Question

Resolva cada racionalização a seguir: ( Com resolução)

a) 6\∛12=

b) 18\ ₇√6=

c) 15\⁶√25=

d) 8\²⁰√2¹⁸=

e) 16\³⁰√2⁴=

Answer 1

$a)\ \frac{6}{ \sqrt[3]{12} } = \frac{6}{ \sqrt[3]{12} } \cdot \frac{\sqrt[3]{18}}{ \sqrt[3]{18}} =\frac{6\sqrt[3]{18}}{ \sqrt[3]{12\cdot18}} =\frac{6\sqrt[3]{18}}{ \sqrt[3]{216}}=\frac{6\sqrt[3]{18}}{6}=\sqrt[3]{18}\\ \\ b) \frac{18}{7 \sqrt{6}} = \frac{18}{7 \sqrt{6}} \cdot \frac{ \sqrt{6}}{ \sqrt{6}} = \frac{18 \sqrt{6}}{7 \sqrt{36}} = \frac{18 \sqrt{6}}{7\cdot6} =\frac{3 \sqrt{6}}{7}$

$c)\ \frac{15}{6 \sqrt{5}} = \frac{15}{6 \sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{15\sqrt{5}}{6\sqrt{25}} = \frac{15 \sqrt{5} }{6\cdot5}=\frac{15 \sqrt{5} }{30}= \frac{ \sqrt{5} }{2}\\ \\d)\ \frac{8}{20 \sqrt{2^{18}}} = \frac{2}{5 \cdot 2^ \frac{18}{2}} = \frac{2}{5 \cdot 2^9} =\frac{1}{5 \cdot 2^8} =\frac{1}{1280} \\ \\ e) \ \frac{16}{30 \sqrt{2^4}}=\frac{16}{30 \cdot 2^{4:2}}= \frac{16}{30 \cdot 2^2}=\frac{16}{30 \cdot 4}= \frac{4}{30}=\frac{2}{15}$