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

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

anonymous
 3 years 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks I got it. thank for your help!!

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

whpalmer4
 3 years 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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