Answer
$$-\dfrac{|x|\sqrt{35}}{5}$$
Work Step by Step
RECALL:
(1) For any real numbers $a \ge 0, b\ge0$, $\sqrt{ab}=\sqrt{a} \cdot \sqrt{b}$
(2) For any real numbers $a \ge 0, b\gt 0$, $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$.
(3) $\sqrt{a^2}=|a|$
Simplify the radicand (expression inside the radical sign) by cancelling the common factor $x$ to obtain:
\begin{align*}
\require{cancel}
-\sqrt{\dfrac{7\cancel{x^3}^{x^2}}{5\cancel{x}}}&=-\sqrt{\dfrac{7x^2}{5}}
\end{align*}
Use rule (2) above to obtain:
\begin{align*}
-\sqrt{\dfrac{7x^2}{5}}&=-\dfrac{\sqrt{7x^2}}{\sqrt{5}}\\\\
\end{align*}
Rationalize the denominator by multiplying $\sqrt5$ to both the numerator and denominator to obtain:
\begin{align*}
-\dfrac{\sqrt{7x^2}}{\sqrt{5}} \cdot \dfrac{\sqrt5}{\sqrt5}&=-\dfrac{\sqrt{35x^2}}{\sqrt{25}}\\\\
&=-\dfrac{\sqrt{35x^2}}{5}
\end{align*}
Using rules (1) and (3) above gives:
\begin{align*}
-\dfrac{\sqrt{35x^2}}{5}&=-\dfrac{\sqrt{35} \cdot \sqrt{x^2}}{5}\\\\
&=-\dfrac{\sqrt{35} \cdot |x|}{5}\\\\
&=-\dfrac{|x|\sqrt{35}}{5}
\end{align*}
