(1+(1/(x^2-1)))/((1/x)-(x/(x+1)))

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- Jul 22nd 2010, 04:38 PMsarkyexpress the fraction in simplest form
(1+(1/(x^2-1)))/((1/x)-(x/(x+1)))

- Jul 22nd 2010, 04:45 PMpickslides
Start by making common denominators in both top and bottom of the fraction. You follow?

- Jul 22nd 2010, 04:49 PMsarky
yup i did do that. the answer i got was x^3/(-x^3+2x^2-x-1) but the answer sheet says its x^3/(-x^3+2x^2-1)

- Jul 22nd 2010, 04:50 PMProve It
Thank you for using brackets, but this is still very hard to read...

Is it $\displaystyle \frac{1 + \frac{1}{x^2 - 1}}{\frac{1}{x} - \frac{x}{x + 1}}$?

If so, start by making some common denominators

$\displaystyle \frac{1 + \frac{1}{x^2 - 1}}{\frac{1}{x} - \frac{x}{x + 1}} = \frac{\frac{x^2 - 1}{x^2 - 1} + \frac{1}{x^2 - 1}}{\frac{x + 1}{x(x + 1)} - \frac{x^2}{x(x + 1)}}$

$\displaystyle = \frac{\frac{x^2}{x^2 - 1}}{\frac{-x^2 + x + 1}{x(x + 1)}}$

$\displaystyle = \frac{\frac{x^2}{(x - 1)(x + 1)}}{\frac{-x^2 + x + 1}{x(x + 1)}}$

$\displaystyle = \frac{x^3(x + 1)}{(x - 1)(x + 1)(-x^2 + x + 1)}$

$\displaystyle = \frac{x^3}{(x - 1)(-x^2 + x + 1)}$. - Jul 22nd 2010, 04:55 PMsarky
yup i've got it now. thanks! :)