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