Question

Alguém pode me ajuda por favor? só preciso das respostas

(a) 6-³
(b)3-⅝
(c)8-³
(d)m10xm8xm4
(e)8²x8³
(f)7⅞x7³
(g)9⅝x9²

Answer 1

a)$6^{-3}=\left(\dfrac{1}{6}\right)^{3}=\dfrac{1}{6^{3}}=\dfrac{1}{216}$


b)

$3^{-\frac{5}{8}}=\left(\dfrac{1}{3}\right)^{\frac{5}{8}}=\dfrac{1}{3^{\frac{5}{8}}}=\dfrac{1}{ \sqrt[8]{3^{5}}}=\dfrac{1}{\sqrt[8]{3^{5}}}\cdot\dfrac{\sqrt[8]{3^{3}}}{\sqrt[8]{3^{3}}}=\dfrac{\sqrt[8]{27}}{\sqrt[8]{3^{5}\cdot 3^{3}}}=\dfrac{\sqrt[8]{27}}{\sqrt[8]{3^{5+3}}}\Rightarrow\\ \\ \\ \dfrac{\sqrt[8]{27}}{\sqrt[8]{3^{8}}}=\dfrac{\sqrt[8]{27}}{3}$


c) $8^{-3}=\left(\dfrac{1}{8}\right)^{3}=\dfrac{1}{8^{3}}=\dfrac{1}{512}$


d) $m^{10}\cdot m^{8}\cdot m^{4}=m^{10+8+4}=m^{22}$

e)$8^{2}\cdot 8^{3}=8^{2+3}=8^{5}=32768$


f) $7^{\frac{7}{8}}\cdot 7^{3}=7^{\frac{7}{8}+3}=7^{\frac{7+24}{8}}=7^{\frac{31}{8}}=\sqrt[8]{7^{31}}$


g) $9^{\frac{5}{8}}\cdot 9^{2}=9^{\frac{5}{8}+2}=9^{\frac{5+16}{8}}=9^{\frac{21}{8}}=\sqrt[8]{9^{21}}$