Answer
$$\dfrac{\sqrt{-35}}{7}$$
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 $a \ge 0, b\gt 0$, $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$
Use rule (2) above to obtain:
\begin{align*}
\sqrt{\dfrac{-5}{7}}&=\dfrac{\sqrt{-5}}{\sqrt7}\\
\end{align*}
Rationalize the denominator by multiplying $\sqrt7$ to both the numerator and denominator to obtain:
\begin{align*}
\dfrac{\sqrt{-5}}{\sqrt7} \cdot \dfrac{\sqrt7}{\sqrt7}&=\dfrac{\sqrt{-35}}{\sqrt{49}}\\\\
&=\dfrac{\sqrt{-35}}{7}
\end{align*}
