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zaphod
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\[\frac{\sqrt{6}}{\sqrt{2}} + \frac{3}{\sqrt{3}} + \frac{\sqrt{15}}{\sqrt{5}} +\frac{\sqrt{18}}{\sqrt{6}}\]

waterineyes
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Remember one thing:
\[\large \color{green}{\frac{\sqrt{x}}{\sqrt{y}} = \sqrt{\frac{x}{y}}}\]

waterineyes
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So for first term apply this formula... and tell me what do you get..??

waterineyes
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@mathslover will suggest you to take the rationalization process I guess..

mathslover
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\[\large{\frac{\sqrt{6}}{\sqrt{2}}=\sqrt{\frac{6}{2}}}\]
\[\large{\sqrt{6}=\sqrt{3}}\]
\[\large{\frac{3}{\sqrt{3}}}\]Now rationalizing the above expression

waterineyes
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Every term will contain 3 after solving Strange WOW..

zaphod
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yes i get it, but then how do i simplify later

mathslover
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sorry i meant : \(\frac{\sqrt{6}}{\sqrt{2}}=\sqrt{3}\)

zaphod
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yes there are three terms like that and the other one 3/root 3

waterineyes
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What do you get show us first..

mathslover
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dw:1342950785620:dw

zaphod
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root 3 + root 3 + root 3 + 3/ root 3

mathslover
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Yes right

waterineyes
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Oh sorry..

waterineyes
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Take the LCD..

zaphod
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then it becomes 3 root 3 + 3/root 3

waterineyes
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LCD = root(3)

zaphod
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i get it thanks :)

zaphod
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we have to rationalize later :)

waterineyes
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Take the LCM and then we will do rationalization..

waterineyes
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Yes you are absolutely correct @zaphod

zaphod
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thanks alot :) guys

mathslover
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Now rationalizing \(\huge{\frac{3}{\sqrt{3}}}\) = \(\huge{\frac{3*\sqrt{3}}{\sqrt{3}*\sqrt{3}}}\)
\[\huge{\frac{\sqrt{3}\cancel{3}^1}{\cancel{3}^1}}\]
\[\huge{\sqrt{3}}\]

waterineyes
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Or if you take rationalization process first then you need not to take LCD..

waterineyes
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dw:1342951138950:dw

waterineyes
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Did you understand how to do this?? @zaphod

zaphod
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i got the answer

waterineyes
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My question is not about the answer..
I said did you get how to do this??

zaphod
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yes i got it

waterineyes
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That is nice..