Answer
\[\frac{-3{{y}^{15}}}{{{x}^{3}}}\].
Work Step by Step
Quotient rule:
When exponential expressions with same base are divided then subtract the exponent in the denominator with the exponent in the numerator of the common base:
\[\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\].
So,
\[\begin{align}
& \frac{24{{x}^{2}}{{y}^{13}}}{-8{{x}^{5}}{{y}^{-2}}}=-\frac{24}{8}\left( {{x}^{2-5}}{{y}^{13-\left( -2 \right)}} \right) \\
& =-3{{x}^{-3}}{{y}^{13+2}} \\
& =-3{{x}^{-3}}{{y}^{15}}
\end{align}\]
Negative exponent rule:
For any real number \[b\]other than zero and \[m\]a natural number \[{{b}^{-m}}=\frac{1}{{{b}^{m}}}\].
Here, \[x\ne 0\]. So,
\[-3{{x}^{-3}}{{y}^{15}}=\frac{-3{{y}^{15}}}{{{x}^{3}}}\]
So, \[\frac{24{{x}^{2}}{{y}^{13}}}{-8{{x}^{5}}{{y}^{-2}}}=\frac{-3{{y}^{15}}}{{{x}^{3}}}\].
