1/v+1 - v/v-4 + v^2+4/v^2-3v-4
For future reference, if you have a series of terms being divided by a series of terms, write the numerator and denominator in parenthesis to show they are ALL being divided.
Your problem,
1/v+1 - v/v-4 + v^2+4/v^2-3v-4
Is more easily understood written as:
1/(v + 1) - v/(v - 4) + (v^2 + 4)/(v^2 - 3v - 4)
We need to find the lowest common denominator. To do this, factor each denominator:
(v + 1) is factored already
(v - 4) is factored already
(v^2 - 3v - 4) = (v - 4)(v + 1) is now in factored form
Next we need to use the LCD to rewrite each fraction:
1/(v + 1)*(v - 4)/(v - 4) - v/(v - 4)*(v + 1)/(v + 1) + (v^2 + 4)/[(v - 4)(v + 1)]
Adding each numerator, we get:
[(v - 4) - v(v + 1) + (v^2 + 4)]/[(v - 4)(v + 1)]
(v - 4 - v^2 - v + v^2 + 4)/[(v - 4)(v + 1)]
(0)/[(v - 4)(v + 1)] = 0
The problem with this is that you cannot legally multiply every term by the LCD. In essence, you are changing the value of the expression by doing so. (That method is only legal when it is an "equation" where there is an equal sign and some terms on the other side of it. In any algibraic operation you do, you MUST ALWAYS make sure you preserve the identity of the original problem (its value or its relationship).)