Đáp án:
$\begin{array}{l}
a)3\sqrt {32} - \dfrac{4}{{\sqrt 2 }} - \sqrt 3 .\sqrt {24} - 6\sqrt {\dfrac{1}{2}} \\
= 3.4\sqrt 2 - 2\sqrt 2 - \sqrt {72} - 3\sqrt 2 \\
= 12\sqrt 2 - 2\sqrt 2 - 6\sqrt 2 - 3\sqrt 2 \\
= \sqrt 2 \\
b)\dfrac{{2\sqrt 3 - \sqrt {21} }}{{2 - \sqrt 7 }} + \dfrac{{12}}{{3 + \sqrt 3 }} + \sqrt {7 + 4\sqrt 3 } \\
= \dfrac{{\sqrt 3 \left( {2 - \sqrt 7 } \right)}}{{2 - \sqrt 7 }} + \dfrac{{12\left( {3 - \sqrt 3 } \right)}}{{{3^2} - 3}} + \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} \\
= \sqrt 3 + 2\left( {3 - \sqrt 3 } \right) + 2 + \sqrt 3 \\
= \sqrt 3 + 6 - 2\sqrt 3 + 2 + \sqrt 3 \\
= 8
\end{array}$
