Question

9. calcule: (* FAÇA OS CALCULOS-NÃO ESQUEÇA: ​

Answer 1

Resposta:

a) $2^{-1}$ = $\frac{1}{2^1}$ = $\frac{1}{2}$

b) $3^{-2}$ = $\frac{1}{3^2}$ = $\frac{1}{3*3}$ = $\frac{1}{9}$

c) $(\frac{5}{3})^{-1}$ = $\frac{3}{5}$

d) $(\frac{2}{4})^{2}$= $\frac{2 * 2}{4 * 4}$ = $\frac{4}{16}$ = $\frac{1}{4}$

e) $(\frac{2}{3})^{-5}$ = $(\frac{3}{2} )^{5}$ = $\frac{3 * 3 * 3 * 3 * 3}{2 * 2 * 2 * 2 * 2}$ = $\frac{243}{32}$

f) $(-\frac{1}{2})^{3}$ = $-\frac{1 * 1 * 1}{2 * 2 * 2}$ = $\frac{1}{8}$

Explicação passo a passo: Basta utilizar as propriedades da potenciação. Espero que te ajudo, valeuuu <3

Answer 2

Resposta:

$\textsf{Segue as respostas abaixo}$

Explicação passo-a-passo:

$\mathsf{a)\ 2^{-1}=\left(\dfrac{1}{2}\right)^1=\dfrac{1}{2} }$

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

$\mathsf{c)\ \left(\dfrac{5}{3}\right)^{-1}=\left(\dfrac{3}{5}\right)^1=\dfrac{3}{5} }$

$\mathsf{ d) \ \left(\dfrac{2}{4}\right)^2=\left(\dfrac{2}{(2)^2}\right)^2=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}}$

$\mathsf{ e) \ \left(\dfrac{2}{3}\right)^{-5}=\left(\dfrac{3}{2}\right)^5=\dfrac{243}{32}}$

$\mathsf{f)\ \left(-\dfrac{1}{2}\right)^3=-\left(\dfrac{1}{2}\right)^3=-\dfrac{1}{8} }$