Đáp án:
$\begin{array}{l}
a)\left( {\dfrac{1}{2}\sqrt {\dfrac{1}{2}} - \dfrac{3}{2}\sqrt {4,5} + \dfrac{2}{5}\sqrt {50} } \right):\dfrac{4}{{15}}\sqrt {\dfrac{1}{8}} \\
= \left( {\dfrac{1}{2}.\dfrac{{\sqrt 2 }}{2} - \dfrac{3}{2}\sqrt {\dfrac{9}{2}} + \dfrac{2}{5}.5\sqrt 2 } \right):\left( {\dfrac{4}{{15}}.\dfrac{1}{{2\sqrt 2 }}} \right)\\
= \left( {\dfrac{{\sqrt 2 }}{4} - \dfrac{3}{2}.\dfrac{{3\sqrt 2 }}{2} + 2\sqrt 2 } \right):\dfrac{{\sqrt 2 }}{{15}}\\
= \left( {\dfrac{{\sqrt 2 }}{4} - \dfrac{{9\sqrt 2 }}{4} + 2\sqrt 2 } \right).\dfrac{{15}}{{\sqrt 2 }}\\
= \left( { - 2\sqrt 2 + 2\sqrt 2 } \right).\dfrac{{15}}{{\sqrt 2 }}\\
= 0\\
b)\sqrt {{{\left( {\sqrt 3 - 3} \right)}^2}} .\sqrt {\dfrac{1}{{3 - \sqrt 3 }}} \\
= \left( {3 - \sqrt 3 } \right).\dfrac{1}{{\sqrt {3 - \sqrt 3 } }}\\
= \sqrt {3 - \sqrt 3 } \\
c)\sqrt {\sqrt 2 - 1} .\sqrt {\sqrt 2 - 1} \\
= \sqrt {\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)} \\
= \sqrt {{2^2} - 1} \\
= \sqrt 1 \\
= 1
\end{array}$
