Well, it is quite easy to rationalize if there are only 1 or 2 sqrt denominators.
For example: c/(sqrt a + sqrt b) --> (c(sqrt a - sqrt b))/(a^2 -b)
But what if there were 3?
for example:
1/(sqrt 2 + sqrt 3 + sqrt 5) =?
Well, it is quite easy to rationalize if there are only 1 or 2 sqrt denominators.
For example: c/(sqrt a + sqrt b) --> (c(sqrt a - sqrt b))/(a^2 -b)
But what if there were 3?
for example:
1/(sqrt 2 + sqrt 3 + sqrt 5) =?
I've never actually had to do one of these but here's how I'd go about it.
Say you have $\displaystyle \frac{1}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$.
Multiply top and bottom by something like $\displaystyle (\sqrt{a} - \sqrt{b} - \sqrt{c})$. This gives you:
$\displaystyle \frac{(\sqrt{a} - \sqrt{b} - \sqrt{c})}{(\sqrt{a} + \sqrt{b} + \sqrt{c})(\sqrt{a} - \sqrt{b} - \sqrt{c})}$
Expanding this gives you
$\displaystyle \frac{(\sqrt{a} - \sqrt{b} - \sqrt{c})}{(a - b - c) - 2\sqrt{bc}}$.
Now this looks complicated but $\displaystyle (a - b - c)$ is now an integer, let's call this $\displaystyle D$, you now have:
$\displaystyle \frac{(\sqrt{a} - \sqrt{b} - \sqrt{c})}{D - 2\sqrt{bc}}$.
I'm sure you can now rationalise this fraction?
Hope this helps