How do you rationalize the denominator and simplify #1 / (sqrt5+sqrt6)#?

1 Answer
Mar 16, 2016

#1/(sqrt(5)+sqrt(6)) = sqrt(6)-sqrt(5)#

Explanation:

It works out a little easier if you swap #sqrt(5)+sqrt(6)# around first.

Multiply both numerator and denominator by the conjugate #sqrt(6)-sqrt(5)# of the denominator:

#1/(sqrt(5)+sqrt(6))#

#=1/(sqrt(6)+sqrt(5))#

#=(sqrt(6)-sqrt(5))/((sqrt(6)-sqrt(5))(sqrt(6)+sqrt(5)))#

#=(sqrt(6)-sqrt(5))/((sqrt(6))^2-(sqrt(5))^2)#

#=(sqrt(6)-sqrt(5))/(6-5)#

#=(sqrt(6)-sqrt(5))/1#

#=sqrt(6)-sqrt(5)#