Question

alguém ajuda? por favor tem a resposta ali mais quero a conta.

Answer 1

a) $\frac{ 2^{3}.4^{2} }{8^{2}}-\frac{1}{2} ^{-1}= \\ \frac{ 2^{3}.(2^{2})^{2} }{(2^{3})^{2}} - (2 ^{-1})^{-1}= \\ \frac{2^{3}.2^{4}}{2^{6}} - 2^{1}=\\ 2^{3+4-6}-2=\\ 2-2=0$

b)$2^{-3}+ (\frac{1}{2})^{2}=\\ \frac{1}{2^{3}}+ \frac{1^{2}}{2^{2}}=\\ \frac{1}{8}+\frac{1}{4}=\\ \frac{1+2}{8}=\frac{3}{8}$

c)$( \frac{1}{2})^{-1} .4^{-2}=\\ (2^{-1})^{-1}.+(2^{2})^{-2}=\\ 2^{1}.2^{-4}=\\ 2^{1+(-4)}=2^{-3}=\\ \frac{1}{2^{3}}= \frac{1}{8}$

d)$\frac{5.2^{12}}{2^{9}}=\\ 5.2^{12-9}=\\ 5.2^{3}=5.8=40$