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Write as a simple fraction in lowest terms

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\frac{\frac{2}{3}x(x^{2}+4)^{1/2}(x^{2}-9)^{-2/3}-x(x^{2}-9)^{1/3}(x^{2}+4)^{-1/2}}{x^{2}+4}[/itex]

    3. The attempt at a solution

    [itex]\frac{x(x^{2}+4)^{-1/2}(x^{2}-9)^{-2/3}(\frac{2}{3}(x^{2}+4)-(x-9))}{x^{2}+4}[/itex]

    Then:

    [itex]\frac{x(\frac{2}{3}(x^{2}+4)-(x-9))}{(x^{2}+4)^{3/2}(x^{2}-9)^{2/3}}[/itex]

    What should I do next? I multiply the numerator but this, it seems leads to a dead-end. Or is there a mistake involved in the aforementioned steps?

    P. S. The book gives the answer

    [itex]\frac{-x^{3}+35x}{3(x^{2}+4)^{3/2}(x^{2}-9)^{2/3}}[/itex]

    P. S. S. Can someone tell me how to write tex instead of itex automatically?
     
    Last edited: Nov 27, 2011
  2. jcsd
  3. Nov 27, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The first term in the numerator has a factor of [itex](x^2+ 4)^{1/2}[/itex] and the second a factor of [itex](x^2+ 4)^{-1/2}[/itex]. -1/2 is the smaller power so note that [itex](x^2+ 4)^{1/2}= (x^2+ 4)(x^2+ 4)^{-1/2}[/itex] and factor out [itex](x^2+ 4)^{-1/2}[/itex]. The first term has a factor of [itex](x^2- 9)^{-2/3}[/itex] and the second a factor of [itex](x^2- 9)^{1/3}[/itex]. -2/3 is the smaller power so note that [itex](x^2- 9)^{1/3}= (x^2- 9)(x^2- 9)^{-2/3}[/itex] and factor out [itex](x^2- 9)^{-2/3}[/itex]. Of course, there is an x in both terms so factor that out:
    [tex]x(x^2+ 4)^{-1/2}(x^2- 9)^{-2/3}\frac{\frac{2}{3}(x^2+ 4)- x^2+ 9}{x^2+ 4}[/tex]
    Of course that [itex]x^2+ 4[/itex] in the denominator can be absorbed into the [itex](x^2+ 4)^{-1/2}[/itex] to give [itex](x^2+ 4)^{-3/2}[/itex].

    I don't do the tex "automatically" but the you can "edit" and manually remove the "i". Sometimes when I realize that I have used a number of "itex"s where I want "tex", I copy the whole thing to the "clipboard", open "Notepad" (standard with Windows), paste into Notepad, use the editing features there, then reverse.
     
    Last edited: Nov 27, 2011
  4. Nov 27, 2011 #3
    I don't understand, you just rewrote what I did, or have I overlooked something?
     
  5. Nov 28, 2011 #4

    Mentallic

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    Homework Helper

    You haven't made a mistake yet, because both answers are equivalent. Just expand the numerator and collect like terms.

    Oh and you made a typo in the numerator, you forgot the square in x2-9 :

    [tex]\frac{x(\frac{2}{3}(x^{2}+4)-(x^2-9))}{(x^{2}+4)^{3/2}(x^{2}-9)^{2/3}}[/tex]
     
  6. Nov 28, 2011 #5
    The typo was the mistake (as usual for me). Thank's for helping...
     
  7. Nov 28, 2011 #6

    Mentallic

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    Homework Helper

    Oh, well, np :biggrin:
     
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