Question

Se  x + 1/x = λ,  encontre  x²ⁿ + 1/x²ⁿ  em função de  n  e  λ,  com  n  natural,  n ≥ 1.

Answer 1

Olá Lukyo.

Organizando a expressão.

$\mathsf{x+\dfrac{1}{x}=\lambda}$

Multiplique ambos os lados por x.

$\mathsf{x^2+1=\lambda x}\\\\\\\mathsf{x^2+\lambda x+1=0}$

Obtemos uma equação do segundo grau em x. Resolvendo, temos.

$\mathsf{x=\dfrac{\lambda\pm\sqrt{\lambda^2-4}}{2}~(i)}$

Encontrando o intervalo de $\mathsf{\lambda}$ para que x seja um número real.

$\mathsf{\Delta=\lambda^2-4\geq0}\\\\\\\mathsf{\lambda^2\geq4}\\\\\\\mathsf{|\lambda|\geq2}$

Temos também que:

$\mathsf{x+\dfrac{1}{x}=\lambda}\\\\\\\mathsf{\dfrac{1}{x}=\lambda-x}$

Substituindo o valor de x encontrado em (i) na equação acima.

$\mathsf{\dfrac{1}{x}=\lambda-\Big(\dfrac{\lambda\pm\sqrt{\lambda^2-4}}{2}\Big)}\\\\\\\\\mathsf{\dfrac{1}{x}=\dfrac{2\lambda-\lambda\mp\sqrt{\lambda^2-4}}{2}}\\\\\\\\\mathsf{\dfrac{1}{x}=\dfrac{\lambda\mp\sqrt{\lambda^2-4}}{2}}$

Logo,

$\mathsf{x^{2n}+\Big(\dfrac{1}{x}\Big)^{2n}=\Big(\dfrac{\lambda\pm\sqrt{\lambda^2-4}}{2}\Big)^{2n}+\Big(\dfrac{\lambda\mp\sqrt{\lambda^2-4}}{2}\Big)^{2n}}$

Note que podemos desprezar o sinal de mais ou menos aqui, pois ambas as parcelas tem um papel simétrico.

$\mathsf{x^{2n}+\Big(\dfrac{1}{x}\Big)^{2n}=\Big(\dfrac{\lambda+\sqrt{\lambda^2-4}}{2}\Big)^{2n}+\Big(\dfrac{\lambda-\sqrt{\lambda^2-4}}{2}\Big)^{2n}}$

Dúvidas? comente.

Answer 2

Explicação passo-a-passo:

$\sf x+\dfrac{1}{x}=\lambda$

$\sf x^2-\lambda x+1=0$

$\sf \Delta=(\lambda)^2-4\cdot1\cdot1$

$\sf \Delta=\lambda^2-4$

$\sf x=\dfrac{-(-\lambda)\pm\sqrt{\lambda^2-4}}{2\cdot1}$

$\sf x=\dfrac{\lambda\pm\sqrt{\lambda^2-4}}{2}$

Devemos ter:

$\sf \lambda^2-4\ge0$

$\sf \lambda^2\ge4$

$\sf \lambda\ge2~ou~\lambda \le -2$

Assim:

$\sf x+\dfrac{1}{x}=\lambda$

$\sf \dfrac{\lambda\pm\sqrt{\lambda^2-1}}{2}+\dfrac{1}{x}=\lambda$

$\sf \dfrac{1}{x}=\lambda-\dfrac{\lambda\pm\sqrt{\lambda^2-4}}{2}$

$\sf \dfrac{1}{x}=\dfrac{2\lambda-\lambda\mp\sqrt{\lambda^2-4}}{2}$

$\sf \dfrac{1}{x}=\dfrac{\lambda\mp\sqrt{\lambda^2-4}}{2}$

Logo:

$\sf x^{2n}+\dfrac{1}{x^{2n}}=\left(\dfrac{\lambda\pm\sqrt{\lambda^2-4}}{2}\right)^{2n}+\left(\dfrac{\lambda\mp\sqrt{\lambda^2-4}}{2}\right)^{2n}$

$\sf \red{x^{2n}+\dfrac{1}{x^{2n}}=\left(\dfrac{\lambda+\sqrt{\lambda^2-4}}{2}\right)^{2n}+\left(\dfrac{\lambda-\sqrt{\lambda^2-4}}{2}\right)^{2n}}$