Đáp án:
$\begin{array}{l}
\frac{{2 + \sqrt 3 }}{{\sqrt 2 + \sqrt {2 + \sqrt 3 } }} + \frac{{2 - \sqrt 3 }}{{\sqrt 2 - \sqrt {2 - \sqrt 3 } }}\\
= \frac{{\sqrt 2 .\left( {2 + \sqrt 3 } \right)}}{{\sqrt 2 .\left( {\sqrt 2 + \sqrt {2 + \sqrt 3 } } \right)}} + \frac{{\sqrt 2 .\left( {2 - \sqrt 3 } \right)}}{{\sqrt 2 .\left( {\sqrt 2 - \sqrt {2 - \sqrt 3 } } \right)}}\\
= \frac{{2\sqrt 2 + \sqrt 6 }}{{2 + \sqrt {4 + 2\sqrt 3 } }} + \frac{{2\sqrt 2 - \sqrt 6 }}{{2 - \sqrt {4 - 2\sqrt 3 } }}\\
= \frac{{2\sqrt 2 + \sqrt 6 }}{{2 + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} + \frac{{2\sqrt 2 - \sqrt 6 }}{{2 - \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}\\
= \frac{{2\sqrt 2 + \sqrt 6 }}{{2 + \sqrt 3 + 1}} + \frac{{2\sqrt 2 - \sqrt 6 }}{{2 - \sqrt 3 + 1}}\\
= \frac{{\left( {2\sqrt 2 + \sqrt 6 } \right).\left( {3 - \sqrt 3 } \right) + \left( {2\sqrt 2 - \sqrt 6 } \right).\left( {3 + \sqrt 3 } \right)}}{{\left( {3 + \sqrt 3 } \right).\left( {3 - \sqrt 3 } \right)}}\\
= \frac{{6\sqrt 2 - 2\sqrt 6 + 3\sqrt 6 - \sqrt {18} + 6\sqrt 2 + 2\sqrt 6 - 3\sqrt 6 - \sqrt {18} }}{{9 - 3}}\\
= \frac{{12\sqrt 2 - 2\sqrt {18} }}{6}\\
= \frac{{12\sqrt 2 - 2.3\sqrt 2 }}{6}\\
= \sqrt 2 \\
Vậy\,\frac{{2 + \sqrt 3 }}{{\sqrt 2 + \sqrt {2 + \sqrt 3 } }} + \frac{{2 - \sqrt 3 }}{{\sqrt 2 - \sqrt {2 - \sqrt 3 } }} = \sqrt 2
\end{array}$
