Concursos Públicos ⇒ (IDHTEC) Operação com Raízes Tópico resolvido

[loma0320]

19. Marque a alternativa correta.

(A) [tex3]\sqrt[3]{2}+\sqrt[3]{2}+\sqrt[3]{2}\in\mathbb Z[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt[3]{2}+\sqrt[3]{2}+\sqrt[3]{2}\in\mathbb Z[/tex3]

(B) [tex3]\sqrt[3]{2}\times\sqrt[3]{2}\times\sqrt[3]{2}\in\mathbb Z[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt[3]{2}\times\sqrt[3]{2}\times\sqrt[3]{2}\in\mathbb Z[/tex3]

(C) [tex3]\sqrt[3]{2}+\sqrt[3]{2}-\sqrt[3]{2}\in\mathbb Z[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt[3]{2}+\sqrt[3]{2}-\sqrt[3]{2}\in\mathbb Z[/tex3]

(D) [tex3]\sqrt[3]{2}-\sqrt[3]{2}-\sqrt[3]{2}\in\mathbb Z[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt[3]{2}-\sqrt[3]{2}-\sqrt[3]{2}\in\mathbb Z[/tex3]

(E) [tex3]\sqrt[3]{2}\div\sqrt[3]{2}\div \sqrt[3]{2}\in\mathbb Z[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt[3]{2}\div\sqrt[3]{2}\div \sqrt[3]{2}\in\mathbb Z[/tex3]

Resposta

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B

[deOliveira]

[tex3](A)\hspace2mm\sqrt[3]2+\sqrt[3]2+\sqrt[3]2=3\sqrt[3]2\not\in\mathbb Z\\(B)\hspace2mm\sqrt[3]2\times\sqrt[3]2\times\sqrt[3]2=\sqrt[3]{2^3}=2\in\mathbb Z\\(C)\hspace2mm\sqrt[3]2+\sqrt[3]2-\sqrt[3]2=\sqrt[3]2\not\in\mathbb Z\\(D)\hspace2mm\sqrt[3]2-\sqrt[3]2-\sqrt[3]2=-3\sqrt[3]2\in\mathbb Z\\(E)\hspace2mm\sqrt[3]2:\sqrt[3]2:\sqrt[3]2=1:\sqrt[3]2=\frac1{\sqrt[3]2}\not\in\mathbb Z[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3](A)\hspace2mm\sqrt[3]2+\sqrt[3]2+\sqrt[3]2=3\sqrt[3]2\not\in\mathbb Z\\(B)\hspace2mm\sqrt[3]2\times\sqrt[3]2\times\sqrt[3]2=\sqrt[3]{2^3}=2\in\mathbb Z\\(C)\hspace2mm\sqrt[3]2+\sqrt[3]2-\sqrt[3]2=\sqrt[3]2\not\in\mathbb Z\\(D)\hspace2mm\sqrt[3]2-\sqrt[3]2-\sqrt[3]2=-3\sqrt[3]2\in\mathbb Z\\(E)\hspace2mm\sqrt[3]2:\sqrt[3]2:\sqrt[3]2=1:\sqrt[3]2=\frac1{\sqrt[3]2}\not\in\mathbb Z[/tex3]

Então temos que a resposta correta é a letra B

[loma0320]

Muito Obrigada!