Colégio Naval 1975 ⇒ Questão 01 - CN - 1975

[petrasMOD]

Achar o valor de: [tex3]6\cdot\(\sqrt[3]{3,375}\cdot \sqrt{1,777...}\cdot\sqrt[5]{32^{-1}} \)[/tex3]
a) [tex3]\sqrt[3]{3}+\sqrt2[/tex3] b) 20 c) [tex3]\sqrt{2}+\sqrt3[/tex3] d) [tex3]17+\sqrt5[/tex3] e) [tex3]\frac{48}{7}[/tex3]

[tex3]\sqrt[3]{3,375 }=\frac{\sqrt[3]{3^3.5^3}}{\sqrt[3]{10^3}}=\frac{3.5}{10} =\frac{3}{2}\\
\sqrt{1,7777...} = \sqrt{\frac{16}{9}} = \frac{4}{3}\\
\sqrt[5]{32^{-1}}=\sqrt[5]{\frac{1}{2^5}} = \frac{1}{2}\\
\therefore 6\cdot \(\frac{3}{2}+\frac{4}{3}+\frac{1}{2} \) = 6\cdot \frac{10}{3} = \boxed{20_{//}}
[/tex3]