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Write as a simple fraction in lowest terms

  • Thread starter mindauggas
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  • #1
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Homework Statement



[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]

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?
 
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Answers and Replies

  • #2
HallsofIvy
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Homework Statement



[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]
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].

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?
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.
 
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  • #3
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I don't understand, you just rewrote what I did, or have I overlooked something?
 
  • #4
Mentallic
Homework Helper
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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]
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]
 
  • #5
127
0
The typo was the mistake (as usual for me). Thank's for helping...
 
  • #6
Mentallic
Homework Helper
3,798
94
The typo was the mistake (as usual for me). Thank's for helping...
Oh, well, np :biggrin:
 

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