Question

The multiplicative inverse of
${\left(\frac{1}{2}\right)}^4÷{\left(\frac{1}{3}\right)}^4+{\left(-\frac{1}{2}\right)}^{{}^3}\text{is}$

Answer 1

$={\left(\frac{1}{2}\right)}^4÷{\left(\frac{1}{3}\right)}^4+{\left(\frac{-1}{2}\right)}^3$

$=\frac{1}{16}÷\frac{1}{81}+\left(\frac{-1}{8}\right)$

$=\frac{1}{16}\times \frac{81}{1}+\frac{-1}{8}$

$=\frac{81}{16}-\frac{1}{8}$

$=\frac{81-2}{16}$

$=\frac{79}{16}$ Ans