Question

Q6. Evaluate.
(1) $\left\{{\left(\frac{1}{2}\right)}^{-3}-{\left(\frac{1}{3}\right)}^{-2}\right\}÷{\left(\frac{1}{5}\right)}^{-2}$ (2) ${\left\{{\left(\frac{2}{3}\right)}^{-2}-{\left(\frac{3}{4}\right)}^{-1}\right\}}^{-2}$
(3) $\left\{{\left(\frac{1}{2}\right)}^{-3}-{\left(\frac{1}{3}\right)}^{-2}\right\}÷{\left(\frac{1}{4}\right)}^{-1}$

Answer 1

Dear student,


$$\begin{gathered} 6.\left(iii\right){\left\{{\left(\frac{1}{2}\right)}^{-3}+{\left(\frac{1}{3}\right)}^{-2}\right\}}^{-1}÷{\left(\frac{1}{4}\right)}^{-1} \\ ={\left\{{\left(2\right)}^3+{\left(3\right)}^2\right\}}^{-1}÷4--[u\sin g{a}^{-m}=\frac{1}{a^m}] \\ ={\left\{8+9\right\}}^{-1}÷4 \\ =\left(17\right){}^{-1}÷4 \\ =\frac{1}{17}÷4--[u\sin g{a}^{-m}=\frac{1}{a^m}] \\ =\frac{1}{17}\times \frac{1}{4} \\ =\frac{1}{68} \\ 7a)Letxbymultipliedby{\left(\frac{-3}{4}\right)}^{2}toget{\left(\frac{-2}{3}\right)}^{-1} \\ \Rightarrow x\times {\left(\frac{-3}{4}\right)}^2={\left(\frac{-2}{3}\right)}^{-1} \\ \Rightarrow x\times {\left(\frac{-3}{4}\right)}^2=\left(\frac{3}{-2}\right)--[u\sin g{a}^{-m}=\frac{1}{a^m}] \\ \Rightarrow x\times \frac{9}{16}=\frac{3}{-2} \\ \Rightarrow x=\frac{3}{-2}\times \frac{16}{9} \\ \Rightarrow x=-\frac{1}{1}\times \frac{8}{3} \\ \Rightarrow \boxed{x=-\frac{8}{3}} \end{gathered}$$

Regards