Realizando os cálculos podemos concluir que o valor da expressão é:
[tex]\boxed{\boxed{\large\displaystyle\text{$\mathsf{\sqrt[12]{3^{5}}}$}}}[/tex]
Conforme a alternativa D.
Primeiramente, devemos simplificar a expressão.
[tex]\large\displaystyle\text{$\mathsf{x=\dfrac{3+\sqrt{6} }{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{3+\sqrt{6} }{5\sqrt{3}-2\sqrt{2^{2}.3}-\sqrt{4^{2}.2}+\sqrt{5^{2}.2}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{3+\sqrt{6} }{5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{3+\sqrt{6} }{\sqrt{3}+\sqrt{2}}}$}[/tex]
Racionalizando o denominador:
[tex]\large\displaystyle\text{$\mathsf{x=\dfrac{3+\sqrt{6} }{\sqrt{3}+\sqrt{2}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{(3+\sqrt{6})~.~(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})~.~(\sqrt{3}-\sqrt{2})}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{3\sqrt{3}-3\sqrt{2}+\sqrt{18}-\sqrt{12}}{3-\sqrt{6}+\sqrt{6}-2}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{3\sqrt{3}-3\sqrt{2}+\sqrt{3^{2}.2}-\sqrt{2^{2}.3}}{1}}$}[/tex]
[tex]\large\displaystyle\text{$\mathsf{x=3\sqrt{3}-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}}$}\\\boxed{\large\displaystyle\text{$\mathsf{x=\sqrt{3}}$}}[/tex]
Sendo x =√3, podemos substituir na expressão:
[tex]\large\displaystyle\text{$\mathsf{\dfrac{x^{2}}{\sqrt[6]{x^{7}}}}$}\\\\\\\large\displaystyle\text{$\mathsf{\dfrac{\sqrt{3}^{2}}{\sqrt[6]{\sqrt{3}^{7}}}}$}\\\\\\\large\displaystyle\text{$\mathsf{\dfrac{\sqrt{3}^{2}}{\sqrt[6]{\sqrt{3}.\sqrt{3}^{6}}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}.\sqrt[6]{\sqrt{3}}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{\dfrac{\sqrt{3}}{\sqrt[6]{\sqrt{3}}}}$}[/tex]
[tex]\large\displaystyle\text{$\mathsf{\dfrac{(\sqrt[6]{\sqrt{3}})^{6}}{\sqrt[6]{\sqrt{3}}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{\dfrac{\sqrt[6]{\sqrt{3}}~.~(\sqrt[6]{\sqrt{3}})^{5}}{\sqrt[6]{\sqrt{3}}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{(\sqrt[6]{\sqrt{3}})^{5}}$}\\\\\\\boxed{\large\displaystyle\text{$\mathsf{\sqrt[12]{3^{5}}}$}}[/tex]
⭐ Espero ter ajudado! ⭐
Veja mais exercícios semelhantes em:
https://brainly.com.br/tarefa/53963461
O valor da expressão é:[tex]\Large \text {$ \dfrac{x^2}{\sqrt[6]{x^7}} = \sqrt [12]{3^{^5}}$}[/tex] ⟹ Alternativa D.
[tex]\large \text {$ x=\dfrac{3+\sqrt{6}}{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}} $}[/tex]
[tex]\large \text {$ 3 = \sqrt 3 \cdot \sqrt 3 $}[/tex]
[tex]\large \text {$ \sqrt 6 = \sqrt 2 \cdot \sqrt 3 $}[/tex]
[tex]\large \text {$ 2 \sqrt{1 2} = 2 \cdot \sqrt {4 \cdot 3} = 4 \cdot \sqrt {3} $}[/tex]
[tex]\large \text {$ \sqrt{32} = \cdot \sqrt {16 \cdot 2} = 4 \cdot \sqrt {2} $}[/tex]
[tex]\large \text {$ \sqrt{50} = \cdot \sqrt {25 \cdot 2} = 5 \cdot \sqrt {2} $}[/tex]
[tex]\large \text {$ x=\dfrac{\sqrt 3 \cdot \sqrt 3+\sqrt2 \cdot \sqrt 3}{5\sqrt{3}-4\sqrt{3}-4 \sqrt{2}+5 \sqrt{2}} $}[/tex]
[tex]\large \text {$ x=\dfrac{\sqrt 3 \cdot (\sqrt 2+\sqrt 3)}{\sqrt{3}+ \sqrt{2}} \quad \Longrightarrow \quad $ \sf Simplifique.}[/tex]
[tex]\large \text {$ x=\sqrt 3 $}[/tex]
[tex]\large\text{$ \dfrac{x^2}{\sqrt[6]{x^7}} = \dfrac{{(\sqrt 3)}^2}{\sqrt[6]{(\sqrt 3)^7}} =\dfrac{3}{\left(3^{^{\frac{1}{2}}} \right)^{\frac{7}{6}}}=\dfrac{3}{3^{^{\frac{7}{12}}}} = 3^{^{\left( 1-\frac {7}{12} \right) }} = 3^{^{\frac {5}{12}}} $}[/tex]
[tex]\Large \text {$ \dfrac{x^2}{\sqrt[6]{x^7}} = \sqrt [12]{3^{^5}}$}[/tex] Alternativa D.
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Realizando os cálculos podemos concluir que o valor da expressão é:
[tex]\boxed{\boxed{\large\displaystyle\text{$\mathsf{\sqrt[12]{3^{5}}}$}}}[/tex]
Conforme a alternativa D.
Resolução do exercício
Primeiramente, devemos simplificar a expressão.
[tex]\large\displaystyle\text{$\mathsf{x=\dfrac{3+\sqrt{6} }{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{3+\sqrt{6} }{5\sqrt{3}-2\sqrt{2^{2}.3}-\sqrt{4^{2}.2}+\sqrt{5^{2}.2}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{3+\sqrt{6} }{5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{3+\sqrt{6} }{\sqrt{3}+\sqrt{2}}}$}[/tex]
Racionalizando o denominador:
[tex]\large\displaystyle\text{$\mathsf{x=\dfrac{3+\sqrt{6} }{\sqrt{3}+\sqrt{2}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{(3+\sqrt{6})~.~(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})~.~(\sqrt{3}-\sqrt{2})}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{3\sqrt{3}-3\sqrt{2}+\sqrt{18}-\sqrt{12}}{3-\sqrt{6}+\sqrt{6}-2}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{3\sqrt{3}-3\sqrt{2}+\sqrt{3^{2}.2}-\sqrt{2^{2}.3}}{1}}$}[/tex]
[tex]\large\displaystyle\text{$\mathsf{x=3\sqrt{3}-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}}$}\\\boxed{\large\displaystyle\text{$\mathsf{x=\sqrt{3}}$}}[/tex]
Sendo x =√3, podemos substituir na expressão:
[tex]\large\displaystyle\text{$\mathsf{\dfrac{x^{2}}{\sqrt[6]{x^{7}}}}$}\\\\\\\large\displaystyle\text{$\mathsf{\dfrac{\sqrt{3}^{2}}{\sqrt[6]{\sqrt{3}^{7}}}}$}\\\\\\\large\displaystyle\text{$\mathsf{\dfrac{\sqrt{3}^{2}}{\sqrt[6]{\sqrt{3}.\sqrt{3}^{6}}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}.\sqrt[6]{\sqrt{3}}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{\dfrac{\sqrt{3}}{\sqrt[6]{\sqrt{3}}}}$}[/tex]
[tex]\large\displaystyle\text{$\mathsf{\dfrac{(\sqrt[6]{\sqrt{3}})^{6}}{\sqrt[6]{\sqrt{3}}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{\dfrac{\sqrt[6]{\sqrt{3}}~.~(\sqrt[6]{\sqrt{3}})^{5}}{\sqrt[6]{\sqrt{3}}}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{(\sqrt[6]{\sqrt{3}})^{5}}$}\\\\\\\boxed{\large\displaystyle\text{$\mathsf{\sqrt[12]{3^{5}}}$}}[/tex]
⭐ Espero ter ajudado! ⭐
Veja mais exercícios semelhantes em:
https://brainly.com.br/tarefa/53963461
O valor da expressão é:
[tex]\Large \text {$ \dfrac{x^2}{\sqrt[6]{x^7}} = \sqrt [12]{3^{^5}}$}[/tex] ⟹ Alternativa D.
[tex]\large \text {$ x=\dfrac{3+\sqrt{6}}{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}} $}[/tex]
[tex]\large \text {$ 3 = \sqrt 3 \cdot \sqrt 3 $}[/tex]
[tex]\large \text {$ \sqrt 6 = \sqrt 2 \cdot \sqrt 3 $}[/tex]
[tex]\large \text {$ 2 \sqrt{1 2} = 2 \cdot \sqrt {4 \cdot 3} = 4 \cdot \sqrt {3} $}[/tex]
[tex]\large \text {$ \sqrt{32} = \cdot \sqrt {16 \cdot 2} = 4 \cdot \sqrt {2} $}[/tex]
[tex]\large \text {$ \sqrt{50} = \cdot \sqrt {25 \cdot 2} = 5 \cdot \sqrt {2} $}[/tex]
[tex]\large \text {$ x=\dfrac{\sqrt 3 \cdot \sqrt 3+\sqrt2 \cdot \sqrt 3}{5\sqrt{3}-4\sqrt{3}-4 \sqrt{2}+5 \sqrt{2}} $}[/tex]
[tex]\large \text {$ x=\dfrac{\sqrt 3 \cdot (\sqrt 2+\sqrt 3)}{\sqrt{3}+ \sqrt{2}} \quad \Longrightarrow \quad $ \sf Simplifique.}[/tex]
[tex]\large \text {$ x=\sqrt 3 $}[/tex]
[tex]\large\text{$ \dfrac{x^2}{\sqrt[6]{x^7}} = \dfrac{{(\sqrt 3)}^2}{\sqrt[6]{(\sqrt 3)^7}} =\dfrac{3}{\left(3^{^{\frac{1}{2}}} \right)^{\frac{7}{6}}}=\dfrac{3}{3^{^{\frac{7}{12}}}} = 3^{^{\left( 1-\frac {7}{12} \right) }} = 3^{^{\frac {5}{12}}} $}[/tex]
[tex]\Large \text {$ \dfrac{x^2}{\sqrt[6]{x^7}} = \sqrt [12]{3^{^5}}$}[/tex] Alternativa D.
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