Đáp án:
$\begin{array}{l}
a)\sqrt {5 - 2\sqrt 6 } + \sqrt {12} + \dfrac{1}{2}\sqrt 8 \\
= \sqrt {3 - 2\sqrt 3 .\sqrt 2 + 2} + 2\sqrt 3 + \dfrac{1}{2}.2\sqrt 2 \\
= \sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} + 2\sqrt 3 + \sqrt 2 \\
= \sqrt 3 - \sqrt 2 + 2\sqrt 3 + \sqrt 2 \\
= 3\sqrt 3 \\
b)\dfrac{{3\sqrt 2 - 6}}{{\sqrt 2 - 1}} + \dfrac{{6\sqrt 2 - 4}}{{\sqrt 2 - 3}}\\
= \dfrac{{3\sqrt 2 \left( {1 - \sqrt 2 } \right)}}{{\sqrt 2 - 1}} + \dfrac{{2\sqrt 2 \left( {3 - \sqrt 2 } \right)}}{{\sqrt 2 - 3}}\\
= - 3\sqrt 2 - 2\sqrt 2 \\
= - 5\sqrt 5 \\
c)\dfrac{{\sqrt 6 - \sqrt 8 }}{{\sqrt 3 - 2}} - \sqrt {3 - 2\sqrt 2 } + \dfrac{1}{{1 + \sqrt 2 }}\\
= \dfrac{{\sqrt 2 \left( {\sqrt 3 - 2} \right)}}{{\sqrt 3 - 2}} - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} + \dfrac{{\sqrt 2 - 1}}{{2 - 1}}\\
= \sqrt 2 - \left( {\sqrt 2 - 1} \right) + \sqrt 2 - 1\\
= \sqrt 2
\end{array}$
