Đáp án:
$\begin{array}{l}
a)\dfrac{{2 + \sqrt 2 }}{{2 - \sqrt 2 }} + \dfrac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }}\\
= \dfrac{{{{\left( {2 + \sqrt 2 } \right)}^2} + {{\left( {2 - \sqrt 2 } \right)}^2}}}{{\left( {2 - \sqrt 2 } \right)\left( {2 + \sqrt 2 } \right)}}\\
= \dfrac{{4 + 4\sqrt 2 + 2 + 4 - 4\sqrt 2 + 2}}{{{2^2} - 2}}\\
= \dfrac{{12}}{2}\\
= 6 = \sqrt {36} \\
4\sqrt 2 = \sqrt {16.2} = \sqrt {32} < \sqrt {36} \\
\Leftrightarrow \dfrac{{2 + \sqrt 2 }}{{2 - \sqrt 2 }} + \dfrac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }} > 4\sqrt 2 \\
b)\sqrt {\dfrac{{2\sqrt 3 + 3}}{{2\sqrt 3 - 3}}} = \sqrt {\dfrac{{{{\left( {2\sqrt 3 + 3} \right)}^2}}}{{{{\left( {2\sqrt 3 } \right)}^2} - {3^2}}}} = \dfrac{{2\sqrt 3 + 3}}{{\sqrt {18 - 9} }}\\
= \dfrac{{2\sqrt 3 + 3}}{{\sqrt 9 }}\\
= \dfrac{{2\sqrt 3 + 3}}{3} = \dfrac{{\sqrt 3 \left( {2 + \sqrt 3 } \right)}}{3}\\
= \dfrac{{\sqrt 3 }}{3}.\left( {2 + \sqrt 3 } \right) < 2 + \sqrt 3 \\
\Leftrightarrow \sqrt {\dfrac{{2\sqrt 3 + 3}}{{2\sqrt 3 - 3}}} < 2 + \sqrt 3 \\
c)\dfrac{{\sqrt {3 + \sqrt 5 } }}{{\sqrt 2 }} = \dfrac{{\sqrt {6 + 2\sqrt 5 } }}{2} = \dfrac{{\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }}{2} = \dfrac{{\sqrt 5 + 1}}{2}\\
Vậy\,\dfrac{{\sqrt {3 + \sqrt 5 } }}{{\sqrt 2 }} = \dfrac{{1 + \sqrt 5 }}{2}
\end{array}$
