Answer
$\dfrac{2\sqrt{a}-3}{2\sqrt{a}-\sqrt{b}}=\dfrac{4a+2\sqrt{ab}-6\sqrt{a}-3\sqrt{b}}{4a-b}$
Work Step by Step
$\dfrac{2\sqrt{a}-3}{2\sqrt{a}-\sqrt{b}}$
Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify if possible:
$\dfrac{2\sqrt{a}-3}{2\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{a}-3}{2\sqrt{a}-\sqrt{b}}\cdot\dfrac{2\sqrt{a}+\sqrt{b}}{2\sqrt{a}+\sqrt{b}}=...$
$...=\dfrac{(2\sqrt{a}-3)(2\sqrt{a}+\sqrt{b})}{(2\sqrt{a})^{2}-(\sqrt{b})^{2}}=...$
$...=\dfrac{(2\sqrt{a})^{2}+2\sqrt{ab}-6\sqrt{a}-3\sqrt{b}}{4a-b}=\dfrac{4a+2\sqrt{ab}-6\sqrt{a}-3\sqrt{b}}{4a-b}$
