Question

Potências Fracionárias com Expoente Negativo​

Answer 1

Lembrando que:

$\rightarrow~\sqrt[b]{a^c}~=~a^{\frac{c}{b}}\\\\\\a)\\\left(\frac{1}{25}\right)^{-\frac{1}{2}}~=~\left(25\right)^{\frac{1}{2}}~=~\sqrt[2]{25}~=~\boxed{5}\\\\\\b)\\\left(-\frac{2}{3}\right)^{-3}~=~\left(-\frac{3}{2}\right)^{3}~=~\left(\frac{-3}{2}\right)^{3}~=~\frac{(-3)^3}{2^3}~=~\frac{-27}{8}~=~\boxed{-\frac{27}{8}}$

$c)\\\left(\sqrt{5}\right)^{-4}~=~\left(\frac{1}{-\sqrt{5}}\right)^4~=~\frac{1^4}{(-\sqrt{5})^4}~=~\frac{1}{\left(-\left(5^{\frac{1}{2}}\right)\right)^4}~~=~\frac{1}{\left(-5\right)^{\frac{4}{2}}}~=~\frac{1}{(-5)^2}~=~\boxed{\frac{1}{25}}$

$d)\\\left(0,5\right)^{-3}~=~\left(\frac{1}{2}\right)^{-3}~=~(2)^3~=~\boxed{8}$