Question

Por favor,alguém ajude?! Obrigada!

Answer 1

Olá Geise,

vamos relembrar uma das propriedades da exponenciação:

$\huge\boxed{a^{-m}~\to~ \dfrac{1}{a^m}}$

_______________________


$\left[\left(- \dfrac{1}{2}\right)^4\div\left(- \dfrac{1}{2}\right)^3\right]\cdot\left(- \dfrac{1}{2}\right)^6+2^{-7}=\left[ \dfrac{1}{16}\div\left(- \dfrac{1}{8}\right)\right]\cdot \dfrac{1}{64}+ \dfrac{1}{2^7}\\\\\\ \left[\left(- \dfrac{1}{2}\right)^4\div\left(- \dfrac{1}{2}\right)^3\right]\cdot\left(- \dfrac{1}{2}\right)+2^{-7}=\left[ \dfrac{1}{16}\cdot\left(- \dfrac{8}{1}\right)\right]\cdot \dfrac{1}{64}+ \dfrac{1}{128}\\.$

$\left[\left(- \dfrac{1}{2} \right)^4\div\left(- \dfrac{1}{2} \right)^3\right]\cdot\left(- \dfrac{1}{2} \right)^6+2^{-7}=\left( -\dfrac{8}{16}\right)\cdot \dfrac{1}{64}+ \dfrac{1}{128}\\\\\\ \left[\left(- \dfrac{1}{2} \right)^4\div\left(- \dfrac{1}{2} \right)^3\right]\cdot\left(- \dfrac{1}{2} \right)^6+2^{-7}=\left(- \dfrac{8\cdot1}{16\cdot64}\right)+ \dfrac{1}{128}$

$\left[\left(- \dfrac{1}{2} \right)^4\div\left(- \dfrac{1}{2} \right)^3\right]\cdot\left(- \dfrac{1}{2} \right)^6+2^{-7}=\left(- \dfrac{8}{1.024} \right)+ \dfrac{1}{128}\\\\\\ \left[\left(- \dfrac{1}{2} \right)^4\div\left(- \dfrac{1}{2} \right)^3\right]\cdot\left(- \dfrac{1}{2} \right)^6+2^{-7}=\left(- \dfrac{8\div8}{1.024\div8} \right)+ \dfrac{1}{128}$

$\left[\left(- \dfrac{1}{2} \right)^4\div\left(- \dfrac{1}{2} \right)^3\right]\cdot\left(- \dfrac{1}{2} \right)^6+2^{-7}= -\dfrac{1}{128}+ \dfrac{1}{128}\\\\\\ \large\boxed{\boxed{\boxed{\left[\left(- \dfrac{1}{2} \right)^4\div\left(- \dfrac{1}{2} \right)^3\right]\cdot\left(- \dfrac{1}{2} \right)^6+2^{-7}=0}}}.\\.$

E portanto alternativa E .

Tenha ótimos estudos ^^