Đáp án:

$\begin{array}{l}
\dfrac{{2 + \sqrt 3 }}{{\sqrt 2  + \sqrt {2 + \sqrt 3 } }} + \dfrac{{2 - \sqrt 3 }}{{\sqrt 2  - \sqrt {2 - \sqrt 3 } }}\\
 = \dfrac{{2\sqrt 2  + \sqrt 6 }}{{2 + \sqrt {4 + 2\sqrt 3 } }} + \dfrac{{2\sqrt 2  - \sqrt 6 }}{{2 - \sqrt {4 - 2\sqrt 3 } }}\\
 = \dfrac{{2\sqrt 2  + \sqrt 6 }}{{2 + \sqrt {{{\left( {\sqrt 3  + 1} \right)}^2}} }} + \dfrac{{2\sqrt 2  - \sqrt 6 }}{{2 - \sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}} }}\\
 = \dfrac{{2\sqrt 2  + \sqrt 6 }}{{2 + \sqrt 3  + 1}} + \dfrac{{2\sqrt 2  - \sqrt 6 }}{{2 - \sqrt 3  + 1}}\\
 = \dfrac{{2\sqrt 2  + \sqrt 6 }}{{3 + \sqrt 3 }} + \dfrac{{2\sqrt 2  - \sqrt 6 }}{{3 - \sqrt 3 }}\\
 = \dfrac{{\left( {2\sqrt 2  + \sqrt 6 } \right).\left( {3 - \sqrt 3 } \right) + \left( {2\sqrt 2  - \sqrt 6 } \right).\left( {3 + \sqrt 3 } \right)}}{{\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}\\
 = \dfrac{{\sqrt 2 .\sqrt 3 \left[ {\left( {2 + \sqrt 3 } \right).\left( {1 - \sqrt 3 } \right) + \left( {2 - \sqrt 3 } \right)\left( {1 + \sqrt 3 } \right)} \right]}}{{9 - 3}}\\
 = \dfrac{{\sqrt 6 .\left( {2 - 2\sqrt 3  + \sqrt 3  - 3 + 2 + 2\sqrt 3  - \sqrt 3  - 3} \right)}}{6}\\
 = \dfrac{{\sqrt 6 .\left( { - 2} \right)}}{6}\\
 = \dfrac{{ - \sqrt 6 }}{3}
\end{array}$