Matemáticas, pregunta formulada por samuelmegatron6, hace 1 año

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Contestado por CarlosMath
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E=\dfrac{x^2}{1-\dfrac{1}{x^2+\dfrac{\dfrac{1}{x}}{x+\dfrac{1}{x}}}}+\dfrac{x^2-2}{1-\dfrac{1}{x^2-\dfrac{\dfrac{1}{x}}{x-\dfrac{1}{x}}}}\\ \\ \\E=\dfrac{x^2}{1-\dfrac{1}{x^2+\dfrac{\dfrac{1}{x}}{\dfrac{x^2+1}{x}}}}+\dfrac{x^2-2}{1-\dfrac{1}{x^2-\dfrac{\dfrac{1}{x}}{\dfrac{x^2-1}{x}}}}\\ \\ \\E=\dfrac{x^2}{1-\dfrac{1}{x^2+\dfrac{1}{x^2+1}}}}+\dfrac{x^2-2}{1-\dfrac{1}{x^2-\dfrac{1}{x^2-1}}}\\ \\ \\

E=\dfrac{x^2}{1-\dfrac{1}{\dfrac{x^4+x^2+1}{x^2+1}}}}+\dfrac{x^2-2}{1-\dfrac{1}{\dfrac{x^4-x^2-1}{x^2-1}}}\\ \\ \\E=\dfrac{x^2}{1-\dfrac{x^2+1}{x^4+x^2+1}}}+\dfrac{x^2-2}{1-\dfrac{x^2-1}{x^4-x^2-1}}\\ \\ \\E=\dfrac{x^2}{\dfrac{x^4}{x^4+x^2+1}}}+\dfrac{x^2-2}{\dfrac{x^4-2x^2}{x^4-x^2-1}}\\ \\\\E=\dfrac{x^4+x^2+1}{x^2}+\dfrac{x^4-x^2-1}{x^2}\\ \\ \\\boxed{E=x^2}

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