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Sendo $2x + \frac{2}{x} = 3$ , determine o valor de $x^{2} + \frac{1}{ x^{2} } + 1$

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Lukyo

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$\large\begin{array}{l} \textsf{Se}\\\\ \mathsf{2x+\dfrac{\,2\,}{x}=3}\\\\\\ \textsf{ent\~ao}\\\\ \mathsf{2\!\left(x+\dfrac{\,1\,}{x}\right)=3}\\\\ \mathsf{x+\dfrac{\,1\,}{x}=\dfrac{\,3\,}{2}} \end{array}$

$\large\begin{array}{l} \mathsf{\left(x+\dfrac{\,1\,}{x}\right)^{\!\!2}=\mathsf{\left(\dfrac{\,3\,}{2}\right)^{\!\!2}}}\\\\ \mathsf{x^2+2\cdot \diagup\!\!\!\! x\cdot \dfrac{\,1\,}{\diagup\!\!\!\! x}+\left(\dfrac{\,1\,}{x}\right)^{\!\!2}=\dfrac{\,9\,}{4}}\\\\ \mathsf{x^2+2+\dfrac{1}{x^2}=\dfrac{\,9\,}{4}}\\\\ \mathsf{x^2+\dfrac{1}{x^2}+1+1=\dfrac{\,9\,}{4}} \end{array}$

$\large\begin{array}{l} \mathsf{x^2+\dfrac{1}{x^2}+1=\dfrac{\,9\,}{4}-1}\\\\ \mathsf{x^2+\dfrac{1}{x^2}+1=\dfrac{\,9\,}{4}-\dfrac{\,4\,}{4}}\\\\ \mathsf{x^2+\dfrac{1}{x^2}+1=\dfrac{9-4}{4}}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{x^2+\dfrac{1}{x^2}+1=\dfrac{5}{4}} \end{array}}\qquad\quad\checkmark \end{array}$

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$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$

Tags: expressão algébrica racional fração quadrado valor álgebra

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