\[\begin{array}{l}
a)\frac{1}{{\sqrt 3 - 1}} - \frac{1}{{\sqrt 3 + 1}}\\
= \frac{{\sqrt 3 + 1 - \left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\\
= \frac{{\sqrt 3 + 1 - \sqrt 3 + 1}}{{3 - 1}}\\
= \frac{2}{2} = 1\\
b)\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 2}} - \frac{1}{{1 + \sqrt 2 }} + \frac{{\sqrt 2 + 1}}{{\sqrt 2 }}\\
= \frac{{\sqrt 2 - 1}}{{\sqrt 2 \left( {1 + \sqrt 2 } \right)}} - \frac{{\sqrt 2 }}{{\sqrt 2 \left( {1 + \sqrt 2 } \right)}} + \frac{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 \left( {1 + \sqrt 2 } \right)}}\\
= \frac{{\sqrt 2 - 1 - \sqrt 2 + 2 + 2\sqrt 2 + 1}}{{\sqrt 2 \left( {1 + \sqrt 2 } \right)}}\\
= \frac{{2\sqrt 2 + 2}}{{\sqrt 2 \left( {1 + \sqrt 2 } \right)}}\\
= \frac{{2\left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 \left( {1 + \sqrt 2 } \right)}}\\
= \sqrt 2 \\
c)\sqrt x - 2 + \frac{{10 - x}}{{\sqrt x + 2}}\;\left( {x \ge 0} \right)\\
= \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right) + 10 - x}}{{\sqrt x + 2}}\\
= \frac{{x - 4 + 10 - x}}{{\sqrt x + 2}}\\
= \frac{6}{{\sqrt x + 2}}\\
d)\frac{{x\sqrt x - y\sqrt y }}{{\sqrt x - \sqrt y }}\;\left( {x \ge 0;y \ge 0;x \ge y} \right)\\
= \frac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{\sqrt x - \sqrt y }}\\
= x + \sqrt {xy} + y
\end{array}\]