Fórum de Matemática
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Se \(x^2=x\sqrt{3}-1\),calcule o valor de \(x^{5}+\frac{1}{x^{5}}\)

a)\(\sqrt{2}\) b)\(\sqrt{3}\) c)\(-\sqrt{3}\) d)\(-\sqrt{2}\)

\(x^2=x\sqrt{3}-1\Rightarrow x\sqrt{3}=x^2+1\)
\(x^4=(x^2)^2=(x\sqrt{3}-1)^2=3x^2-2(x\sqrt{3})+1=3x^2-2(x^2+1)+1=(x^2)-1=(x\sqrt{3}-1)-1=x\sqrt{3}-2\)
\(x^5=x^4\cdot x=(x\sqrt{3}-2)x=x^2\sqrt{3}-2x=3x-\sqrt{3}-2x=x-\sqrt{3}\)

\(x^5+\frac{1}{x^5}=x-\sqrt{3}+\frac{1}{x-\sqrt{3}}=\frac{x^2-2x\sqrt{3}+4}{x-\sqrt{3}}=\frac{x\sqrt{3}-1-2x\sqrt{3}+4}{x-\sqrt{3}}=\frac{-x\sqrt{3}+3}{x-\sqrt{3}}=-\sqrt{3}\)