#### Answer

$\dfrac{1}{\sqrt{a+h}+\sqrt{a}}$

#### Work Step by Step

Multiplying by the conjugate of the numerator, the rationalized-numerator form of the given expression, $
\dfrac{\sqrt{a+h}-\sqrt{a}}{h}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{a+h}-\sqrt{a}}{h}\cdot\dfrac{\sqrt{a+h}+\sqrt{a}}{\sqrt{a+h}+\sqrt{a}}
\\\\=
\dfrac{(\sqrt{a+h})^2-(\sqrt{a})^2}{h(\sqrt{a+h})+h(\sqrt{a})}
\\\\=
\dfrac{a+h-a}{h\sqrt{a+h}+h\sqrt{a}}
\\\\=
\dfrac{h}{h(\sqrt{a+h}+\sqrt{a})}
\\\\=
\dfrac{\cancel{h}}{\cancel{h}(\sqrt{a+h}+\sqrt{a})}
\\\\=
\dfrac{1}{\sqrt{a+h}+\sqrt{a}}
.\end{array}