Question

calcule as potências com expoentes negativos e fracionários.​

Answer 1

Explicação passo-a-passo:

olá darling 36

a)

$( \frac{5}{7}) - 2 \\ \\ \frac{5 - 14}{7} \\ \\ \frac{ - 9}{7} \\ \\ - \frac{9}{7}$

b)

$( \frac{2}{3} ) - 4 \\ \\ \frac{2 - 12}{3} \\ \\ \frac{ - 10}{3} \\ \\ - \frac{10}{3}$

c)

$( \frac{2}{5} ) - 3 \\ \\ \frac{2 - 15}{5} \\ \\ \frac{ - 13}{5} \\ \\ - \frac{13}{5}$

d)

$( \frac{ 6}{3} ) - 1 \\ \\ 2 - 1 \\ \\ 1$

Espero te ajudado

Answer 2

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador https://brainly.com.br/tarefa/38364077

                                               

Potência de expoente negativo

$\huge\boxed{\boxed{\boxed{\boxed{\sf\bigg(\dfrac{a}{b}\bigg)^{-n}=\bigg(\dfrac{b}{a}\bigg)^n}}}}$

$\tt a)~\sf\bigg(\dfrac{5}{7}\bigg)^{-2}=\bigg(\dfrac{7}{5}\bigg)^2=\dfrac{49}{25}\\\tt b)~\sf\bigg(\dfrac{2}{3}\bigg)^{-4}=\bigg(\dfrac{3}{2}\bigg)^4=\dfrac{81}{16}\\\tt c)~\sf\bigg(\dfrac{2}{5}\bigg)^{-3}=\bigg(\dfrac{5}{2}\bigg)^3=\dfrac{125}{8}\\\tt d)~\sf\bigg(\dfrac{6}{3}\bigg)}^{-1}=\bigg(\dfrac{3}{6}\bigg)^1=\dfrac{3\div3}{6\div3}=\dfrac{1}{2}$