Answer
$\dfrac{\sqrt{r}+\sqrt{3}}{r-3}$
Work Step by Step
Rationalize the denominator by multiplying $\sqrt{r}+\sqrt{3}$ to both the numerator and the denominator to have:
$=\dfrac{1}{\sqrt{r}-\sqrt{3}} \cdot \dfrac{\sqrt{r}+\sqrt{3}}{\sqrt{r}+\sqrt{3}}
\\=\dfrac{\sqrt{r}+\sqrt{3}}{r-3}$
