Question

Find the value of:

$\left(1\right)(2^0+3^{-1})\times 3^2$

$\left(ii\right)(2^{-1}\times 3^{-1})÷2^{-3}$

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

Answer 1

$\left(1\right)(2^0+3^{-1})\times 3^2=(1+\frac{1}{3})\times 9$

$=\frac{3+1}{3}\times 9=\frac{4}{3}\times 9=12$

$\left(ii\right)(2^{-1}\times 3^{-1})÷2^{-3}$

$=(\frac{1}{2}\times \frac{1}{3})÷\frac{1}{2^3}=\frac{1}{6}÷\frac{1}{8}$

$=\frac{1}{6}\times \frac{8}{1}=\frac{4}{3}$

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

$=(2{)}^2+(3{)}^2+(4{)}^2$

=4+9+16=29