Question

calcule as seguintes potencias.​

Answer 1

Olá!

A)

$3^4\\\\=81$

B)

$2^5\\\\=32$

C)

$1^4\\\\=1$

D)

$0^6\\\\=0$

E)

$\left(-2\right)^4\\\\=2^4\\\\=16$

F)

$\left(\frac{3}{4}\right)^3\\\\=\frac{3^3}{4^3}\\\\=\frac{27}{4^3}\\\\=\frac{27}{64}$

G)

$\left(-\frac{2}{3}\right)^3\\\\\left(-a\right)^n=-a^n\\\\\left(-\frac{2}{3}\right)^3=-\left(\frac{2}{3}\right)^3\\\\=-\left(\frac{2}{3}\right)^3\\\\\left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}\\\\\left(\frac{2}{3}\right)^3=\frac{2^3}{3^3}\\\\=-\frac{2^3}{3^3}\\\\2^3=8\\\\3^3=27\\\\=-\frac{8}{27}$

H)

$5^0\\\\=1$

I)

$\left(2,\:43\right)^0\\\\2.43^0\\\\=1$

J)

$\left(-0.5\right)^0\\\\=1$

K)

$17^1\\\\=17$

L)

$\left(1,\:45\right)^1\\\\1.45^1\\\\=1,45$

M)

$\left(-5\right)^1\\\\=-5$

N)

$\left(-\frac{4}{7}\right)^1\\\\=-\frac{4}{7}$

O)

$3^{-1}\\\\=\frac{1}{3}$

P)

$\left(-3\right)^{-2}\\\\^{-b}=\frac{1}{a^b}\\\\\left(-3\right)^{-2}=\frac{1}{\left(-3\right)^2}\\\\=\frac{1}{\left(-3\right)^2}\\\\\left(-3\right)^{-2}=9\\\\=\frac{1}{9}$

Q)

$2^{-4}\\\\a^{-b}=\frac{1}{a^b}\\\\2^{-4}=\frac{1}{2^4}\\\\=\frac{1}{2^4}\\\\2^4=16\\\\=\frac{1}{16}$

R)

$\left(\frac{2}{3}\right)^{-2}\\\\a^{-b}=\frac{1}{a^b}\\\\\left(\frac{2}{3}\right)^{-2}=\frac{1}{\left(\frac{2}{3}\right)^2}\\\\=\frac{1}{\left(\frac{2}{3}\right)^2}\\\\\left(\frac{2}{3}\right)^{2}=\frac{4}{9}\\\\=\frac{1}{\frac{4}{9}}$

S)

$\left(-\frac{2}{3}\right)^{-4}\\\\a^{-b}=\frac{1}{a^b}\\\\\left(-\frac{2}{3}\right)^{-4}=\frac{1}{\left(-\frac{2}{3}\right)^4}\\\\=\frac{1}{\left(-\frac{2}{3}\right)^4}\\\\\left(-\frac{2}{3}\right)^{4}=\frac{16}{81}\\\\=\frac{1}{\frac{16}{81}}\\\\=\frac{81}{16}$

T)

$\left(-\frac{3}{4}\right)^{-3}\\\\a^{-b}=\frac{1}{a^b}\\\\\left(-\frac{3}{4}\right)^{-3}=\frac{1}{\left(-\frac{3}{4}\right)^3}\\\\=\frac{1}{\left(-\frac{3}{4}\right)^3}\\\\\left(-\frac{3}{4}\right)^{3}= \frac{27}{64} \\\\=\frac{1}{-\frac{27}{64}}\\\\=\frac{1}{-\frac{27}{64}}= -\frac{64}{27}\\\\=-\frac{64}{27}$

U)

$\left(\frac{1}{5}\right)^{-1}\\\\a^{-1}=\frac{1}{a}\\\\\left(\frac{1}{5}\right)^{-1}=\frac{1}{\frac{1}{5}}\\\\=\frac{1}{\frac{1}{5}}\\\\=\frac{1}{\frac{1}{5}}=5\\\\=5$

V)

$\left(\frac{1}{3}\right)^{-2}\\\\a^{-b}=\frac{1}{a^b}\\\\\left(\frac{1}{3}\right)^{-2}=\frac{1}{\left(\frac{1}{3}\right)^2}\\\\=\frac{1}{\left(\frac{1}{3}\right)^2}\\\\\left(\frac{1}{3}\right)^{-2}= \frac{1}{9} \\\\=\frac{1}{\frac{1}{9}}\\\\=\frac{1}{\frac{1}{9}}=9\\\\=9$

W)

$\left(-0.75\right)^{-2}\\\\a^{-b}=\frac{1}{a^b}\\\\\left(-0.75\right)^{-2}=\frac{1}{\left(-0.75\right)^2}\\\\=\frac{1}{\left(-0.75\right)^2}\\\\\left(-0.75\right)^{-2}=0,5625\\\\=\frac{1}{0.5625}$

Espero ter ajudado, tenha bons estudos!