Answer
$\dfrac{2\sqrt{x}-6}{x-9}$
Work Step by Step
Rationalizing the denominator of $
\dfrac{2}{\sqrt{x}+3}
$, results to
\begin{array}{l}
\dfrac{2}{\sqrt{x}+3}
\cdot
\dfrac{\sqrt{x}-3}{\sqrt{x}-3}
\\\\=
\dfrac{2\sqrt{x}-6}{(\sqrt{x})^2-(3)^2}
\\\\=
\dfrac{2\sqrt{x}-6}{x-9}
.\end{array}