anonymous
  • anonymous
How would I find the least common denominator of these fractions (posted below)...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{ \sqrt[3]{x} }{ 2\sqrt{x} } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} }\]
anonymous
  • anonymous
yaaa.. no
anonymous
  • anonymous
lol That's exactly what I thought.

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anonymous
  • anonymous
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...
anonymous
  • anonymous
srry xD
anonymous
  • anonymous
@triciaal @geny55
anonymous
  • anonymous
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. :)
anonymous
  • anonymous
@pooja195 @perl
anonymous
  • anonymous
@Ashleyisakitty @west_coast
anonymous
  • anonymous
i can help
anonymous
  • anonymous
It would be appreciated. :)
anonymous
  • anonymous
ok, paste both fractions in separate comments so i can see them clearer
anonymous
  • anonymous
|dw:1442863551641:dw|
anonymous
  • anonymous
Okay just a second. (and nice picture lol)
anonymous
  • anonymous
TY TY :)
anonymous
  • anonymous
\[\frac{ \sqrt[3]{x} }{ 2\sqrt{x} }\]
anonymous
  • anonymous
ehhhhhhhh................. idk thiss
anonymous
  • anonymous
\[\frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} }\]
anonymous
  • anonymous
Eh. It's okay.
anonymous
  • anonymous
@Nnesha @Preetha @kiamousekia @iGreen
anonymous
  • anonymous
would you like help with this \[\frac{ \sqrt[3]{x} }{ 2\sqrt{x} } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} }\]
anonymous
  • anonymous
Yes, please.
anonymous
  • anonymous
$$\Large \frac{ \sqrt[3]{x} }{ 2\sqrt{x} }\cdot \frac{\sqrt x }{\sqrt x } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} } \cdot \frac{\sqrt[3] {x^2} }{\sqrt[3] {x^2}} $$
anonymous
  • anonymous
First i would rationalize the denominators
anonymous
  • anonymous
Is what you posted above the rationalization?
phi
  • phi
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
anonymous
  • anonymous
So \[6x^{1/6}\] would be the common denominator ?
phi
  • phi
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, \( x^1 \cdot x^1 = x^2\)
anonymous
  • anonymous
Ah okay. Just to check, is this a problem that I have to multiply the numerators by the same values as the denominators?
anonymous
  • anonymous
Because I have heard both yes and no from different people depending on the equations.
phi
  • phi
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.
phi
  • phi
\[ \frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } + \frac{ x^\frac{1}{2} +3 }{ 3x^\frac{1}{3} } \] multiply the first fraction by 1 (in the form 3 x^(1/3) / (3 x^(1/3)) \[ \frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } \cdot \frac{ 3x^\frac{1}{3} }{ 3x^\frac{1}{3} }+ \frac{ x^\frac{1}{2} +3 }{ 3x^\frac{1}{3} } \]
phi
  • phi
and multiply the second fraction by 1 in the form 2x^(1/2) / 2x^(1/2) \[ \frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } \cdot \frac{ 3x^\frac{1}{3} }{ 3x^\frac{1}{3} }+ \frac{ (x^\frac{1}{2} +3 )}{ 3x^\frac{1}{3} }\cdot \frac{ 2x^\frac{1}{2} }{ 2x^\frac{1}{2} }\]
phi
  • phi
I assume you know how to add the exponents 1/2 + 1/3 that is the same as 3/6 + 2/6 = 5/6
phi
  • phi
so you get 6 x^(5/6) as the common denominator
anonymous
  • anonymous
Yes, I understand. Thank you. :) I am solving it out now.
anonymous
  • anonymous
I got\[\frac{ 3x^{2/3}+2\sqrt{x} (\sqrt{x}+3) }{ 6x^{5/6} }\]
phi
  • phi
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!)
anonymous
  • anonymous
And that would be the final answer correct?
phi
  • phi
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.
anonymous
  • anonymous
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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