Question

Potenciação, alguém consegue?

Answer 1

Olá Isadora,

use as propriedades da exponenciação:

$a^{m+n}~\to~a^m*a^n\\\\ a^{-m}~\to~ \dfrac{1}{a^m}$

________________

$\dfrac{2^{n-4}+2^{n-2}+3^{n-1}}{2^{n-2}+2^{n-1}}~~\to~~ \dfrac{2^n*2^{-4}+2^n*2^{-2}+2^n*2^{-1}}{2^n*2^{-2}+2^n*2^{-1}}$

Pondo  $2^n$  em evidência, teremos:

$\dfrac{2^n(2^{-4}+2^{-2}+2^{-1})}{2^n(2^{-2}+2^{-1})}~~\to~~ \dfrac{\not2^n\left( \dfrac{1}{2^4}+ \dfrac{1}{2^2}+ \dfrac{1}{2^1}\right) }{\not2^n\left( \dfrac{1}{2^2}+ \dfrac{1}{2^1} \right) }~~\to~~ \dfrac{ \dfrac{1}{16}+ \dfrac{1}{4}+ \dfrac{1}{2} }{ \dfrac{1}{4}+ \dfrac{1}{2} }\\\\\\ ~~\to~~ \dfrac{ \dfrac{13}{16} }{ \dfrac{3}{4} }~~\to~~ \dfrac{13}{16}: \dfrac{3}{4}~~\to~~ \dfrac{13}{16}* \dfrac{4}{3}~~\to~~ \dfrac{52}{48}~~\to~~ \dfrac{52:4}{48:4}= \dfrac{13}{12}$

Ou seja, nenhuma das alternativas.

Tenha ótimos estudos =))