Đáp án:
$\begin{array}{l}
a)A = 3\sqrt 3 + 4\sqrt {12} - 5\sqrt {27} \\
= 3\sqrt 3 + 4.\sqrt {4.3} - 5.\sqrt {9.3} \\
= 3\sqrt 3 + 4.2\sqrt 3 - 5.3\sqrt 3 \\
= 3\sqrt 3 + 8\sqrt 3 - 15\sqrt 3 \\
= - 4\sqrt 3 \\
c)D = \dfrac{1}{2}\sqrt {48} - 2\sqrt {75} - \dfrac{{\sqrt {33} }}{{\sqrt {11} }} + 5\sqrt {1\dfrac{1}{3}} \\
= \dfrac{1}{2}.4\sqrt 3 - 2.5\sqrt 3 - \sqrt 3 + 5\sqrt {\dfrac{4}{3}} \\
= 2\sqrt 3 - 10\sqrt 3 - \sqrt 3 + 5.\dfrac{{2\sqrt 3 }}{3}\\
= - 11\sqrt 3 + \dfrac{{10\sqrt 3 }}{3}\\
= \dfrac{{ - 23\sqrt 3 }}{3}\\
d)\dfrac{{15}}{{\sqrt 6 + 1}} + \dfrac{4}{{\sqrt 6 - 2}} + \dfrac{{12}}{{\sqrt 6 - 3}} - \sqrt 6 \\
= \dfrac{{15\left( {\sqrt 6 - 1} \right)}}{{\left( {\sqrt 6 + 1} \right)\left( {\sqrt 6 - 1} \right)}} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{{6 - 4}} + \dfrac{{12\left( {\sqrt 6 + 3} \right)}}{{6 - 9}} - \sqrt 6 \\
= \dfrac{{15\left( {\sqrt 6 - 1} \right)}}{{6 - 1}} + 2\left( {\sqrt 6 + 2} \right) + \dfrac{{12\left( {\sqrt 6 + 3} \right)}}{{ - 3}} - \sqrt 6 \\
= 3\left( {\sqrt 6 - 1} \right) + 2\sqrt 6 + 4 - 4\sqrt 6 - 12 - \sqrt 6 \\
= - 11
\end{array}$
