Đáp án:
$\begin{array}{l}
a)\dfrac{4}{{\sqrt 3 + 1}} + \dfrac{1}{{\sqrt 3 - 2}} + \dfrac{6}{{\sqrt 3 - 3}}\\
= \dfrac{{4\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} + \dfrac{{\sqrt 3 + 2}}{{3 - 4}} + \dfrac{6}{{\sqrt 3 \left( {1 - \sqrt 3 } \right)}}\\
= \dfrac{{4\left( {\sqrt 3 - 1} \right)}}{2} + \dfrac{{\sqrt 3 + 2}}{{ - 1}} + \dfrac{{2\sqrt 3 \left( {1 + \sqrt 3 } \right)}}{{1 - 3}}\\
= 2\sqrt 3 - 2 - \sqrt 3 - 2 - \sqrt 3 \left( {1 + \sqrt 3 } \right)\\
= \sqrt 3 - 4 - \sqrt 3 - 3\\
= - 7\\
b)\dfrac{4}{{3 + \sqrt 5 }} - \dfrac{8}{{1 + \sqrt 5 }} + \dfrac{{15}}{{\sqrt 5 }}\\
= \dfrac{{4\left( {3 - \sqrt 5 } \right)}}{{9 - 5}} - \dfrac{{8\left( {1 - \sqrt 5 } \right)}}{{1 - 5}} + 3\sqrt 5 \\
= 3 - \sqrt 5 + 2\left( {1 - \sqrt 5 } \right) + 3\sqrt 5 \\
= 5
\end{array}$
