Đáp án:
$\begin{array}{l}
\dfrac{{10 + 2\sqrt {10} }}{{\sqrt 5 + \sqrt 2 }} + \dfrac{8}{{1 - \sqrt 5 }}\\
= \dfrac{{\sqrt 2 .\sqrt {10} \left( {\sqrt 5 + \sqrt 2 } \right)}}{{\sqrt 5 + \sqrt 2 }} + \dfrac{{8\left( {1 + \sqrt 5 } \right)}}{{1 - 5}}\\
= \sqrt 2 .\sqrt 2 .\sqrt 5 + \dfrac{{8\left( {1 + \sqrt 5 } \right)}}{{ - 4}}\\
= 2\sqrt 5 - 2 - 2\sqrt 5 \\
= - 2\\
\dfrac{{2\sqrt 8 - \sqrt {12} }}{{\sqrt {18} - \sqrt {48} }} - \dfrac{{\sqrt 5 + \sqrt {27} }}{{\sqrt {30} + \sqrt {162} }}\\
= \dfrac{{2.2\sqrt 2 - 2.\sqrt 3 }}{{3.\sqrt 2 - 4\sqrt 3 }} - \dfrac{{\sqrt 5 + \sqrt {27} }}{{\sqrt 5 .\sqrt 6 + \sqrt {27.6} }}\\
= \dfrac{{\sqrt 2 .\left( {4 - \sqrt 2 .\sqrt 3 } \right)}}{{\sqrt 3 .\left( {\sqrt 3 .\sqrt 2 - 4} \right)}} - \dfrac{{\sqrt 5 + \sqrt {27} }}{{\sqrt 6 \left( {\sqrt 5 + \sqrt {27} } \right)}}\\
= \dfrac{{ - \sqrt 2 }}{{\sqrt 3 }} - \dfrac{1}{{\sqrt 6 }}\\
= \dfrac{{ - 2\sqrt 2 .\sqrt 3 }}{6} - \dfrac{{\sqrt 6 }}{6}\\
= \dfrac{{ - 3\sqrt 6 }}{6}\\
= - \dfrac{{\sqrt 6 }}{2}
\end{array}$
