Question 239400
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You really need to use parentheses more because I am having to guess at what you mean.


I think you are trying to represent


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(\frac{x+1}{x-1}\ +\  1\right)^5}{\left(\frac{x+1}{x-1}\ -\ 1\right)}]


If this is correct, you should have represented it as:


[((x+1)/(x-1)) + 1]^5
_______________     
[((x+1)/(x-1)) - 1]


Continuing on that presumption, begin with the numerator.  The LCD in the numerator is *[tex \Large x\ -\ 1], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(\frac{x+1+(x-1)}{x-1}\right)^5}{\left(\frac{x+1}{x-1}\ -\ 1\right)}]


The LCD in the denominator is also *[tex \Large x\ -\ 1], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(\frac{x+1+(x-1)}{x-1}\right)^5}{\left(\frac{x+1-(x-1)}{x-1}\right)}]


Collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(\frac{2x}{x-1}\right)^5}{\left(\frac{2}{x-1}\right)}]


Factor *[tex \Large x^5] from the numerator of the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(\frac{2}{x-1}\right)^5}{\left(\frac{2}{x-1}\right)}x^5]


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{a^m}{a^n}\ =\ a^{m-n}]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{2}{x-1}\right)^4\,x^5]


There are a couple of different ways to go from here.  If this were an intermediate calculation, I would leave it like it is since this expression is probably easier to manipulate than any expanded version.  But you can easily expand it if you like:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{16x^5}{(x-1)^4}]


Or go one step further and use the Binomial Theorem to expand the denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{16x^5}{x^4-4x^3+6x^2-4x+1}]


Which is 'simplest' becomes a matter of artistic license at this point.  Your choice.


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