## jairo.9 Group Title 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? one year ago one year ago

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1. whpalmer4 Group Title

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

2. jairo.9 Group Title

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)???

3. whpalmer4 Group Title

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

4. whpalmer4 Group Title

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

5. jairo.9 Group Title

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

6. jairo.9 Group Title

Thanks I got it. thank for your help!!

7. whpalmer4 Group Title

What technique did you use?

8. whpalmer4 Group Title

$\frac{2}{z+2}-\frac{1}{z+1}$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}$