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 one year ago
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
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?

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whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{z}{(z+1)}  \frac{z}{(z+2)}\]Make a common denominator: \[\frac{z}{(z+1)}*\frac{(z+2)}{(z+2)}  \frac{z}{(z+2)}*\frac{(z+1)}{(z+1)} = \frac{z^2 + 2z  z^2 z}{(z+1)(z+2)} = \frac{z}{(z+1)(z+2)} \]

jairo.9
 one year ago
Best ResponseYou've already chosen the best response.0Thanks 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)???

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1I just showed you, didn't I? I went in the other direction, starting with the result and going back to the start.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1If that makes you uncomfortable, you could use partial fractions to go from \[\frac{z}{(z+1)(z+2)} \rightarrow \frac{z}{(z+1)}\frac{z}{(z+2)}\]

jairo.9
 one year ago
Best ResponseYou've already chosen the best response.0but partial fractions gives me 2/(z+2)1/(z+1) and I couldn't get z/(z+1)z/(z+2)

jairo.9
 one year ago
Best ResponseYou've already chosen the best response.0Thanks I got it. thank for your help!!

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1What technique did you use?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{2}{z+2}\frac{1}{z+1}\]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! :) \[\frac{z}{(z+1)(z+2)} = \frac{z +z^2  z^2}{(z+1)(z+2)} = \frac{z(z+2)z(z+1)}{(z+1)(z+2)} =\]\[ \frac{z(z+2)}{(z+1)(z+2)} \frac{z(z+1)}{(z+1)(z+2)} = \frac{z}{z+2}\frac{z}{z+1}\]
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