• Matéria: Matemática
  • Autor: jorgeivansantos
  • Perguntado 5 anos atrás

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Anexos:

Respostas

respondido por: HoundSpacePigXels
1

Resposta:

\LARGE\bold{x^{14}}

Explicação passo-a-passo:

\LARGE\left(\frac{\sqrt[3]{\frac{x}{\sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt{x}}}}}}}}{\:\sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt{\frac{x}{\sqrt[3]{x}}}}}}}}\right)^{-72} \LARGE=\frac{1}{\left(\frac{\sqrt[3]{\frac{x}{\sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt{x}}}}}}}}{\sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt{\frac{x}{\sqrt[3]{x}}}}}}}}\right)^{72}} \LARGE=\frac{\frac{1}{\left(\sqrt[3]{\frac{x}{\sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt{x}}}}}}}\right)}^{72}}{\left(\sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt{\frac{x}{\sqrt[3]{x}}}}}}}\right)^{72}}

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Numerador

\LARGE\left(\sqrt[3]{\frac{x}{\sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt{x}}}}}}}\right)^{72}=\left(\frac{x}{\sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt{x}}}}}}\right)^{\frac{1}{3}\cdot \:72} \LARGE=\left(\frac{x}{\sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt{x}}}}}}\right)^{24}=\left(\frac{x}{\sqrt{x^{\frac{5}{6}}}}\right)^{24} \LARGE=\frac{x^{24}}{\left(\sqrt{x^{\frac{5}{6}}}\right)^{24}}=\frac{x^{24}}{\left(x^{\frac{5}{6}}\right)^{12}}=\frac{x^{24}}{x^{10}}=x^{24-10}=x^{14}

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Denominador

\LARGE\left(\sqrt{\frac{x}{\sqrt[3]{\frac{x}{\sqrt{\frac{x}{\sqrt[3]{x}}}}}}}\right)^{72}=\left(\frac{x}{\sqrt[3]{\frac{x}{\sqrt{\frac{x}{\sqrt[3]{x}}}}}}\right)^{\frac{1}{2}\cdot \:72} \LARGE=\left(\frac{x}{\sqrt[3]{\frac{x}{\sqrt{\frac{x}{\sqrt[3]{x}}}}}}\right)^{36}=\left(x^{\frac{2}{3}}\sqrt[6]{x^{\frac{2}{3}}}\right)^{36} \LARGE=\left(x^{\frac{2}{3}}\right)^{36}\left(\sqrt[6]{x^{\frac{2}{3}}}\right)^{36}=x^{24}\left(\sqrt[6]{x^{\frac{2}{3}}}\right)^{36}=x^{24}\left(x^{\frac{2}{3}}\right)^6=x^{24}x^4=x^{24+4}=x^{28}

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Substituindo

\LARGE\frac{x^{14}}{x^{28}}=\frac{1}{x^{14}}=\frac{1}{\frac{1}{x^{14}}}=x^{14}

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