#### Answer

$-\dfrac{7r\sqrt{2rs}}{s^3}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
-\sqrt{\dfrac{98r^3}{s^5}}
,$ multiply the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index.
$\bf{\text{Solution Details:}}$
Multiplying the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index results to
\begin{array}{l}\require{cancel}
-\sqrt{\dfrac{98r^3}{s^5}\cdot\dfrac{s}{s}}
\\\\=
-\sqrt{\dfrac{98r^3s}{s^6}}
.\end{array}
Writing the radicand as an expression that contains a factor that is a perfect power of the index results to
\begin{array}{l}\require{cancel}
-\sqrt{\dfrac{49r^2\cdot2rs}{s^6}}
\\\\=
-\sqrt{\left(\dfrac{7r}{s^3}\right)^2\cdot2rs}
.\end{array}
Extracting the root of the factor that is a perfect power of the index results to
\begin{array}{l}\require{cancel}
-\dfrac{7r}{s^3}\sqrt{2rs}
\\\\=
-\dfrac{7r\sqrt{2rs}}{s^3}
.\end{array}