Question 520162
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Recall that an expression with a negative exponent moves from numerator to denominator or vice versa, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ +\ \frac{1}{x\ +\ 2}}{1\ -\ \frac{1}{x\ +\ 2}}]


Start with the numerator expression: The common denominator is *[tex \Large x\ +\ 2], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\frac{x^2\ +\ 2x\ +\ 1}{x\ +\ 2}}{1\ -\ \frac{1}{x\ +\ 2}}] 


Then same procedure for the denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\frac{x^2\ +\ 2x\ +\ 1}{x\ +\ 2}}{\frac{x\ +\ 1}{x\ +\ 2}}]


To divide fractions, invert and multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{x^2\ +\ 2x\ +\ 1}{x\ +\ 2}\right)\left(\frac{x\ +\ 2}{x\ +\ 1}\right)] 


Eliminate identical factors in the numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2\ +\ 2x\ +\ 1}{x\ +\ 1}] 


Factor the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\ +\ 1\right)\left(x\ +\ 1\right)}{x\ +\ 1}]


Eliminate identical factors in the numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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