+x^2-2x}=\dfrac{1}{x+1}\)