Question 151059
{{{(sqrt(2))/(sqrt(6)-sqrt(3))}}} Start with the given expression.



Take note that the conjugate of the denominator {{{sqrt(6)-sqrt(3)}}} is {{{sqrt(6)+sqrt(3)}}}. If we multiply the fraction by {{{(sqrt(6)+sqrt(3))/(sqrt(6)+sqrt(3))}}}, this will eliminate the square roots in the denominator




{{{((sqrt(2))/(sqrt(6)-sqrt(3)))((sqrt(6)+sqrt(3))/(sqrt(6)+sqrt(3)))}}} Multiply the fraction by {{{(sqrt(6)+sqrt(3))/(sqrt(6)+sqrt(3))}}}



{{{((sqrt(2))(sqrt(6)+sqrt(3)))/((sqrt(6)-sqrt(3))(sqrt(6)+sqrt(3)))}}} Combine the fractions.



{{{((sqrt(2))(sqrt(6)+sqrt(3)))/(3)}}} Foil the denominator to get {{{(sqrt(6)-sqrt(3))(sqrt(6)+sqrt(3))=sqrt(6)sqrt(6)+sqrt(6)sqrt(3)-sqrt(3)sqrt(6)-sqrt(3)sqrt(3)=6-3=3}}}



{{{(sqrt(2)sqrt(6)+sqrt(2)sqrt(3))/(3)}}} Distribute



{{{(sqrt(12)+sqrt(6))/(3)}}} Combine and multiply the square roots.



{{{(2*sqrt(3)+sqrt(6))/(3)}}} Simplify the square root.




So after rationalizing the denominator, the expression {{{(sqrt(2))/(sqrt(6)-sqrt(3))}}} simplifies to {{{(2*sqrt(3)+sqrt(6))/(3)}}}




In other words, {{{(sqrt(2))/(sqrt(6)-sqrt(3))=(2*sqrt(3)+sqrt(6))/(3)}}}