Question

Calcule as potências
a) (3/7)-2=
b) (2/5)-1=
c) (1/3)-3=
d) (-5/4)-3
e) (-1/3)-2

​

Answer 1

Resposta:

resposta abaixo na explicação

Explicação passo-a-passo:

a) (3/7)-2= > $(\frac{3}{7} )^{-2} => \frac{1}{(\frac{3}{7})^{2} } => \frac{1}{\frac{9}{49}^{} } => \frac{1}{1} .\frac{49}{9} = \frac{49}{9}$

b) (2/5)-1= > $(\frac{2}{5} )^{-1} => \frac{1}{(\frac{2}{5})^{1} } => \frac{1}{\frac{2}{5}^{} } => \frac{1}{1} .\frac{5}{2} = \frac{5}{2}$

c) (1/3)-3=> $(\frac{1}{3} )^{-3} => \frac{1}{(\frac{1}{3})^{3} } => \frac{1}{\frac{1}{27}^{} } => \frac{1}{1} .\frac{27}{1} = 27$

d) (-5/4)-3  => $(-\frac{5}{4} )^{-3} => \frac{1}{(-\frac{5}{4})^{3} } => \frac{1}{-\frac{125}{64}^{} } => \frac{1}{1} .-\frac{64}{125} = -\frac{64}{125}$

e) (-1/3)-2=> $(-\frac{1}{3} )^{-2} => \frac{1}{(-\frac{1}{3})^{2} } => \frac{1}{\frac{1}{9}^{} } => \frac{1}{1} .\frac{9}{1} = 9$