Đáp án:
$\begin{array}{l}
B1)\\
a)\sqrt {\dfrac{{27}}{{242}}} = \sqrt {\dfrac{{9.3}}{{2.121}}} = \dfrac{{3\sqrt 3 }}{{11\sqrt 2 }} = \dfrac{{3\sqrt 6 }}{{22}}\\
b)ab\sqrt {\dfrac{{18}}{{ab}}} = \sqrt {ab} .3\sqrt 2 = 3\sqrt {2ab} \\
c)b\sqrt {\dfrac{a}{{ - b}}} = \sqrt {{b^2}.\dfrac{a}{{ - b}}} = \sqrt { - ab} \\
B2)\\
a)\dfrac{2}{{3\sqrt 5 }} = \dfrac{{2.\sqrt 5 }}{{3.5}} = \dfrac{{2\sqrt 5 }}{{15}}\\
b)\dfrac{{\sqrt 2 - \sqrt 6 }}{{6\sqrt 2 }} = \dfrac{{\sqrt 2 :\sqrt 2 - \sqrt 6 :\sqrt 2 }}{6} = \dfrac{{1 - \sqrt 3 }}{6}\\
c)\dfrac{{1 + \sqrt 2 }}{{1 - \sqrt 2 }} = \dfrac{{{{\left( {1 + \sqrt 2 } \right)}^2}}}{{\left( {1 - \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}} = \dfrac{{3 + 2\sqrt 2 }}{{1 - 2}}\\
= - 3 - 2\sqrt 2
\end{array}$
