Question 683330
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5x}{x^2\,-\,4}\ +\ \frac{9}{x\,-\,2}\ =\ \frac{8x}{x^2\,-\,4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5x}{x^2\,-\,4}\ +\ \frac{9}{x\,-\,2}\ -\ \frac{8x}{x^2\,-\,4}\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-3x}{x^2\,-\,4}\ +\ \frac{9}{x\,-\,2}\ =\ 0] 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-3x}{x^2\,-\,4}\ +\ \frac{9}{x\,-\,2}\left(\frac{x\,+\,2}{x\,+\,2}\right)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{6x\,+\,18}{x^2\,-\,4}\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x\ +\ 18\ =\ 0]


Solve for *[tex \LARGE x]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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