Question 28478
Rationalise the denominator:
{{{2/((sqrt(6)-sqrt(5)))}}} Multiply both the numerator and the denominator by the conjugate of{{{(sqrt(6)-sqrt(5))}}} which is{{{(sqrt(6)+sqrt(5))}}}
{{{2(sqrt(6)+sqrt(5))/(sqrt(6)-sqrt(5))(sqrt(6)+sqrt(5))}}} Simplify.
{{{2(sqrt(6)+sqrt(5))/(6-5)}}}={{{2(sqrt(6)+sqrt(5))}}} or {{{2sqrt(6)+2sqrt(5)}}}

Explanation:{{{(a-b)(a+b) = a^2-b^2}}} In your problem,{{{a = sqrt(6)}}} and {{{b = sqrt(5)}}} so {{{a^2 = (sqrt(6))^2}}} = 6. And {{{b^2 = (sqrt(5))^2}}} = 5
Multiply {{{sqrt(6)-sqrt(5)}}} by its conjugate {{{sqrt(6)+sqrt(5)}}} using the FOIL method, it works out like this.

{{{(sqrt(6)-sqrt(5))(sqrt(6)+sqrt(5)) = 6 + (sqrt(6))(sqrt(5)) - (sqrt(5))(sqrt(6)) - 5}}} but, as you can see (I hope), the centre terms drop out leaving you with (6 - 5 = 1)