How do you simplify $\sqrt{\dfrac{{{x}^{9}}}{25}}$?

Hint: We first explain the process of root finding in case of fractions. The root of the whole fraction happens by finding the root of the denominator and numerator separately as \[\sqrt{\dfrac{p}{q}}=\dfrac{\sqrt{p}}{\sqrt{q}}\]. We find the root values and place to find the solution. We try to form the indices formula for the value 2. We find the prime factorisation of 25.

Complete step-by-step answer:
We have to solve the answer of $\sqrt{\dfrac{{{x}^{9}}}{25}}$. We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 2 we get \[{{a}^{\dfrac{1}{2}}}=\sqrt{a}\]. The main part in the root is a fraction.
For a general fraction $\dfrac{p}{q}$, the root of the whole thing works as separately taken roots of denominator and the numerator.
So, \[\sqrt{\dfrac{p}{q}}=\dfrac{\sqrt{p}}{\sqrt{q}}\].
This means to find the answer of $\sqrt{\dfrac{{{x}^{9}}}{25}}$, we need to find \[\sqrt{{{x}^{9}}}\] and $\sqrt{25}$.
Now we find the root values.
For \[\sqrt{{{x}^{9}}}=\sqrt{{{\left( {{x}^{3}} \right)}^{2}}}={{x}^{3}}\] and for $\sqrt{25}$, the prime factorisation of 25 will be
$\begin{align}
  & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore, $\sqrt{25}=\sqrt{5\times 5}=5$.
We now place the values and get $\sqrt{\dfrac{{{x}^{9}}}{25}}=\dfrac{\sqrt{{{x}^{9}}}}{\sqrt{25}}=\dfrac{{{x}^{3}}}{5}$.
The simplified value of the square root $\sqrt{\dfrac{{{x}^{9}}}{25}}$ is $\dfrac{{{x}^{3}}}{5}$.

Note: The thing we need to remember is that in normal cases of finding roots we find two values for square roots. But in this case, we only found and used positive results. We didn’t consider the negative part as the sign of the root was already provided as positive. If it was told to find the root of $\dfrac{{{x}^{9}}}{25}$, then we would have taken both positive and negative values.