a) \(\displaystyle \sqrt {{4 \over 5}} = \sqrt {{{4.5} \over {5.5}}} = {{\sqrt {4.5} } \over {\sqrt {{5^2}} }} = {{2\sqrt 5 } \over 5}\)
b) \(\displaystyle \sqrt {{3 \over {125}}} = \sqrt {{{3.125} \over {125.125}}} = {{\sqrt {3.125} } \over {\sqrt {{{125}^2}} }} = {{5\sqrt {15} } \over {125}} = {{\sqrt {15} } \over {25}}\)
c) \(\sqrt {\dfrac{3}{{2{a^3}}}} = \dfrac{{\sqrt 3 }}{{\sqrt {2{a^3}} }} = \dfrac{{\sqrt 3 }}{{\sqrt {{a^2}.2a} }} = \dfrac{{\sqrt 3 }}{{\left| a \right|\sqrt {2a} }} = \dfrac{{\sqrt 3 }}{{a\sqrt {2a} }}\) \( = \dfrac{{\sqrt 3 .\sqrt {2a} }}{{a\sqrt {2a} .\sqrt {2a} }} = \dfrac{{\sqrt {6a} }}{{2{a^2}}}\)
