Đáp án:
$\begin{array}{l}
a)\left[ {\frac{2}{{xy}}:{{\left( {\frac{1}{x} - \frac{1}{y}} \right)}^2}} \right] - \frac{{{x^2} + {y^2}}}{{{{\left( {x - y} \right)}^2}}}\\
= \frac{2}{{xy}}:\frac{{{{\left( {y - x} \right)}^2}}}{{{x^2}{y^2}}} - \frac{{{x^2} + {y^2}}}{{{{\left( {x - y} \right)}^2}}}\\
= \frac{2}{{xy}}.\frac{{{x^2}{y^2}}}{{{{\left( {x - y} \right)}^2}}} - \frac{{{x^2} + {y^2}}}{{{{\left( {x - y} \right)}^2}}}\\
= \frac{{2xy}}{{{{\left( {x - y} \right)}^2}}} - \frac{{{x^2} + {y^2}}}{{{{\left( {x - y} \right)}^2}}}\\
= \frac{{ - {{\left( {x - y} \right)}^2}}}{{{{\left( {x - y} \right)}^2}}}\\
= - 1\\
b)\left( {1 - \frac{2}{{x + 1}}} \right):\left( {1 - \frac{{{x^2} - 2}}{{{x^2} - 1}}} \right)\\
= \frac{{x + 1 - 2}}{{x + 1}}:\frac{{{x^2} - 1 - {x^2} + 2}}{{{x^2} - 1}}\\
= \frac{{x - 1}}{{x + 1}}.\frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{1}\\
= {\left( {x - 1} \right)^2}
\end{array}$
