Colégio Naval 1978 ⇒ Questão 07 - CN - 1978 Tópico resolvido

[petrasMOD]

7) O valor mais aproximado de

[tex3]\frac{16 ^{-0,75}+\sqrt[5]{ 0,00243}} {\frac{2}{3}+4,333\dots}[/tex3]

é:
a) 0,045 b) 0,125 c) 0,315 d) 0,085 e) 0,25

[petrasMOD]

[tex3]16^{-0,75}=16^{\frac{3}{4}}=(2^4)^{-\frac{3}{4}}=2^{-3}=\frac{1}{2^3}=\frac{1}{8}\\

\sqrt[5]{0,00243}=\left(\frac{243}{100000}\right)^{\frac{1}{5}}=\left(\frac{3^5}\\{10^5}\right)^{\frac{1}{5}}=\frac{3}{10}
\\4,333\ldots=4,\overline{3}=\frac{43-4}{9}=\frac{39}{9}\\
\therefore \frac{16^{-0,75}+\sqrt[5]{0,00243}}{\frac{2}{3}+4,333..}=\frac{\frac{1}{8}+\frac{3}{10}}{\frac{2}{3}+\frac{39}{9}}=\frac{\frac{5+12}{40}}{\frac{6+39}{9}}=\frac{\frac{17}{40}}{\frac{45}{9}}=\frac{17}{40}\times\frac{1}{5}=\frac{17}{200}=\boxed{0,085_{//}}[/tex3]