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Thread: Any guideline??

  1. #1
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    Any guideline??

    The problem
    $\displaystyle (1 - x^-2) : (1 - x^-1)$

    My aproach:

    = $\displaystyle (\frac{1}{1} - \frac{1}{x^2}) : (\frac{1}{1} - \frac{1}{x})$

    = $\displaystyle (\frac{x^2}{x^2} - \frac{1}{x^2}) : (\frac{x}{x} - \frac{1}{x})$

    = $\displaystyle \frac{x^2 - 1}{x^2} : \frac{x - 1}{x}$

    = $\displaystyle \frac{x^2 - 1}{x^2} * \frac{x}{x - 1}$

    = $\displaystyle \frac{x^3 - x}{x^3 - x^2}$

    But according to the book the answer should be = $\displaystyle \frac{x + 1}{x}$


    Mmm...-2 and -1 on the original problem should be exponents.
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  2. #2
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    Quote Originally Posted by Alienis Back View Post
    The problem
    $\displaystyle (1 - x^-2) : (1 - x^-1)$

    My aproach:

    = $\displaystyle (\frac{1}{1} - \frac{1}{x^2}) : (\frac{1}{1} - \frac{1}{x})$

    = $\displaystyle (\frac{x^2}{x^2} - \frac{1}{x^2}) : (\frac{x}{x} - \frac{1}{x})$

    = $\displaystyle \frac{x^2 - 1}{x^2} : \frac{x - 1}{x}$

    = $\displaystyle \frac{x^2 - 1}{x^2} * \frac{x}{x - 1}$

    = $\displaystyle \frac{x^3 - x}{x^3 - x^2}$

    But according to the book the answer should be = $\displaystyle \frac{x + 1}{x}$


    Mmm...-2 and -1 on the original problem should be exponents.
    You haven't gone far enough.

    $\displaystyle \frac{x^3 - x}{x^3 - x^2}$

    $\displaystyle = \frac{x(x^2 - 1)}{x^2(x - 1)}$

    $\displaystyle = \frac{x^2 - 1}{x(x - 1)}$

    $\displaystyle = \frac{(x + 1)(x - 1)}{x(x - 1)}$

    $\displaystyle = \frac{x + 1}{x}$.
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  3. #3
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    What did you do here?

    Quote Originally Posted by Prove It View Post
    You haven't gone far enough.


    $\displaystyle = \frac{x(x^2 - 1)}{x^2(x - 1)}$

    $\displaystyle = \frac{x^2 - 1}{x(x - 1)}$.

    I mean $\displaystyle x:x^2 = x^1-2 = x^-1$...mmm...
    Last edited by Alienis Back; Apr 22nd 2009 at 05:54 PM. Reason: can anyone tell me how to write 1-2 as exponent
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  4. #4
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    Quote Originally Posted by Alienis Back View Post
    What did you do here?




    I mean $\displaystyle x:x^2 = x^1-2 = x^-1$...mmm...
    [tex]Remember that $\displaystyle x^2 = x\cdot x$

    So $\displaystyle \frac{x}{x^2} = \frac{x}{x\cdot x} = \frac{1}{x}$.


    Therefore

    $\displaystyle \frac{x(x^2 - 1)}{x^2(x - 1)} = \frac{x(x^2 - 1)}{x\cdot x(x - 1)} = \frac{1(x^2 - 1)}{x(x - 1)}$
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