Đáp án:
$\begin{array}{l}
a)\sqrt {\dfrac{3}{{50}}} = \dfrac{{\sqrt 3 }}{{5\sqrt 2 }} = \dfrac{{\sqrt 6 }}{{10}}\\
b)\sqrt {\dfrac{5}{{98}}} = \dfrac{{\sqrt 5 }}{{7\sqrt 2 }} = \dfrac{{\sqrt {10} }}{{14}}\\
c)\sqrt {\dfrac{{{{\left( {1 - \sqrt 3 } \right)}^2}}}{{27}}} \\
= \dfrac{{\sqrt 3 - 1}}{{3\sqrt 3 }} = \dfrac{{\sqrt 3 \left( {\sqrt 3 - 1} \right)}}{9} = \dfrac{{3 - \sqrt 3 }}{9}\\
d)\dfrac{a}{b}\sqrt {\dfrac{b}{a}} = \dfrac{{\sqrt a }}{{\sqrt b }} = \dfrac{{\sqrt {ab} }}{b}\\
e)\sqrt {\dfrac{1}{b} + \dfrac{1}{{{b^2}}}} = \sqrt {\dfrac{{b + 1}}{{{b^2}}}} = \dfrac{{\sqrt {b + 1} }}{{\left| b \right|}}\\
f)3xy\sqrt {\dfrac{2}{{xy}}} = 3\sqrt {xy} .\sqrt 2 = 3\sqrt {2xy} \\
g)\dfrac{1}{{3\sqrt {20} }} = \dfrac{1}{{3.2\sqrt 5 }} = \dfrac{{\sqrt 5 }}{{30}}\\
h)\dfrac{{2\sqrt 2 + 2}}{{5\sqrt 2 }} = \dfrac{{\sqrt 2 \left( {2 + \sqrt 2 } \right)}}{{5\sqrt 2 }} = \dfrac{{2 + \sqrt 2 }}{5}\\
i)\dfrac{{y + b\sqrt y }}{{b\sqrt y }} = \dfrac{{\sqrt y \left( {\sqrt y + b} \right)}}{{b\sqrt y }} = \dfrac{{b + \sqrt y }}{b}\\
k)\dfrac{p}{{2\sqrt p - 1}} = \dfrac{{p\left( {2\sqrt p + 1} \right)}}{{\left( {2\sqrt p - 1} \right)\left( {2\sqrt p + 1} \right)}}\\
= \dfrac{{2p\sqrt p + p}}{{4p - 1}}\\
l)\dfrac{3}{{\sqrt {10} + 7}} = \dfrac{{3\left( {7 - \sqrt {10} } \right)}}{{49 - 10}} = \dfrac{{7 - \sqrt {10} }}{{13}}\\
m)\dfrac{y}{{\sqrt x - \sqrt y }} = \dfrac{{y\left( {\sqrt x + \sqrt y } \right)}}{{x - y}} = \dfrac{{y\sqrt x + y\sqrt y }}{{x - y}}\\
n)\dfrac{{2ab}}{{\sqrt a - \sqrt b }} = \dfrac{{2ab\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}
\end{array}$
