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\[\frac{\sqrt{6}}{\sqrt{2}} + \frac{3}{\sqrt{3}} + \frac{\sqrt{15}}{\sqrt{5}} +\frac{\sqrt{18}}{\sqrt{6}}\]
Remember one thing: \[\large \color{green}{\frac{\sqrt{x}}{\sqrt{y}} = \sqrt{\frac{x}{y}}}\]
So for first term apply this formula... and tell me what do you get..??

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@mathslover will suggest you to take the rationalization process I guess..
\[\large{\frac{\sqrt{6}}{\sqrt{2}}=\sqrt{\frac{6}{2}}}\] \[\large{\sqrt{6}=\sqrt{3}}\] \[\large{\frac{3}{\sqrt{3}}}\]Now rationalizing the above expression
Every term will contain 3 after solving Strange WOW..
yes i get it, but then how do i simplify later
sorry i meant : \(\frac{\sqrt{6}}{\sqrt{2}}=\sqrt{3}\)
yes there are three terms like that and the other one 3/root 3
What do you get show us first..
|dw:1342950785620:dw|
root 3 + root 3 + root 3 + 3/ root 3
Yes right
Oh sorry..
Take the LCD..
then it becomes 3 root 3 + 3/root 3
LCD = root(3)
i get it thanks :)
we have to rationalize later :)
Take the LCM and then we will do rationalization..
Yes you are absolutely correct @zaphod
thanks alot :) guys
Now rationalizing \(\huge{\frac{3}{\sqrt{3}}}\) = \(\huge{\frac{3*\sqrt{3}}{\sqrt{3}*\sqrt{3}}}\) \[\huge{\frac{\sqrt{3}\cancel{3}^1}{\cancel{3}^1}}\] \[\huge{\sqrt{3}}\]
Or if you take rationalization process first then you need not to take LCD..
|dw:1342951138950:dw|
Did you understand how to do this?? @zaphod
i got the answer
My question is not about the answer.. I said did you get how to do this??
yes i got it
That is nice..

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