O valor de x na expressão x!/3!.(x-3)!=55x/3.... Pergunta de ideia deProfessorVanderlan

ProfessorVanderlan @ProfessorVanderlan

December 2019 1 65 Report

O valor de x na expressão x!/3!.(x-3)!=55x/3

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niltonjr2001 $\mathrm{\dfrac{x!}{3!(x-3)!}=\dfrac{55x}{3}\ \to\ \dfrac{x(x-1)(x-2)(x-3)!}{3.2(x-3)!}=\dfrac{55x}{3}}\\\\\\ \mathrm{\dfrac{(x-1)(x-2)}{2}=55\ \to\ (x-1)(x-2)=110}\\\\ \mathrm{x^2-2x-x+2=110\ \to\ x^2-3x-108=0}\\\\ \textbf{Atrav\'es da f\'ormula quadr\'atica, teremos que:}\\\\ \mathrm{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-3)\pm\sqrt{(-3)^2-4.1.(-108)}}{2.1}}\\\\ \mathrm{x=\dfrac{3\pm\sqrt{9+432}}{2}=\dfrac{3\pm\sqrt{441}}{2}=\dfrac{3\pm21}{2}}$

$\mathrm{x_1=\dfrac{3-21}{2}=\dfrac{-18}{2}=-9\ \to\ \nexists\ x!}\\\\ \mathrm{x_2=\dfrac{3+21}{2}=\dfrac{24}{2}=12\ \to\ \exists\ x!}\\\\ \mathbf{Resposta}\ \to\ \boxed{\mathrm{x=12}}$

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