Question 328444
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You know, parentheses are rather cheap.  Use them liberally so that we don't have to guess what you mean.  I'm going to go out on a limb here and interpret your expression as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{s\ +\ d}\ -\ \frac{d}{s\ -\ d}]


To the extent that is a correct interpretation, you should have rendered it as (d/(s + d)) - (d/(s - d)).


Let's look at your supposition that this expression evaluates to 0.  If that is true, then *[tex \LARGE \frac{d}{s\ +\ d}] MUST be equal to *[tex \LARGE \frac{d}{s\ -\ d}] which can only be true if *[tex \LARGE s\ +\ d\ =\ s\ -\ d] which, in turn, can only be true if *[tex \LARGE d\ =\ 0].  But since nothing in your problem statement would allow us to restrict the value of *[tex \LARGE d] to zero, your supposition about the value of your original expression is incorrect.


The process of adding two fractions involves applying the Lowest Common Denominator.  Since the two denominators have no common factors, the LCD must be the product of the two denominators, namely *[tex \LARGE s^2\ -\ d^2] 


Applying the LCD we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d(s\ -\ d)}{(s\ +\ d)(s\ -\ d)}\ -\ \frac{d(s\ +\ d)}{(s\ +\ d)(s\ -\ d)}]


Combining terms (remembering to distribute the minus sign between the two fractions):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{ds\ -\ d^2\ -\ ds\ -\ d^2}{s^2\ -\ d^2}\ \ =\ \frac{-2d^2}{s^2\ -\ d^2}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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