How do you rationalize the denominator of ##1/(sqrt(2)+sqrt(3)+sqrt(5))##?

This is a very good question!

If it were simply ##1/(sqrta+sqrtb)##, we would use the conjugate,

and we would multiply by ##(sqrta-sqrtb)/(sqrta-sqrtb)##.

Let's try something like that and see if it works. (This is what we do with problems of kinds we have not seen before. Try something and see if it works.)

##1/(sqrt(2)+sqrt(3)+sqrt(5)) = 1/(sqrt(2)+(sqrt(3)+sqrt(5)))##

## = 1/([sqrt(2)+(sqrt(3)+sqrt(5))]) ([sqrt(2)-(sqrt(3)+sqrt(5))])/([sqrt(2)-(sqrt(3)+sqrt(5))])##

## = ([sqrt(2)-(sqrt(3)+sqrt(5))]) / ([sqrt(2) + (sqrt(3)+sqrt(5))][sqrt(2)-(sqrt(3)+sqrt(5))])##

## = ([sqrt(2)-(sqrt(3)+sqrt(5))]) / (2 -(sqrt(3)+sqrt(5))^2 )##

## = ([sqrt(2)-(sqrt(3)+sqrt(5))]) / (2 -(3+2sqrt(15)+5) )##

## = ([sqrt(2)-(sqrt(3)+sqrt(5))]) / (-6 - 2sqrt(15)) ##

Did that help? (Yes, it did. We now have a more familiar looking problem.

## = ([sqrt(2)-(sqrt(3)+sqrt(5))] [-6 + 2sqrt(15)]) / ([-6 - 2sqrt(15)][-6 + 2sqrt(15)])##

## = ([sqrt(2)-(sqrt(3)+sqrt(5))] [-6 + 2sqrt(15)]) / (36-4(15))##

## = ([sqrt(2)-(sqrt(3)+sqrt(5))] [-6 + 2sqrt(15)]) / -24##

## = ([sqrt(2)-(sqrt(3)+sqrt(5))] [3 - sqrt(15)]) / 12##

Multiply the numerator if you like, to get:

## = ([sqrt(2)-sqrt(3)-sqrt(5)] [3 - sqrt(15)]) / 12##

## =( [sqrt(2)-sqrt(3)-sqrt(5)] [3 - sqrt(15)]) / 12##

## =(3sqrt(2)-3sqrt(3)-3sqrt(5)- sqrt(30)+sqrt45+sqrt75) / 12##

## =(3sqrt(2)-3sqrt(3)-3sqrt(5)- sqrt(30)+3sqrt5+5sqrt3) / 12##

## =(3sqrt(2)+2sqrt(3)- sqrt(30)) / 12##