Question

Expressão matemática (foto)

Answer 1

$\dfrac{2^{n+3}\cdot 2-2^{n-1}\cdot 7}{5\cdot 2^{n-4}}\\ \\ \\ =\dfrac{2^{n-4+7}\cdot 2-2^{n-4+3}\cdot 7}{5\cdot 2^{n-4}}\\ \\ \\ =\dfrac{(2^{n-4}\cdot 2^{7})\cdot 2-(2^{n-4}\cdot 2^{3})\cdot 7}{5\cdot 2^{n-4}}\\ \\ \\ =\dfrac{2^{n-4}\cdot (2^{7}\cdot 2-2^{3}\cdot 7)}{5\cdot 2^{n-4}}$


Simplificando o fator comum $2^{n-4}$ no numerador e no denominador, chegamos a

$=\dfrac{2^{7}\cdot 2-2^{3}\cdot 7}{5}\\ \\ \\ =\dfrac{128\cdot 2-8\cdot 7}{5}\\ \\ \\ =\dfrac{256-56}{5}\\ \\ \\ =\dfrac{200}{5}\\ \\ \\ =40$


$\boxed{ \begin{array}{c} \dfrac{2^{n+3}\cdot 2-2^{n-1}\cdot 7}{5\cdot 2^{n-4}}=40 \end{array} }$