Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{2}{{\sqrt 6 - 2}} + \dfrac{2}{{\sqrt 6 + 2}} + \dfrac{5}{{\sqrt 6 }}\\
= \dfrac{{2.\left( {\sqrt 6 + 2} \right) + 2.\left( {\sqrt 6 - 2} \right)}}{{\left( {\sqrt 6 - 2} \right).\left( {\sqrt 6 + 2} \right)}} + \dfrac{5}{{\sqrt 6 }}\\
= \dfrac{{2\sqrt 6 + 4 + 2\sqrt 6 - 4}}{{{{\sqrt 6 }^2} - {2^2}}} + \dfrac{5}{{\sqrt 6 }}\\
= \dfrac{{4\sqrt 6 }}{2} + \dfrac{5}{{\sqrt 6 }}\\
= 2\sqrt 6 + \dfrac{5}{{\sqrt 6 }}\\
= \dfrac{{2\sqrt 6 .\sqrt 6 + 5}}{{\sqrt 6 }}\\
= \dfrac{{17}}{{\sqrt 6 }}\\
b,\\
\dfrac{1}{{\sqrt 3 + \sqrt 2 - \sqrt 5 }} - \dfrac{1}{{\sqrt 3 + \sqrt 2 + \sqrt 5 }}\\
= \dfrac{{\left( {\sqrt 3 + \sqrt 2 + \sqrt 5 } \right) - \left( {\sqrt 3 + \sqrt 2 - \sqrt 5 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 - \sqrt 5 } \right).\left( {\sqrt 3 + \sqrt 2 + \sqrt 5 } \right)}}\\
= \dfrac{{2\sqrt 5 }}{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2} - {{\sqrt 5 }^2}}}\\
= \dfrac{{2\sqrt 5 }}{{3 + 2.\sqrt 3 .\sqrt 2 + 2 - 5}}\\
= \dfrac{{2\sqrt 5 }}{{2\sqrt 6 }}\\
= \sqrt {\dfrac{5}{6}}
\end{array}\)
