Question

Simplify: $[{\left(\frac{-8}{16}\right)}^{-1}\times {\left(\frac{-16}{5}\right)}^{-1}]÷{\left(\frac{16}{5}\right)}^{-1}$

Answer 1

$[{\left(\frac{-8}{16}\right)}^{-1}\times {\left(\frac{-16}{5}\right)}^{-1}]÷{\left(\frac{16}{5}\right)}^{-1}$

Since, $(\frac{a}{b}{)}^{-1}=\frac{b}{a}$ and $a,b\ne 0$

$=[\left(\frac{16}{-8}\right)\times \left(\frac{5}{-16}\right)]÷\left(\frac{5}{16}\right)$

$=\frac{16}{-8}\times \frac{5}{-16}\times \frac{16}{5}$

$=\frac{16}{8}$

$=2$

$\therefore [{\left(\frac{-8}{16}\right)}^{-1}\times {\left(\frac{-16}{5}\right)}^{-1}]÷{\left(\frac{16}{5}\right)}^{-1}=2$