a) Ta có:
\(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}\)
\(=\dfrac{\sqrt{a}}{\sqrt b}+\sqrt{ab}+\dfrac{a}{b}.\dfrac{\sqrt{b}}{\sqrt a}\)
\(=\dfrac{\sqrt{a}.\sqrt b}{(\sqrt b)^2}+\sqrt{ab}+\dfrac{a}{b}.\dfrac{\sqrt{b}.\sqrt a}{(\sqrt a)^2}\)
\(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}+\dfrac{a}{b}.\dfrac{\sqrt{ab}}{a}\)
\(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}+\dfrac{\sqrt{ab}}{b}\)
\(={\left(\dfrac{\sqrt{ab}}{b}+\dfrac{\sqrt{ab}}{b} \right)}+\sqrt{ab}\)
\(=\dfrac{2\sqrt{ab}}{b}+\sqrt{ab}\)
\(=\dfrac{2\sqrt{ab}}{b}+\dfrac{b\sqrt{ab}}{b}\)
\(=\dfrac{2+b}{b}\sqrt{ab}\).
b) Ta có:
\(\sqrt{\dfrac{m}{1-2x+x^{2}}}.\sqrt{\dfrac{4m-8mx+4mx^{2}}{81}}\)
\(=\sqrt{\dfrac{m}{1-2x+x^{2}}}.\sqrt{\dfrac{4m(1-2x+x^{2})}{81}}\)
\(=\sqrt{\dfrac{m}{1-2x+x^{2}}.\dfrac{4m(1-2x+x^{2})}{81}}\)
\(=\sqrt{\dfrac{m}{1}.\dfrac{4m}{81}}=\sqrt{\dfrac{4m^{2}}{81}}\)
\(=\sqrt{\dfrac{(2m)^2}{9^2}}=\dfrac{|2m|}{9}=\dfrac{2m}{9}\).
(vì \(m >0\) nên \(|2m|=2m\).)
