Answer
$\dfrac{\sqrt{a}+1}{2\sqrt{a}-\sqrt{b}}=\dfrac{2a+\sqrt{ab}+2\sqrt{a}+\sqrt{b}}{4a-b}$
Work Step by Step
$\dfrac{\sqrt{a}+1}{2\sqrt{a}-\sqrt{b}}$
Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify if possible:
$\dfrac{\sqrt{a}+1}{2\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{a}+1}{2\sqrt{a}-\sqrt{b}}\cdot\dfrac{2\sqrt{a}+\sqrt{b}}{2\sqrt{a}+\sqrt{b}}=...$
$...=\dfrac{(\sqrt{a}+1)(2\sqrt{a}+\sqrt{b})}{(2\sqrt{a})^{2}-(\sqrt{b})^{2}}=\dfrac{2\sqrt{a^{2}}+\sqrt{ab}+2\sqrt{a}+\sqrt{b}}{4a-b}=...$
$...=\dfrac{2a+\sqrt{ab}+2\sqrt{a}+\sqrt{b}}{4a-b}$