Đáp án:
$\begin{array}{l}
a)\sqrt {\dfrac{5}{7}} = \dfrac{{\sqrt {35} }}{7}\\
\sqrt {\dfrac{7}{{20}}} = \dfrac{{\sqrt 7 }}{{2\sqrt 5 }} = \dfrac{{\sqrt {35} }}{{10}}\\
b)\dfrac{{12}}{{5\sqrt 3 }} = \dfrac{{4\sqrt 3 }}{5}\\
\dfrac{2}{{\sqrt 2 }} = \sqrt 2 \\
c)\dfrac{{2 - \sqrt 3 }}{{3\sqrt 5 }} = \dfrac{{\left( {2 - \sqrt 3 } \right).\sqrt 5 }}{{15}} = \dfrac{{2\sqrt 5 - \sqrt {15} }}{{\sqrt {15} }}\\
\dfrac{1}{{\sqrt 3 + \sqrt 7 }} = \dfrac{{\sqrt 7 - \sqrt 3 }}{{7 - 3}} = \dfrac{{\sqrt 7 - \sqrt 3 }}{4}\\
d)\dfrac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} = \dfrac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{{5 - 1}} = \dfrac{{6 - 2\sqrt 5 }}{4} = \dfrac{{3 - \sqrt 5 }}{2}\\
\dfrac{{2\sqrt {10} - 5}}{{4 - \sqrt {10} }} = \dfrac{{\left( {2\sqrt {10} - 5} \right)\left( {4 + \sqrt {10} } \right)}}{6}\\
= \dfrac{{8\sqrt {10} + 20 - 20 - 5\sqrt {10} }}{6}\\
= \dfrac{{3\sqrt {10} }}{6}\\
= \dfrac{{\sqrt {10} }}{2}\\
e)\dfrac{x}{{2\sqrt x - 1}} = \dfrac{{x\left( {2\sqrt x + 1} \right)}}{{4x - 1}} = \dfrac{{2x\sqrt x + x}}{{4x - 1}}\\
\dfrac{1}{{\sqrt x - \sqrt y }} = \dfrac{{\sqrt x + \sqrt y }}{{x - y}}
\end{array}$
