Question

$\left[\right(\frac{1}{2}{)}^{-1}+(\frac{2}{3}{)}^2-(\frac{3}{4}{)}^0{]}^{-2}$ is equal to

• A. $\frac{81}{169}$
• B. $\frac{-81}{169}$
• C. $\frac{81}{169}$
• D. $\frac{169}{81}$

Answer 1

The correct option is A $\frac{81}{169}$

Given, $\left[\right(\frac{1}{2}{)}^{-1}+(\frac{2}{3}{)}^2-(\frac{3}{4}{)}^0{]}^{-2}$

We know that, $a^{-n}=\frac{1}{a^n}$.

$=[2+(\frac{2}{3}{)}^2-1{]}^{-2}$

$\Rightarrow$ $[2+(\frac{4}{9})-1{]}^{-2}$

$\Rightarrow$ $[\frac{18+4-9}{9}{]}^{-2}=[\frac{13}{9}{]}^{-2}$

$\Rightarrow$ $[\frac{9}{13}{]}^2$

$\Rightarrow$ $\frac{81}{169}$