Cứu tui vs

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
a,\ \frac{2}{\sqrt{6} -2} +\frac{2}{\sqrt{6} +2} +\frac{5}{\sqrt{6}}\\
=\frac{2\left(\sqrt{6} +2\right)}{\left(\sqrt{6} -2\right)\left(\sqrt{6} +2\right)} +\frac{2\left(\sqrt{6} -2\right)}{\left(\sqrt{6} -2\right)\left(\sqrt{6} +2\right)} +\frac{5\sqrt{6}}{6}\\
=\frac{2\left(\sqrt{6} +2\right)}{6-4} +\frac{2\left(\sqrt{6} -2\right)}{4} +\frac{5\sqrt{6}}{6}\\
=\sqrt{6} +2+\sqrt{6} -2+\frac{5\sqrt{6}}{6}\\
=\frac{12\sqrt{6} +5\sqrt{6}}{6}\\
=\frac{17\sqrt{6}}{6}
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
b,\ \frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6} -\sqrt{2}}\\
=\frac{2\sqrt{3-\sqrt{3+\sqrt{12+4\sqrt{3} +1}}}}{\sqrt{6} -\sqrt{2}}\\
=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left( 2\sqrt{3} +1\right)^{2}}}}}{\sqrt{6} -\sqrt{2}}\\
=\frac{2\sqrt{3-\sqrt{3+2\sqrt{3} +1}}}{\sqrt{6} -\sqrt{2}}\\
=\frac{2\sqrt{3-\sqrt{\left(\sqrt{3} +1\right)^{2}}}}{\sqrt{6} -\sqrt{2}}\\
=\frac{2\sqrt{3-\sqrt{3} -1}}{\sqrt{2} .\left(\sqrt{3} -1\right)}\\
=\frac{\sqrt{2} .\sqrt{2-\sqrt{3}}}{\sqrt{3} -1}\\
=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{3} -1}\\
=\frac{\sqrt{3-2\sqrt{3} +1}}{\sqrt{3} -1}\\
=\frac{\sqrt{\left(\sqrt{3} -1\right)^{2}}}{\sqrt{3} -1}\\
=\frac{\sqrt{3} -1}{\sqrt{3} -1}\\
=1
\end{array}$