a) Ta có \(\sqrt {\dfrac{{2{a^2}{b^4}}}{{50}}} = \sqrt {\dfrac{{{a^2}{b^4}}}{{25}}} = \dfrac{{\sqrt {{a^2}{b^4}} }}{{\sqrt {25} }} = \dfrac{{\sqrt {{a^2}} .\sqrt {{b^4}} }}{5} = \dfrac{{|a|{b^2}}}{5}\)
b) Ta có \(\dfrac{{\sqrt {2a{b^2}} }}{{\sqrt {162} }} = \sqrt {\dfrac{{2a{b^2}}}{{162}}} = \sqrt {\dfrac{{a{b^2}}}{{81}}} = \dfrac{{\sqrt {a{b^2}} }}{{\sqrt {81} }} = \dfrac{{\sqrt a .\sqrt {{b^2}} }}{9} = \dfrac{{\left| b \right|\sqrt a }}{9}\)
