1/v+1 + v/(9-v) + v^2+9/v^2-8v-9
please help!
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!
Is your question, $\displaystyle =\frac {1}{v+1} + \frac {v}{9-v} + \frac {v^2 +9}{v^2-8v-9}$
$\displaystyle =\frac {1}{v+1} - \frac {v}{v-9} + \frac {v^2 +9}{(v-9)(v+1)}$
$\displaystyle =\frac {v-9-v(v+1)+v^2+9}{(v-9)(v+1)}$
$\displaystyle =\frac {v-9-v^2-v+v^2+9}{(v-9)(v+1)}$
$\displaystyle =\frac {0}{(v-9)(v+1)}$
$\displaystyle =0$