#### Answer

$\sqrt{r}+3$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{r-9}{\sqrt{r}-3}
,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to simplify the result.
$\bf{\text{Solution Details:}}$
Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{r-9}{\sqrt{r}-3} \cdot\dfrac{\sqrt{r}+3}{\sqrt{r}+3}
\\\\=
\dfrac{(r-9)(\sqrt{r}+3)}{(\sqrt{r}-3)(\sqrt{r}+3)}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{(r-9)(\sqrt{r}+3)}{(\sqrt{r})^2-(3)^2}
\\\\=
\dfrac{(r-9)(\sqrt{r}+3)}{r-9}
\\\\=
\dfrac{(\cancel{r-9})(\sqrt{r}+3)}{\cancel{r-9}}
\\\\=
\sqrt{r}+3
.\end{array}