Đáp án:
$\begin{array}{l}
a)\dfrac{{5 + 2\sqrt 5 }}{{\sqrt 5 + \sqrt 2 }} = \dfrac{{\left( {5 + 2\sqrt 5 } \right)\left( {\sqrt 5 - \sqrt 2 } \right)}}{{5 - 2}}\\
= \dfrac{{5\sqrt 5 - 5\sqrt 2 + 10 - 2\sqrt {10} }}{3}\\
b)\sqrt {\dfrac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }}} = \sqrt {\dfrac{{{{\left( {2 - \sqrt 3 } \right)}^2}}}{{{2^2} - 3}}} = \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} = 2 - \sqrt 3 \\
c)\left( {\dfrac{2}{{\sqrt 3 - 1}} + \dfrac{3}{{\sqrt 3 - 2}} + \dfrac{{15}}{{3 - \sqrt 3 }}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \left( {\dfrac{{2\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} + \dfrac{{3\left( {\sqrt 3 + 2} \right)}}{{3 - {2^2}}} + \dfrac{{15\left( {3 + \sqrt 3 } \right)}}{{{3^2} - 3}}} \right).\dfrac{{5 - \sqrt 3 }}{{25 - 3}}\\
= \left( {\sqrt 3 + 1 - 3\sqrt 3 - 6 + \dfrac{{15 + 5\sqrt 3 }}{2}} \right).\dfrac{{5 - \sqrt 3 }}{{22}}\\
= \dfrac{{\sqrt 3 + 5}}{2}.\dfrac{{5 - \sqrt 3 }}{{22}}\\
= \dfrac{{{5^2} - 3}}{{2.22}}\\
= \dfrac{1}{2}\\
d)\left( {\dfrac{{\sqrt {14} - \sqrt 7 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt {15} - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 7 - \sqrt 5 }}\\
= \left( {\dfrac{{\sqrt 7 \left( {\sqrt 2 - 1} \right)}}{{1 - \sqrt 2 }} + \dfrac{{\sqrt 5 \left( {\sqrt 3 - 1} \right)}}{{1 - \sqrt 3 }}} \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= \left( { - \sqrt 7 - \sqrt 5 } \right)\left( {\sqrt 7 - \sqrt 5 } \right)\\
= - \left( {\sqrt 7 + \sqrt 5 } \right)\left( {\sqrt 7 - \sqrt 5 } \right)\\
= - \left( {7 - 5} \right)\\
= - 2
\end{array}$
