Re: V Maratona de Matemática IME/ITA
Enviado: 12 Jun 2016, 17:37
Solução do Problema 30:
Lembrando que [tex3](a-b)^3=a^3-3a^2b+3ab^2-b^3[/tex3]
[tex3]\sqrt{9+\frac{125}{7}}=a[/tex3]
[tex3]x^3=\left(\sqrt[3]{(3+a)}-\sqrt[3]{(-3+a)}\right)^3[/tex3]
[tex3]x^3=(3+a)-3\sqrt[3]{(3+a)^2(-3-a)}+3\sqrt[3]{(3+a)(-3+a)^2}-(-3+a)[/tex3]
[tex3]x^3=6-3{\sqrt[3]{(3+a)(-3-a)}}\underbrace{\left(\sqrt[3]{(3+a)}-\sqrt[3]{(-3-a)} \right)}_{\text{x}}[/tex3]
[tex3]x^3=6-3\sqrt[3]{\frac{125}{27}}x[/tex3]
[tex3]x^3=6-5x \rightarrow \ \ x^3+5x-6=0[/tex3]
[tex3](x-1)(x^2+x+6)=0[/tex3]
[tex3]x=\left\{1,\,\frac{-1\pm \sqrt{-23} }{2}\right\} \rightarrow x \in \mathbb{R} \rightarrow \boxed{x=1}[/tex3]
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Problema 31
(IME 1986) Determine o valor de [tex3]\log_{\sqrt{0,333...}}\sqrt{0,037037...}[/tex3]
3
Lembrando que [tex3](a-b)^3=a^3-3a^2b+3ab^2-b^3[/tex3]
[tex3]\sqrt{9+\frac{125}{7}}=a[/tex3]
[tex3]x^3=\left(\sqrt[3]{(3+a)}-\sqrt[3]{(-3+a)}\right)^3[/tex3]
[tex3]x^3=(3+a)-3\sqrt[3]{(3+a)^2(-3-a)}+3\sqrt[3]{(3+a)(-3+a)^2}-(-3+a)[/tex3]
[tex3]x^3=6-3{\sqrt[3]{(3+a)(-3-a)}}\underbrace{\left(\sqrt[3]{(3+a)}-\sqrt[3]{(-3-a)} \right)}_{\text{x}}[/tex3]
[tex3]x^3=6-3\sqrt[3]{\frac{125}{27}}x[/tex3]
[tex3]x^3=6-5x \rightarrow \ \ x^3+5x-6=0[/tex3]
[tex3](x-1)(x^2+x+6)=0[/tex3]
[tex3]x=\left\{1,\,\frac{-1\pm \sqrt{-23} }{2}\right\} \rightarrow x \in \mathbb{R} \rightarrow \boxed{x=1}[/tex3]
____________________________________________________
Problema 31
(IME 1986) Determine o valor de [tex3]\log_{\sqrt{0,333...}}\sqrt{0,037037...}[/tex3]
Resposta
3