Đáp án:
$\begin{array}{l}
+ )\dfrac{{1 - 3x + 3{x^2} - {x^3}}}{{{x^2} - x}}\\
= \dfrac{{{{\left( {1 - x} \right)}^3}}}{{x\left( {x - 1} \right)}}\\
= \dfrac{{ - {{\left( {x - 1} \right)}^3}}}{{x\left( {x - 1} \right)}}\\
= \dfrac{{ - {{\left( {x - 1} \right)}^2}}}{x}\\
+ )\dfrac{{{x^2} - 2x + 1}}{{ - x}}\\
= \dfrac{{{{\left( {x - 1} \right)}^2}}}{{ - x}}\\
= \dfrac{{ - {{\left( {x - 1} \right)}^2}}}{x}\\
Vậy\,\dfrac{{1 - 3x + 3{x^2} - {x^3}}}{{{x^2} - x}} = \dfrac{{{x^2} - 2x + 1}}{{ - x}}
\end{array}$