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    Hi,

    I don't understand the last part of part C, and how you get the final part of the answer.

    Quick help is appreciated.
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    (Original post by TheKevinFang)
    Hi,

    I don't understand the last part of part C, and how you get the final part of the answer.

    Quick help is appreciated.
    \displaystyle H+\left[2e^2H\times \frac{1}{1-e^2}\right] = \frac{H(1-e^2)}{1-e^2} + \frac{2He^2}{1-e^2} = \frac{H-He^2 + 2He^2}{1-e^2} = ,,,
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    (Original post by notnek)
    \displaystyle H+\left[2e^2H\times \frac{1}{1-e^2}\right] = \frac{H(1-e^2)}{1-e^2} + \frac{2He^2}{1-e^2} = \frac{H-He^2 + 2He^2}{1-e^2} = ,,,
    Sorry I don't understand what this is trying to show?
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    https://onedrive.live.com/redir?resi...nt=photo%2cJPG

    This question, I'm also getting (1+2e/1-e) x (2h/g)^1/2? Why is it (1+e/1-e) x (2h/g)^1/2???
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    (Original post by TheKevinFang)
    Sorry I don't understand what this is trying to show?
    I'm showing how they derived \displaystyle \frac{H(1+e^2)}{1-e^2}. I'll finish it off if you like:

    \displaystyle H+\left[2e^2H\times \frac{1}{1-e^2}\right] = \frac{H(1-e^2)}{1-e^2} + \left[\frac{2He^2}{1-e^2}\right] = \frac{H-He^2 + 2He^2}{1-e^2}

    \displaystyle = \frac{H+He^2}{1-e^2}= \frac{H(1+e^2)}{1-e^2}
 
 
 
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