.Problema de Gincana: Qual o valor?

Problema

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Sabe-se que
[tex]\qquad\qquad \, \, \, \, \, \, \, \, \, \, \, \, \, \, \dfrac{x + y}{x − y}+\dfrac{x − y}{x + y}= 3[/tex].

Determine o valor de
[tex]\qquad\qquad \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \dfrac{x^2 + y^2}{x^2 − y^2}+\dfrac{x^2 − y^2}{x^2 + y^2}[/tex].

Solução

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Observe que:
[tex]\qquad \begin{align*} 3= \dfrac{x + y}{x − y}+\dfrac{x − y}{x + y} & = \dfrac{\left(x + y\right)^2+\left(x − y\right)^2}{x^2 − y^2}
\\&=\dfrac{\left(x^2+2xy+y^2\right)+\left(x^2 − 2xy + y^2\right)}{x^2 − y^2}
\\& =\dfrac{2\left(x^2+y^2\right)}{x^2 − y^2}.
\end{align*}[/tex]
Assim
[tex] \qquad \dfrac{x^2+y^2}{x^2 − y^2}=\dfrac{3}{2}[/tex]
e
[tex] \qquad \dfrac{x^2-y^2}{x^2 + y^2}=\dfrac{2}{3}[/tex],
donde
[tex]\qquad \dfrac{x^2 + y^2}{x^2 − y^2}+\dfrac{x^2 − y^2}{x^2 + y^2}= \dfrac{3}{2}+\dfrac{2}{3}[/tex].
Dessa forma,
[tex]\qquad \boxed{\dfrac{x^2 + y^2}{x^2 − y^2}+\dfrac{x^2 − y^2}{x^2 + y^2}= \dfrac{13}{6}}[/tex].

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Solução elaborada pelos Moderadores do Blog.