\(\left(\dfrac{2x-1}{x+1}-\dfrac{x^2-x+1}{x^2+x}\right)\cdot\dfrac{x^2}{x-1}\)
\(đkxđ:\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\\x\ne0\end{matrix}\right.\)
\(\left(\dfrac{2x-1}{x+1}-\dfrac{x^2-x+1}{x^2+x}\right)\cdot\dfrac{x^2}{x-1}\\ =\left(\dfrac{2x-1}{x+1}-\dfrac{x^2-x+1}{x\left(x+1\right)}\right)\cdot\dfrac{x^2}{x-1}\\ =\left(\dfrac{x\left(2x-1\right)-\left(x^2-x+1\right)}{x\left(x+1\right)}\right)\cdot\dfrac{x^2}{x-1}\\ =\dfrac{2x^2-x-x^2+x-1}{x\left(x+1\right)}\cdot\dfrac{x^2}{x-1}\\ =\dfrac{\left(x^2-1\right)\cdot x^2}{x\left(x+1\right)\cdot\left(x-1\right)}\\ =x\)
