Answer
$\dfrac{1}{\sqrt {x+h-2}+\sqrt {x-2}}$
Work Step by Step
Step $1$.
Given $f(x)=\sqrt {x-2}$, we have $f(x+h)=\sqrt {x+h-2}$
Step $2$.
$f(x+h)-f(x)=\sqrt {x+h-2}-\sqrt {x-2}$
Step $3$.
(Use rationalizing the numerator as suggested.)
$\begin{align*}
\dfrac{f(x+h)-f(x)}{h}&=\dfrac{\sqrt {x+h-2}-\sqrt {x-2}}{h}\\
&=\dfrac{\sqrt {x+h-2}-\sqrt {x-2}}{h}\cdot\dfrac{\sqrt {x+h-2}+\sqrt {x-2}}{\sqrt {x+h-2}+\sqrt {x-2}}\\
&=\dfrac{x+h-2-x+2}{h(\sqrt {x+h-2}+\sqrt {x-2})}\\
&=\dfrac{h}{h(\sqrt {x+h-2}+\sqrt {x-2})}\\
&=\dfrac{1}{\sqrt {x+h-2}+\sqrt {x-2}}
\end{align*}$
