Đáp án:
$\dfrac{2x^2-x}{x-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}=x-1$
Giải thích các bước giải:
$\begin{split}\dfrac{2x^2-x}{x-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}&=\dfrac{2x^2-x}{x-1}-\dfrac{x+1}{x-1}+\dfrac{2-x^2}{x-1}\\&=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\\&=\dfrac{x^2-2x+1}{x-1}\\&=\dfrac{(x-1)^2}{x-1}\\&=x-1\end{split}$