Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{2{x^2} - x}}{{x - 1}} + \frac{{x + 1}}{{1 - x}} + \frac{{2 - x}}{{x - 1}}\\
= \frac{{2{x^2} - x}}{{x - 1}} - \frac{{x + 1}}{{x - 1}} + \frac{{2 - x}}{{x - 1}}\\
= \frac{{2{x^2} - x - \left( {x + 1} \right) + 2 - x}}{{x - 1}}\\
= \frac{{2{x^2} - 3x + 1}}{{x - 1}}\\
= \frac{{\left( {2{x^2} - 2x} \right) - \left( {x - 1} \right)}}{{x - 1}}\\
= \frac{{2x\left( {x - 1} \right) - \left( {x - 1} \right)}}{{x - 1}}\\
= \frac{{\left( {2x - 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)}}\\
= 2x - 1
\end{array}\)
