Đáp án:
$\begin{array}{l}
a)\dfrac{{1 - {x^2}}}{{{x^2} - 2x}}:\dfrac{{x + 1}}{x}\\
= \dfrac{{\left( {1 + x} \right)\left( {1 - x} \right)}}{{x\left( {x - 2} \right)}}.\dfrac{x}{{x + 1}}\\
= \dfrac{{1 - x}}{{x - 2}}\\
b)\dfrac{2}{{x + 2}} + \dfrac{{ - 4}}{{2 - x}} + \dfrac{{5x + 2}}{{4 - {x^2}}}\\
= \dfrac{2}{{x + 2}} + \dfrac{{ - 4}}{{2 - x}} + \dfrac{{5x + 2}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}\\
= \dfrac{{2\left( {2 - x} \right) - 4\left( {x + 2} \right) + 5x + 2}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}\\
= \dfrac{{4 - 2x - 4x - 8 + 5x + 2}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}\\
= \dfrac{{ - x - 2}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}\\
= \dfrac{1}{{x - 2}}\\
c)\dfrac{1}{{xy - {y^2}}} - \dfrac{1}{{{x^2} - xy}}\\
= \dfrac{1}{{y\left( {x - y} \right)}} - \dfrac{1}{{x\left( {x - y} \right)}}\\
= \dfrac{{x - y}}{{xy\left( {x - y} \right)}}\\
= \dfrac{1}{{xy}}
\end{array}$
