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anonymous

  • one year ago

How would I find the least common denominator of these fractions (posted below)...

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  1. anonymous
    • one year ago
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    \[\frac{ \sqrt[3]{x} }{ 2\sqrt{x} } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} }\]

  2. anonymous
    • one year ago
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    yaaa.. no

  3. anonymous
    • one year ago
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    lol That's exactly what I thought.

  4. anonymous
    • one year ago
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    at first i thought you be like 1/18 and 1/4 or something like that and I be like I know EXACTLY how to do that :) but ya... no...

  5. anonymous
    • one year ago
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    srry xD

  6. anonymous
    • one year ago
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    @triciaal @geny55

  7. anonymous
    • one year ago
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    See that's my problem, and I even know the answer (thanks to calculators). Problem is I have to show work, and I can't do that since I have no idea how to. But yeah. Don't worry about it. :)

  8. anonymous
    • one year ago
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    @pooja195 @perl

  9. anonymous
    • one year ago
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    @Ashleyisakitty @west_coast

  10. anonymous
    • one year ago
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    i can help

  11. anonymous
    • one year ago
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    It would be appreciated. :)

  12. anonymous
    • one year ago
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    ok, paste both fractions in separate comments so i can see them clearer

  13. anonymous
    • one year ago
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    |dw:1442863551641:dw|

  14. anonymous
    • one year ago
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    Okay just a second. (and nice picture lol)

  15. anonymous
    • one year ago
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    TY TY :)

  16. anonymous
    • one year ago
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    \[\frac{ \sqrt[3]{x} }{ 2\sqrt{x} }\]

  17. anonymous
    • one year ago
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    ehhhhhhhh................. idk thiss

  18. anonymous
    • one year ago
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    \[\frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} }\]

  19. anonymous
    • one year ago
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    Eh. It's okay.

  20. anonymous
    • one year ago
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    @Nnesha @Preetha @kiamousekia @iGreen

  21. anonymous
    • one year ago
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    would you like help with this \[\frac{ \sqrt[3]{x} }{ 2\sqrt{x} } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} }\]

  22. anonymous
    • one year ago
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    Yes, please.

  23. anonymous
    • one year ago
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    $$\Large \frac{ \sqrt[3]{x} }{ 2\sqrt{x} }\cdot \frac{\sqrt x }{\sqrt x } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} } \cdot \frac{\sqrt[3] {x^2} }{\sqrt[3] {x^2}} $$

  24. anonymous
    • one year ago
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    First i would rationalize the denominators

  25. anonymous
    • one year ago
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    Is what you posted above the rationalization?

  26. phi
    • one year ago
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    it might be easier to see if you know how to write the radicals as exponents \[ \frac{ \sqrt[3]{x} }{ 2\sqrt{x} } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} } \\ \frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } + \frac{ x^\frac{1}{2} +3 }{ 3x^\frac{1}{3} } \] the common denominator is the product of the two denominators you can also rationalize the denominators as jayz suggests. notice that the answer can be written in many different forms

  27. anonymous
    • one year ago
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    So \[6x^{1/6}\] would be the common denominator ?

  28. phi
    • one year ago
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    you add exponents so 6 x^(5/6) you can remember this rule by remembering x *x = x^2 and x is x^1 in other words, \( x^1 \cdot x^1 = x^2\)

  29. anonymous
    • one year ago
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    Ah okay. Just to check, is this a problem that I have to multiply the numerators by the same values as the denominators?

  30. anonymous
    • one year ago
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    Because I have heard both yes and no from different people depending on the equations.

  31. phi
    • one year ago
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    once you know what the common denominator is, you know what you should multiply the "bottom" by (in the the first fraction, for example) we can't just multiply the bottom by something (that will change it) but if we multiply top and bottom by the same thing (for example x/x) that is the same as multiplying by 1 , so it does not change the fraction.

  32. phi
    • one year ago
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    \[ \frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } + \frac{ x^\frac{1}{2} +3 }{ 3x^\frac{1}{3} } \] multiply the first fraction by 1 (in the form 3 x^(1/3) / (3 x^(1/3)) \[ \frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } \cdot \frac{ 3x^\frac{1}{3} }{ 3x^\frac{1}{3} }+ \frac{ x^\frac{1}{2} +3 }{ 3x^\frac{1}{3} } \]

  33. phi
    • one year ago
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    and multiply the second fraction by 1 in the form 2x^(1/2) / 2x^(1/2) \[ \frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } \cdot \frac{ 3x^\frac{1}{3} }{ 3x^\frac{1}{3} }+ \frac{ (x^\frac{1}{2} +3 )}{ 3x^\frac{1}{3} }\cdot \frac{ 2x^\frac{1}{2} }{ 2x^\frac{1}{2} }\]

  34. phi
    • one year ago
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    I assume you know how to add the exponents 1/2 + 1/3 that is the same as 3/6 + 2/6 = 5/6

  35. phi
    • one year ago
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    so you get 6 x^(5/6) as the common denominator

  36. anonymous
    • one year ago
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    Yes, I understand. Thank you. :) I am solving it out now.

  37. anonymous
    • one year ago
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    I got\[\frac{ 3x^{2/3}+2\sqrt{x} (\sqrt{x}+3) }{ 6x^{5/6} }\]

  38. phi
    • one year ago
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    yes, and that can be written as \[ \frac{ 3x^\frac{2}{3} + 2x + 6 x^\frac{1}{2}}{6x^\frac{5}{6}} \] and lots of other ways. for example: \[ \frac{1}{2x^\frac{1}{6}} + \frac{x^\frac{1}{6}}{3} +\frac{1}{x^\frac{1}{3}} \] (all ways look ugly!)

  39. anonymous
    • one year ago
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    And that would be the final answer correct?

  40. phi
    • one year ago
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    I think as good as any is \[ \frac{ 3x^\frac{2}{3} + 2x + 6 x^\frac{1}{2}}{6x^\frac{5}{6}} \] (that is your answer, but "distributing" the \( 2 \sqrt{x}\) on the other hand some people would rather have the denominator 6x (rather than 6 x^(5/6) ) so they would multiply top and bottom by x^(1/6)) But I would leave it the first way. Or show your teacher the different ways and ask if it matters.

  41. anonymous
    • one year ago
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    The way you did it makes sense to me, so I'm going with that one. Thank you so much for your help! :D

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