Đáp án:
$\begin{array}{l}
a)3\sqrt {12} - 5\sqrt {27} + \sqrt {48} \\
= 3.\sqrt {{2^2}.3} - 5\sqrt {{3^2}.3} + \sqrt {{4^2}.3} \\
= 3.2\sqrt 3 - 5.3\sqrt 3 + 4\sqrt 3 \\
= 6\sqrt 3 - 15\sqrt 3 + 4\sqrt 3 \\
= - 5\sqrt 3 \\
b)\sqrt {14 + 6\sqrt 5 } + \sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} \\
= \sqrt {9 + 2.3\sqrt 5 + 5} + \left| {3 - \sqrt 5 } \right|\\
= \sqrt {{{\left( {3 + \sqrt 5 } \right)}^2}} + 3 - \sqrt 5 \\
= 3 + \sqrt 5 + 3 - \sqrt 5 \\
= 6\\
c)\left( {\sqrt 6 + \sqrt 2 } \right).\sqrt {2 - \sqrt 3 } \\
= \left( {\sqrt 3 + 1} \right).\sqrt 2 .\sqrt {2 - \sqrt 3 } \\
= \left( {\sqrt 3 + 1} \right).\sqrt {4 - 2\sqrt 3 } \\
= \left( {\sqrt 3 + 1} \right).\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
= \left( {\sqrt 3 + 1} \right).\left( {\sqrt 3 - 1} \right)\\
= 3 - 1\\
= 2\\
d)\frac{2}{{\sqrt 3 - 1}} - \frac{{3 + \sqrt 3 }}{{\sqrt 3 + 1}}\\
= \frac{{2\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} - \frac{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{{\sqrt 3 + 1}}\\
= \frac{{2\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} - \sqrt 3 \\
= \frac{{2\left( {\sqrt 3 + 1} \right)}}{2} - \sqrt 3 \\
= \sqrt 3 + 1 - \sqrt 3 \\
= 1
\end{array}$
