Matemática, perguntado por geiselucid, 1 ano atrás

Por favor,alguém ajude?! Obrigada!

Anexos:

Soluções para a tarefa

Respondido por korvo
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 ^^

geiselucid: Obrigada Korvo !!!☆☆☆☆
korvo: nds ^^
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