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Math Help - simplify a rational fraction

  1. #1
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    simplify a rational fraction

    I need to simplify the following fraction

    [Math]\frac{2}{x^2-4}+\frac{1}{2x-x^2}[/tex]

    The answer should be [Math]\frac{1}{x(x+2)}[/tex]

    How do i preceed??
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  2. #2
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    Quote Originally Posted by strunz View Post
    I need to simplify the following fraction

    [Math]\frac{2}{x^2-4}+\frac{1}{2x-x^2}[/tex]

    The answer should be [Math]\frac{1}{x(x+2)}[/tex]

    How do i preceed??
    Use the fact that x^2- 4= (x- 2)(x+ 2) and that 2x- x^2= x(2- x)= -x(x-2)

    That makes your fraction \frac{2}{(x- 2)(x+ 2)}- \frac{1}{x(x-2)}

    The "common denominator" is (x- 2)(x+ 2)(x). Multiply numerator and denominator of the first fraction by x and the of the second fraction by (x+ 2).
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  3. #3
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    \frac{2}{x^2-4}+\frac{1}{2x-x^2}=

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

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

    \frac{2x}{(x+2)(x-2)(x)}-\frac{x+2}{(x)(x-2)(x+2)}=

    \frac{2x-(x+2)}{(x)(x-2)(x+2)}=

    \frac{2x}{(x)(x-2)}

    correct plz
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  4. #4
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    Quote Originally Posted by strunz View Post
    \frac{2}{x^2-4}+\frac{1}{2x-x^2}=

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

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

    \frac{2x}{(x+2)(x-2)(x)}-\frac{x+2}{(x)(x-2)(x+2)}=

    \frac{2x-(x+2)}{(x)(x-2)(x+2)}=

    \frac{2x}{(x)(x-2)}

    correct plz
    Hi!

    Your mistake is in the last fraction:

    \frac{2x-(x+2)}{x(x+2)(x-2)} \neq \frac{2x}{x(x-2)}

    Can you see why?
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  5. #5
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    Yes. I see now. Thx
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