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Thread: Operations on Real Numbers and Rational Equations

  1. #1
    Mad
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    Operations on Real Numbers and Rational Equations

    $\displaystyle [(x^2n + x^n)/(2x - 2)] / [(4x^n + 4)/(x^n+1 - x^n)]$


    NOTE: $\displaystyle (x^2n + x^n)$ : The power isn't 2, it's 2n.
    $\displaystyle (x^n+1 - x^n)$ : The power isn't n, it's n+1.

    Sorry, I don't know how to get it to work properly.


    $\displaystyle [(x-1)/(x+1) - (x+1)/(x-1)] / [(x-1)/(x+1) + (x+1)/(x-1)]$



    Last two problems.
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  2. #2
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    [(x^2n + x^n)/(2x - 2)] / [(4x^n + 4)/(x^n+1 - x^n)]
    $\displaystyle \frac {\frac {x^{2n}+x^n}{2x-2}}{\frac {4x^n+4}{x^{n+1}-x^n}}$

    to get more than one thingy in a superscript, surround them with curly brackets eg x^{2n}.

    Just factorise all of the brackets then cancel. Remember that $\displaystyle x^{2n} = x^n*x^n$ and $\displaystyle \frac {a/b}{c/d} = \frac {ad}{bc}$
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  3. #3
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    $\displaystyle
    \frac{{\frac{{x^{2n} + x^n }}
    {{2x - 2}}}}
    {{\frac{{4x^n + 4}}
    {{x^{n + 1} - x^n }}}} = \frac{{x^{2n} + x^n }}
    {{2x - 2}} \cdot \frac{{x^{n + 1} - x^n }}
    {{4x^n + 4}} = \frac{{x^n \left( {x^n + 1} \right)}}
    {{2\left( {x - 1} \right)}} \cdot \frac{{x^n \left( {x - 1} \right)}}
    {{4\left( {x^n + 1} \right)}} = \frac{{x^n x^n }}
    {{\left( 2 \right)\left( 4 \right)}} = \frac{{x^{2n} }}
    {8}
    $




    $\displaystyle
    \boxed{
    {\text{Let }}a = x - 1\text{;
    }
    {\text{ }}b = x + 1 }
    $:


    $\displaystyle \frac{{\frac{a}
    {b} - \frac{b}
    {a}}}
    {{\frac{a}
    {b} + \frac{b}
    {a}}} = \frac{{\frac{{a^2 - b^2 }}
    {{ab}}}}
    {{\frac{{a^2 + b^2 }}
    {{ab}}}}$$\displaystyle = \frac{{a^2 - b^2 }}
    {{ab}} \cdot \frac{{ab}}
    {{a^2 + b^2 }} = \frac{{a^2 - b^2 }}
    {{a^2 + b^2 }} = \frac{{\left( {a + b} \right)\left( {a - b} \right)}}
    {{a^2 + b^2 }} $

    $\displaystyle \frac{{\left( {a + b} \right)\left( {a - b} \right)}}
    {{a^2 + b^2 }} = $$\displaystyle \frac{{\left( {\left( {x - 1} \right) + \left( {x + 1} \right)} \right)\left( {\left( {x - 1} \right) - \left( {x + 1} \right)} \right)}}
    {{\left( {x - 1} \right)^2 + \left( {x + 1} \right)^2 }} = $$\displaystyle \frac{{\left( {2x} \right)\left( { - 2} \right)}}
    {{\left( {x^2 - 2x + 1} \right) + \left( {x^2 + 2x + 1} \right)}} =$$\displaystyle \frac{{ - 4x}}
    {{2x^2 + 2}} = \frac{{ - 2x}}
    {{x^2 + 1}}$
    Last edited by xifentoozlerix; Mar 6th 2008 at 12:14 AM.
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