How do you simplify #(sqrt 6 - sqrt 5) / (sqrt 6 + sqrt 5)#?

Your goal here is to rationalize the denominator by multiplying it by its conjugate.

The conjugate of a binomial can be determined by changing the sign of the second term. In your case, you would have

#sqrt(6) + sqrt(5) -> underbrace(sqrt(6) color(red)(-) sqrt(5))_(color(blue)("conjugate"))#

So, multiply the fraction by # 1 = (sqrt(6) - sqrt(5))/(sqrt(6) - sqrt(5))# to get

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

The denominator takes the form

#color(blue)( (a-b)(a+b) = a^2 - b^2)#

and can thus be written as

#(sqrt(6) + sqrt(5))(sqrt(6) - sqrt(5)) = (sqrt(6))^2 - (sqrt(5))^2 = 6 - 5 = 1#

The numerator takes the form

#color(blue)( (a-b)^2 = a^2 - 2ab + b^2)#

and can be written as

#(sqrt(6) - sqrt(5))(sqrt(6) - sqrt(5)) = (sqrt(6))^2 - 2sqrt(30) + (sqrt(5))^2#

#=6 - 2sqrt(30) + 5#

#= 11 - 2sqrt(30)#

The expression becomes

#((sqrt(6) - sqrt(5))(sqrt(6) - sqrt(5)))/((sqrt(6) + sqrt(5))(sqrt(6) - sqrt(5))) = (11 - 2sqrt(30))/1 = color(green)(11 - 2sqrt(30))#