Questão 2462347

\(\frac{8}{5} \cdot \frac{7}{4} = \frac{14}{5}\)

\(\sqrt{a^2 + b^2} = 2(a + b)\)

\(\sqrt[3]{125}\) + \(\sqrt{144}\) = 37

\(\sqrt[3]{216}\) + \(\sqrt{25}\) + \(\sqrt{10^4}\) = 21