Question

Fill in the blanks.
$(2^{-1}\times 3^{-1}{)}^2\times {\left(\frac{-3}{8}\right)}^{-1}$=________

Answer 1

Given that,

$\begin{array}{c}={\left(\frac{1}{2}\times \frac{1}{3}\right)}^2\times \left(\frac{1}{\frac{-3}{8}}\right)\\ ={\left(\frac{1}{6}\right)}^2\times \left(\frac{8}{-3}\right)\\ =\frac{1}{36}\times \frac{8}{-3}\\ =\frac{2}{-27}\end{array}$

Then,

We get $\frac{-2}{27}$