Answer
$\dfrac{4a-2\sqrt{ab}-6\sqrt{a}+3\sqrt{b}}{4a-b}$
Work Step by Step
Rationalizing the denominator of $
\dfrac{2\sqrt{a}-3}{2\sqrt{a}+\sqrt{b}}
$ results to
\begin{array}{l}
\dfrac{2\sqrt{a}-3}{2\sqrt{a}+\sqrt{b}}
\cdot
\dfrac{2\sqrt{a}-\sqrt{b}}{2\sqrt{a}-\sqrt{b}}
\\\\=
\dfrac{(2\sqrt{a})(2\sqrt{a})+(2\sqrt{a})(-\sqrt{b})+(-3)(2\sqrt{a})+(-3)(-\sqrt{b})}{(2\sqrt{a})^2-(\sqrt{b})^2}
\\\\=
\dfrac{4a-2\sqrt{ab}-6\sqrt{a}+3\sqrt{b}}{4a-b}
.\end{array}