2017-I-02

Ans: $\dfrac{m^{22}}{n^3}$
$\begin{array}{cl}
& \dfrac{(m^4n^{-1})^3}{(m^{-2})^5} \\
= & \dfrac{m^{12}n^{-3}}{m^{-10}} \\
= & \dfrac{m^{12 – (-10)}}{n^3} \\
= & \dfrac{m^{22}}{n^3}
\end{array}$