Question

Resolva as seguintes racionalizações:

a) ∛6\⁴√12
b) √25\∛125
c) ⁴√81.625

Answer 1

Para racionalizar expressões envolvendo raízes com índices diferentes, eu prefiro trabalhar com expoentes fracionários:


a) $\mathsf{\dfrac{\,^3\!\!\!\sqrt{6}}{\,^4\!\!\!\sqrt{12}}}$

$\mathsf{=\dfrac{6^{1/3}}{12^{1/4}}}$


Reduzindo os expoentes ao mesmo denominador comum:

$\mathsf{=\dfrac{6^{4/12}}{12^{3/12}}}$

Multiplicando o numerador e o denominador por $\mathsf{12^{9/12}:}$

$\mathsf{=\dfrac{6^{4/12}\cdot 12^{9/12}}{12^{3/12}\cdot 12^{9/12}}}\\\\\\ \mathsf{=\dfrac{(6^{4}\cdot 12^{9})^{1/12}}{12^{(3/12)+(9/12)}}}\\\\\\ \mathsf{=\dfrac{[6^{4}\cdot (2\cdot 6)^{9}]^{1/12}}{12}}\\\\\\ \mathsf{=\dfrac{[6^{4}\cdot 2^{9}\cdot 6^{9}]^{1/12}}{12}}\\\\\\ \mathsf{=\dfrac{[6^{4+9}\cdot 2^{9}]^{1/12}}{12}}\\\\\\ \mathsf{=\dfrac{[6^{13}\cdot 2^{9}]^{1/12}}{12}}\\\\\\ \mathsf{=\dfrac{[6^{12+1}\cdot 2^{9}]^{1/12}}{12}}\\\\\\ \mathsf{=\dfrac{[6^{12}\cdot (6\cdot 2^{9})]^{1/12}}{12}}$

$\mathsf{=\dfrac{\,^{12}\!\!\!\!\sqrt{6^{12}\cdot (6\cdot 2^{9})}}{12}}\\\\\\ \mathsf{=\dfrac{\,^{12}\!\!\!\!\sqrt{6^{12}}\cdot \,^{12}\!\!\!\!\sqrt{6\cdot 2^{9}}}{12}}\\\\\\ \mathsf{=\dfrac{6\cdot \,^{12}\!\!\!\!\sqrt{2\cdot 3\cdot 2^{9}}}{12}}\\\\\\ \mathsf{=\dfrac{\diagup\!\!\!\! 6\,^{12}\!\!\!\!\sqrt{2^{1+9}\cdot 3}}{\diagup\!\!\!\! 6\cdot 2}}\\\\\\ \mathsf{=\dfrac{\,^{12}\!\!\!\!\sqrt{2^{10}\cdot 3}}{2}}$


ou caso queira desenvolver as potências,

$\mathsf{=\dfrac{\,^{12}\!\!\!\!\sqrt{3\,072}}{2}}$

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b) $\mathsf{\dfrac{\sqrt{25}}{\,^{3}\!\!\!\sqrt{125}}}$

$\mathsf{=\dfrac{\sqrt{5^{2}}}{\,^{3}\!\!\!\sqrt{5^{3}}}}\\\\\\ \mathsf{=\dfrac{5}{5}}\\\\\\ \mathsf{=1}$

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c) $\mathsf{\,^4\!\!\!\sqrt{81\cdot 125}}$

$\mathsf{=\,^4\!\!\!\sqrt{3^{4}\cdot 5^{4}}}\\\\ \mathsf{=\,^4\!\!\!\!\sqrt{(3\cdot 5)^{4}}}\\\\ \mathsf{=3\cdot 5}\\\\ \mathsf{=15}$