Question 282102
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My advice to you:  Start using parentheses -- lots of parentheses.  I think you were trying to render:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{n\ +\ 5\ -\ \frac{12}{n\,+\,1}}{\frac{n\,+\,9}{n\,+\,1}\ -\ \frac{5}{n}]


I'll proceed on that assumption:


The LCD in the denominator is indeed *[tex \Large n(n\,+\,1)].  The LCD in the numerator is simply *[tex \Large n\,+\,1].  So, first apply the LCDs:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{n(n\,+\,1)\ +\ 5(n\,+\,1)\ -\ 12}{\frac{n(n\,+\,9)}{n(n\,+\,1)}\ -\ \frac{5(n\,+\,1)}{n(n\,+\,1)}]


Distribute to eliminate parentheses:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{n^2\,+\,n\ +\ 5n\,+\,5\ -\ 12}{\frac{n^2\,+\,9n\ -\ 5n\ -\ 5}{n(n\,+\,1)}}]


Collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{n^2\,+\,6n\,-\,7}{\frac{n^2\,+\,4n\ -\ 5}{n(n\,+\,1)}}]


Factor the two trinomials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(n\,+\,7)(n\,-\,1)} {\frac{(n\,+\,5)(n\,-\,1)}{n(n\,+\,1)}}]


Invert and multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(n\,+\,7)(n\,-\,1)}{1}\ \cdot\ {\frac{n(n\,+\,1)}{(n\,+\,5)(n\,-\,1)}]


Eliminate factors common to both numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(n\,+\,7)}{1}\ \cdot\ {\frac{n(n\,+\,1)}{(n\,+\,5)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{n(n\,+\,7)(n\,+\,1)}{(n\,+\,5)}]


Remember to exclude any value that would cause any denominator to equal zero during the entire process.  Hence the following values are excluded:


*[tex \Large n\ \neq\ -5], *[tex \Large n\ \neq\ -1], *[tex \Large n\ \neq\ 0], and *[tex \Large n\ \neq\ 1]



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