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jairo.9

  • 2 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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  1. whpalmer4
    • 2 years ago
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    \[\frac{z}{(z+1)} - \frac{z}{(z+2)}\]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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    Thanks I got it. thank for your help!!

  7. whpalmer4
    • 2 years ago
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    What technique did you use?

  8. whpalmer4
    • 2 years ago
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    \[\frac{2}{z+2}-\frac{1}{z+1}\]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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