Question

I have this fraction z/((z+1)(z+2)) and I don´t know how to get this equivalence z/(z+1)-z/(z+2) anyone could help me?

Answer 1

\[\frac{z}{(z+1)} - \frac{z}{(z+2)}\]Make a common denominator:
\[\frac{z}{(z+1)}*\frac{(z+2)}{(z+2)} - \frac{z}{(z+2)}*\frac{(z+1)}{(z+1)} = \frac{z^2 + 2z - z^2 -z}{(z+1)(z+2)} = \frac{z}{(z+1)(z+2)} \]

Answer 2

Thanks for your help, but I don't know how did you get this z(z+1)−z(z+2) from z(z+1)(z+2)???

Answer 3

I just showed you, didn't I? I went in the other direction, starting with the result and going back to the start.

Answer 4

If that makes you uncomfortable, you could use partial fractions to go from \[\frac{z}{(z+1)(z+2)} \rightarrow \frac{z}{(z+1)}-\frac{z}{(z+2)}\]

Answer 5

but partial fractions gives me 2/(z+2)-1/(z+1) and I couldn't get z/(z+1)-z/(z+2)

Answer 6

Thanks I got it. thank for your help!!

Answer 7

What technique did you use?

Answer 8

\[\frac{2}{z+2}-\frac{1}{z+1}\]Make a common denominator
\[\frac{2}{(z+2)}*\frac{(z+1)}{(z+1)} - \frac{1}{(z+1)}*\frac{(z+2)}{(z+2)} = \frac{2(z+1) - (z+2)}{(z+1)(z+2)} = \frac{z}{(z+1)(z+2)}\]Now here's the trick: add antimatter! :-)
\[\frac{z}{(z+1)(z+2)} = \frac{z +z^2 - z^2}{(z+1)(z+2)} = \frac{z(z+2)-z(z+1)}{(z+1)(z+2)} =\]\[ \frac{z(z+2)}{(z+1)(z+2)} -\frac{z(z+1)}{(z+1)(z+2)} = \frac{z}{z+2}-\frac{z}{z+1}\]