Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{2 + \sqrt 3 }}{{\sqrt 6 - \sqrt 3 + \sqrt 2 - 1}}\\
= \dfrac{{2 + \sqrt 3 }}{{\left( {\sqrt 6 - \sqrt 3 } \right) + \left( {\sqrt 2 - 1} \right)}}\\
= \dfrac{{2 + \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 2 - 1} \right) + \left( {\sqrt 2 - 1} \right)}}\\
= \dfrac{{2 + \sqrt 3 }}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\\
= \dfrac{{\left( {2 + \sqrt 3 } \right)\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right).\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}\\
= \dfrac{{2\sqrt 3 - 2 + 3 - \sqrt 3 }}{{\left( {\sqrt 2 - 1} \right).\left( {3 - 1} \right)}}\\
= \dfrac{{\sqrt 3 + 1}}{{2.\left( {\sqrt 2 - 1} \right)}}\\
= \dfrac{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 2 + 1} \right)}}{{2.\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}\\
= \dfrac{{\sqrt 6 + \sqrt 3 + \sqrt 2 + 1}}{{2.\left( {2 - 1} \right)}}\\
= \dfrac{{\sqrt 6 + \sqrt 3 + \sqrt 2 + 1}}{2}
\end{array}\)
