Question

Alguém me ajuda, por favor?

a) $\frac{7}{ \sqrt[3]{7} }$
b) $\frac{5}{ \sqrt[3]{2} }$
c) $\frac{2}{ \sqrt[5]{3} ^{4} }$
d) $\frac{4}{5-\sqrt{3} }$

Answer 1

Olá.

Como temos raízes no denominador, teremos de racionalizar essa fração.
Da A até a C, vamos basicamente multiplicar todos os termos da fração por um valor que retire a raiz do denominador.

Vamos aos cálculos.

A
$\Large\mathsf{\dfrac{7}{\sqrt[3]{7}}=}\\\\\\ \Large\mathsf{\dfrac{7\cdot\sqrt[3]{7}\cdot\sqrt[3]{7}}{\sqrt[3]{7}\cdot\sqrt[3]{7}\cdot\sqrt[3]{7}}=}\\\\\\ \Large\mathsf{\dfrac{7\cdot\sqrt[3]{7^2}}{\sqrt[3]{7^3}}=}\\\\\\ \Large\mathsf{\dfrac{7\cdot\sqrt[3]{7^2}}{7}=}\\\\\\ \Large\mathsf{\dfrac{\not7\cdot\sqrt[3]{7^2}}{\not7}=}\\\\\\ \Large\mathsf{\sqrt[3]{7^2}=}\\\\ \Large\boxed{\boxed{\mathsf{\sqrt[3]{49}~\approxeq~3,65931}}}$


B
$\Large\mathsf{\dfrac{5}{\sqrt[3]{2}}=}\\\\\\ \Large\mathsf{\dfrac{5\cdot\sqrt[3]{2}\cdot\sqrt[3]{2}}{\sqrt[3]{2}\cdot\sqrt[3]{2}\cdot\sqrt[3]{2}}=}\\\\\\ \Large\mathsf{\dfrac{5\cdot\sqrt[3]{2^2}}{\sqrt[3]{2^3}}=}\\\\\\ \mathsf{\Large\dfrac{5\cdot\sqrt[3]{2^2}}{2}=}\\\\\\ \boxed{\boxed{\Large\mathsf{\dfrac{5\cdot\sqrt[3]{4}}{2}=~\approxeq~3,9685}}}$


C
$\Large\mathsf{\dfrac{2}{\sqrt[5]{3^4}}=}\\\\\\ \Large\mathsf{\dfrac{2\cdot\sqrt[5]{3}}{\sqrt[5]{3^4}\cdot\sqrt[5]{3^4}}=}\\\\\\ \Large\mathsf{\dfrac{2\cdot\sqrt[5]{3}}{\sqrt[5]{3^5}}=}\\\\\\ \Large\mathsf{\dfrac{2\cdot\sqrt[5]{3}}{3}=}\\\\\\ \boxed{\boxed{\Large\mathsf{\dfrac{2\sqrt[5]{3}}{3}\approxeq~0,830487}}}$


D
$\Large\mathsf{\dfrac{4}{5-\sqrt[2]{3}}=}\\\\\\ \Large\mathsf{\dfrac{4\cdot\left(5+\sqrt[2]{3}\right)}{5-\sqrt[2]{3}\cdot\left(5+\sqrt[2]{3}\right)}=}\\\\\\ \Large\mathsf{\dfrac{4\cdot\left(5+\sqrt[2]{3}\right)}{25+5\sqrt[2]{3}-5\sqrt[2]{3}-\sqrt[2]{3^2}}=}\\\\\\ \Large\mathsf{\dfrac{4\cdot\left(5+\sqrt[2]{3}\right)}{25-3}=}\\\\\\ \Large\mathsf{\dfrac{4\cdot\left(5+\sqrt[2]{3}\right)}{22}=}\\\\\\ \boxed{\boxed{\Large\mathsf{\dfrac{2\cdot\left(5+\sqrt[2]{3}\right)}{11}~\approxeq~1,22401}}}$

Qualquer dúvida, deixe nos comentários.
Bons estudos.