Question

how to do Q 2 and Q6

​Q9. Simplify :

(ii) $\frac{{\left({\displaystyle \frac{3}{5}}\right)}^3\times {\left({\displaystyle \frac{1}{7}}\right)}^3}{{\left({\displaystyle \frac{3}{5}}\right)}^2\times {\left({\displaystyle \frac{1}{7}}\right)}^4}$ (vi) $\left(2^{-1}÷5^{-1}\right)\times {\left(\frac{-5}{8}\right)}^{-1}$

Answer 1

$$\begin{gathered} \left(ii\right)\frac{{\left({\displaystyle \frac{3}{5}}\right)}^3\times {\left({\displaystyle \frac{1}{7}}\right)}^3}{{\left({\displaystyle \frac{3}{5}}\right)}^2\times {\left({\displaystyle \frac{1}{7}}\right)}^4} \\ ={\left(\frac{3}{5}\right)}^{3-2}\times {\left(\frac{1}{7}\right)}^{3-4}\left[as,a^m÷a^n=a^{m-n}\right] \\ ={\left(\frac{3}{5}\right)}^1\times {\left(\frac{1}{7}\right)}^{-1} \\ =\frac{3}{5}\times 7\left[as,a^{-m}=\frac{1}{a^m}\right] \\ =\frac{21}{5} \\ \left(vi\right)\left(2^{-1}÷5^{-1}\right)\times {\left(\frac{-5}{8}\right)}^{-1} \\ =\left[\frac{1}{2}÷\frac{1}{5}\right]\times \left(\frac{-8}{5}\right) \\ =\left(\frac{1}{2}\times 5\right)\times \left(\frac{-8}{5}\right) \\ =\frac{5}{2}\times \left(\frac{-8}{5}\right) \\ =-4 \end{gathered}$$