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Math Help - Any guideline??

  1. #1
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    Any guideline??

    The problem
    (1 - x^-2) : (1 - x^-1)

    My aproach:

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

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

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

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

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

    But according to the book the answer should be = \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
    (1 - x^-2) : (1 - x^-1)

    My aproach:

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

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

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

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

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

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


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

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

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

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

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

     = \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.


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

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

    I mean  x:x^2 = x^1-2 = x^-1...mmm...
    Last edited by Alienis Back; April 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  x:x^2 = x^1-2 = x^-1...mmm...
    [tex]Remember that x^2 = x\cdot x

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


    Therefore

    \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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