1/v+1 + v/(9-v) + v^2+9/v^2-8v-9

please help!

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- Sep 8th 2008, 11:08 AMbrittneyvAlgebra Factor and Simplify
1/v+1 + v/(9-v) + v^2+9/v^2-8v-9

please help! - Sep 8th 2008, 11:38 AMTKHunny
Do two things,

1) Rewrite it with appropriate parentheses so that it makes sense. Why did you use parentheses in the denominator of the second term but not in the first or third terms? The second term is the good one. Make the other two look like that.

2) The second term may be a bit confusing. Rewrite it just a bit and unconfuse us all. Rather than

+ v/(9-v)

Rewrite it as

- v/(v-9)

Make sure you know why these are equivalent.

Now find common denominators and add. I suspect very strongly that you should be able to factor the denominator in the third term.

Note: Surprising result! - Sep 8th 2008, 11:49 AMShyam
Is your question, $\displaystyle =\frac {1}{v+1} + \frac {v}{9-v} + \frac {v^2 +9}{v^2-8v-9}$

$\displaystyle =\frac {1}{v+1} - \frac {v}{v-9} + \frac {v^2 +9}{(v-9)(v+1)}$

$\displaystyle =\frac {v-9-v(v+1)+v^2+9}{(v-9)(v+1)}$

$\displaystyle =\frac {v-9-v^2-v+v^2+9}{(v-9)(v+1)}$

$\displaystyle =\frac {0}{(v-9)(v+1)}$

$\displaystyle =0$ - Sep 8th 2008, 12:56 PMTKHunny
Right, NOW we know if brittneyv can do it. (Shake)

Quiz Time. Contrary to shyam's result, the original expression is**NOT**ALWAYS zero (0).

Can brittneyv tell me why?