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##

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