2017-I-02

16-06-2021 app.cch No Comments

Ans:

\frac{m^{22}}{n^{3}}

\begin{array}{cl} \frac{(m^{4} n^{- 1})^{3}}{(m^{- 2})^{5}} \\ = & \frac{m^{12} n^{- 3}}{m^{- 10}} \\ = & \frac{m^{12 - (- 10)}}{n^{3}} \\ = & \frac{m^{22}}{n^{3}} \end{array}

2017, HKDSE-MATH, Paper 1 Laws of Indices