Question

Determine o valor de z:

$z= \left[\begin{array}{ccc}(1+ \frac{1}{0,5}) ^{-1} .(1- \frac{1}{0,5}) ^{-1} \\\\\end{array}\right] ^{-2}$

Answer 1

Olá Mari!

$\\ \mathsf{z = \left [ \left ( 1 + \frac{1}{0,5} \right )^{- 1} \cdot \left ( 1 - \frac{1}{0,5} \right )^{- 1} \right ]^{- 2}} \\\\\\ \mathsf{z = \left [ \left ( 1 + \frac{1}{\frac{1}{2}} \right )^{- 1} \cdot \left ( 1 - \frac{1}{\frac{1}{2}} \right )^{- 1} \right ]^{- 2}} \\\\\\ \mathsf{z = \left [ \left ( 1 + 1 \cdot \frac{2}{1} \right )^{- 1} \cdot \left ( 1 - 1 \cdot \frac{2}{1} \right )^{- 1} \right ]^{- 2}} \\\\\\ \mathsf{z = \left [ \left ( 3 \right )^{- 1} \cdot \left ( - 1 \right )^{- 1} \right ]^{- 2}}$

$\\ \mathsf{z = \left [ \left ( \frac{1}{3} \right )^{+ 1} \cdot \left ( \frac{1}{- 1} \right )^{+ 1} \right ]^{- 2}} \\\\\\ \mathsf{z = \left [ \frac{1}{3} \cdot (- 1) \right ]^{- 2}} \\\\\\ \mathsf{z = \left ( - \frac{1}{3} \right )^{- 2}} \\\\\\ \mathsf{z = \left ( - \frac{3}{1} \right )^{+ 2}} \\\\\\ \boxed{\mathsf{z = 9}}$