Question 233254
{{{(sqrt(10)-sqrt(5))/(sqrt(2)-sqrt(5))}}}
Your denominator is a binomial (it has two terms). To rationalize a binomial denominator we need to take advantage of: {{{(a+b)(a-b) = a^2 - b^2}}}. On the left we have two binomials, (a+b) and (a-b). Binomials like this are called conjugates. And the pattern says that when conjugates are multiplied we get a binomial made up of <i>perfect square terms</i>!<br>
So with your denominator, {{{sqrt(2)-sqrt(5)}}}, we need to multiply by its conjugate, {{{sqrt(2)+sqrt(5)}}}, to get a denominator made up of perfect square terms. And, of course, we will have to multiply the numerator by the same thing:
{{{((sqrt(10)-sqrt(5))/(sqrt(2)-sqrt(5)))((sqrt(2)+sqrt(5))/(sqrt(2)+sqrt(5))) = (sqrt(20) - sqrt(10) +sqrt(50) - (sqrt(5))^2)/((sqrt(2))^2 - (sqrt(5))^2) = (sqrt(4*5) - sqrt(10) + sqrt(25*2) - 5)/(2-5) = (sqrt(4)*sqrt(5) - sqrt(10) + sqrt(25)*sqrt(2) - 5)/(-3) = (2sqrt(5) - sqrt(10) + 5sqrt(2) - 5)/(-3)}}}