a) Ta có \(\sqrt {28{a^4}{b^2}} = \sqrt {{{7.2}^2}.{{\left( {{a^2}} \right)}^2}{b^2}} = 2{a^2}\left| b \right|\sqrt 7 \) mà \(b \ge 0 \Rightarrow \left| b \right| = b\) nên \(\sqrt {28{a^4}{b^2}} = 2{a^2}b\sqrt 7 \)
b) Ta có \(\sqrt {72{a^2}{b^4}} = \sqrt {{2^2}{{.2.3}^2}.{a^2}.{{\left( {{b^2}} \right)}^2}} = 2.3.\left| a \right|.{b^2}\sqrt 2 \) mà \(a < 0 \Rightarrow \left| a \right| = - a\)
nên \(\sqrt {72{a^2}{b^4}} = - 6a{b^2}\sqrt 2 .\)
