#### Answer

$\dfrac{a-b}{a-2\sqrt{ab}+b}$

#### Work Step by Step

Multiplying by the conjugate of the numerator, the rationalized-numerator form of the given expression, $
\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}
\\\\=
\dfrac{(\sqrt{a})^2-(\sqrt{b})^2}{\sqrt{a}(\sqrt{a})+\sqrt{a}(-\sqrt{b})-\sqrt{b}(\sqrt{a})-\sqrt{b}(-\sqrt{b})}
\\\\=
\dfrac{a-b}{\sqrt{a(a)}-\sqrt{a(b)}-\sqrt{b(a)}+\sqrt{b(b)}}
\\\\=
\dfrac{a-b}{\sqrt{(a)^2}-\sqrt{ab}-\sqrt{ab}+\sqrt{(b)^2}}
\\\\=
\dfrac{a-b}{a-2\sqrt{ab}+b}
.\end{array}