Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{26}}{{5 + 2\sqrt 3 }} = \dfrac{{26.\left( {5 - 2\sqrt 3 } \right)}}{{\left( {5 + 2\sqrt 3 } \right)\left( {5 - 2\sqrt 3 } \right)}} = \dfrac{{26.\left( {5 - 2\sqrt 3 } \right)}}{{{5^2} - {{\left( {2\sqrt 3 } \right)}^2}}} = \dfrac{{26.\left( {5 - 2\sqrt 3 } \right)}}{{13}} = 2.\left( {5 - 2\sqrt 3 } \right)\\
b,\\
\dfrac{{2\sqrt 6 - 3}}{{4 - \sqrt 6 }} = \dfrac{{2.\sqrt 2 .\sqrt 3 - {{\sqrt 3 }^2}}}{{2.2 - \sqrt 2 .\sqrt 3 }} = \dfrac{{\sqrt 3 .\left( {2\sqrt 2 - \sqrt 3 } \right)}}{{\sqrt 2 .\left( {2\sqrt 2 - \sqrt 3 } \right)}} = \dfrac{{\sqrt 3 }}{{\sqrt 2 }} = \dfrac{{\sqrt 6 }}{2}\\
c,\\
\dfrac{6}{{2\sqrt 3 + \sqrt 2 }} = \dfrac{{6.\left( {2\sqrt 3 - \sqrt 2 } \right)}}{{\left( {2\sqrt 3 + \sqrt 2 } \right)\left( {2\sqrt 3 - \sqrt 2 } \right)}} = \dfrac{{6.\left( {2\sqrt 3 - \sqrt 2 } \right)}}{{{{\left( {2\sqrt 3 } \right)}^2} - {{\sqrt 2 }^2}}} = \dfrac{{6\left( {2\sqrt 3 - \sqrt 2 } \right)}}{{10}} = \dfrac{{3.\left( {2\sqrt 3 - \sqrt 2 } \right)}}{5}\\
d,\\
\dfrac{5}{{\sqrt 7 - \sqrt 2 }} = \dfrac{{5.\left( {\sqrt 7 + \sqrt 2 } \right)}}{{\left( {\sqrt 7 - \sqrt 2 } \right)\left( {\sqrt 7 + \sqrt 2 } \right)}} = \dfrac{{5.\left( {\sqrt 7 + \sqrt 2 } \right)}}{{7 - 2}} = \sqrt 7 + \sqrt 2 \\
e,\\
\dfrac{4}{{\sqrt 3 + \sqrt 2 }} = \dfrac{{4.\left( {\sqrt 3 - \sqrt 2 } \right)}}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}} = \dfrac{{4.\left( {\sqrt 3 - \sqrt 2 } \right)}}{{3 - 2}} = 4.\left( {\sqrt 3 - \sqrt 2 } \right)
\end{array}\)
