Answer
$\dfrac{15\sqrt{r}-5\sqrt{s}}{9r-s}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{5}{3\sqrt{r}+\sqrt{s}}
,$ 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{5}{3\sqrt{r}+\sqrt{s}}\cdot\dfrac{3\sqrt{r}-\sqrt{s}}{3\sqrt{r}-\sqrt{s}}
\\\\=
\dfrac{5(3\sqrt{r}-\sqrt{s})}{(3\sqrt{r}+\sqrt{s})(3\sqrt{r}-\sqrt{s})}
.\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{5(3\sqrt{r}-\sqrt{s})}{(3\sqrt{r})^2-(\sqrt{s})^2}
\\\\=
\dfrac{5(3\sqrt{r}-\sqrt{s})}{9r-s}
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{5(3\sqrt{r})+5(-\sqrt{s})}{9r-s}
\\\\=
\dfrac{15\sqrt{r}-5\sqrt{s}}{9r-s}
.\end{array}
