a) \(với\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}=\dfrac{2a+2a\sqrt{a}}{1-a}\)
b)\(vớia>b>0\)
\(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left(2\sqrt{a}+\sqrt{b}\right)}{4a-b}=\dfrac{12a\sqrt{a}+6a\sqrt{b}}{4a-b}\)