Đáp án:
$\begin{array}{l}
a)M = 7\sqrt {\dfrac{1}{2}} + \dfrac{{\sqrt {44} }}{{\sqrt {22} }} - \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} \\
= \dfrac{{7\sqrt 2 }}{2} + \sqrt {\dfrac{{44}}{{22}}} - \left| {2 - \sqrt 2 } \right|\\
= \dfrac{7}{2}\sqrt 2 + \sqrt 2 - \left( {2 - \sqrt 2 } \right)\\
= \dfrac{9}{2}.\sqrt 2 - 2 + \sqrt 2 \\
= \dfrac{{11\sqrt 2 }}{2} - 2\\
= \dfrac{{11\sqrt 2 - 4}}{2}\\
b)\dfrac{2}{{\sqrt 3 + 1}} = \dfrac{{2\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} = \sqrt 3 - 1\\
\dfrac{{\sqrt {10} - \sqrt 2 }}{{\sqrt 5 - 1}} = \dfrac{{\sqrt 2 \left( {\sqrt 5 - 1} \right)}}{{\sqrt 5 - 1}} = \sqrt 2 \\
3)\dfrac{2}{{\sqrt 3 - 1}} + \dfrac{3}{{\sqrt 3 - 2}}\\
= \dfrac{{2\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} + \dfrac{{3\left( {\sqrt 3 + 2} \right)}}{{3 - 4}}\\
= \sqrt 3 + 1 - 3\left( {\sqrt 3 + 2} \right)\\
= \sqrt 3 + 1 - 3\sqrt 3 - 6\\
= - 2\sqrt 3 - 5
\end{array}$
