a) \(ĐK:a>0\)
\(=\sqrt{\dfrac{3a}{27a}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
b) \(=\dfrac{\sqrt{15}-\sqrt{6}}{\dfrac{7\sqrt{3}}{3}\left(\sqrt{15}-\sqrt{6}\right)}=\dfrac{1}{\dfrac{7\sqrt{3}}{3}}=\dfrac{\sqrt{3}}{7}\)
c) \(=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{5}}{2}\)
d) \(=\dfrac{\sqrt{a}\left(\sqrt{ab}+1\right)-\sqrt{b}\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}=\dfrac{\left(\sqrt{ab}+1\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\)
a: \(\sqrt{\dfrac{3a}{5}}\cdot\sqrt{\dfrac{5}{27a}}=\sqrt{\dfrac{3a}{27a}}=\dfrac{1}{3}\)
b: \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{245}-\sqrt{98}}=\dfrac{\sqrt{3}}{7}\)
