Đáp án:
$\begin{array}{l}
1)\dfrac{{12}}{{5\sqrt 6 }} = \dfrac{{12\sqrt 6 }}{{5.6}} = \dfrac{{2\sqrt 6 }}{5}\\
2)\dfrac{3}{{2 + \sqrt 6 }} = \dfrac{{3\left( {\sqrt 6 - 2} \right)}}{{6 - 4}} = \dfrac{{3\sqrt 6 - 6}}{2}\\
3)\dfrac{3}{{2\sqrt 3 - 3\sqrt 2 }} = \dfrac{3}{{\sqrt 6 \left( {\sqrt 2 - \sqrt 3 } \right)}}\\
= \dfrac{{\sqrt 3 \left( {\sqrt 2 + \sqrt 3 } \right)}}{{\sqrt 2 \left( {2 - 3} \right)}}\\
= \dfrac{{\sqrt 6 \left( {\sqrt 2 + \sqrt 3 } \right)}}{{ - 1}}\\
= - 2\sqrt 3 - 3\sqrt 2 \\
a)\dfrac{3}{{2 + \sqrt 3 }} + \dfrac{{13}}{{4 - \sqrt 3 }} + \dfrac{6}{{\sqrt 3 }}\\
= \dfrac{{3\left( {2 - \sqrt 3 } \right)}}{{{2^2} - 3}} + \dfrac{{13\left( {4 + \sqrt 3 } \right)}}{{{4^2} - 3}} + 2\sqrt 3 \\
= 6 - 3\sqrt 3 + 4 + \sqrt 3 + 2\sqrt 3 \\
= 10\\
b)\dfrac{2}{{3 - \sqrt {10} }} - \dfrac{{36}}{{4 + \sqrt {10} }} - \dfrac{{40}}{{\sqrt {10} }}\\
= \dfrac{{2\left( {3 + \sqrt {10} } \right)}}{{9 - 10}} - \dfrac{{36\left( {4 - \sqrt {10} } \right)}}{{16 - 10}} - 4\sqrt {10} \\
= - 6 - 2\sqrt {10} - 6\left( {4 - \sqrt {10} } \right) - 4\sqrt {10} \\
= - 6 - 6\sqrt {10} - 24 + 6\sqrt {10} \\
= - 30\\
c)\dfrac{2}{{1 - \sqrt 3 }} + \dfrac{3}{{\sqrt 2 + \sqrt 5 }} - \dfrac{{22}}{{\sqrt {12} - 2\sqrt 5 }}\\
= \dfrac{{2\left( {1 + \sqrt 3 } \right)}}{{1 - 3}} + \dfrac{{3\left( {\sqrt 5 - \sqrt 2 } \right)}}{{5 - 2}} - \dfrac{{22\left( {\sqrt 3 + \sqrt 5 } \right)}}{{2\left( {\sqrt 3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}}\\
= - 1 - \sqrt 3 + \sqrt 5 - \sqrt 2 - \dfrac{{11\left( {\sqrt 3 + \sqrt 5 } \right)}}{{3 - 5}}\\
= - 1 - \sqrt 3 + \sqrt 5 - \sqrt 2 + \dfrac{{11}}{2}\sqrt 3 + \dfrac{{11}}{2}\sqrt 5 \\
= - 1 - \sqrt 2 + \dfrac{9}{2}\sqrt 3 + \dfrac{9}{2}\sqrt 5
\end{array}$
