Question

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

Answer 1

The given value is :

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

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

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

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

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

Hence, multiplicative inverse $=\frac{1}{a}$ $=\frac{16}{79}$.