`frac{6}{sqrt 2- sqrt 3}`

solve the equations: frac{2}{sqrt x} + frac{3}{sqrt y} =2 and frac{4}{sqrt x} - frac{9}{sqrt y} =-1

frac{1}{sqrt7-4} frac{1}{sqrt11-7}

Simplify : (sqrt(2))/(sqrt(6)-sqrt(2))- (sqrt(3))/(sqrt(6)+sqrt(2))

Prove that tan 7 (1^(@))/(2) = sqrt2 - sqrt3 -sqrt4 + sqrt6 = (sqrt3 - sqrt2) (sqrt2 -1).

Rationalize the denominator: (2 + sqrt 3)/(2 - sqrt 3) + (2 - sqrt 3)/(2 + sqrt 3) + (sqrt 3 - 1)/(sqrt 3 + 1)

Prove that cot 7 ""(1^(@))/(2) = sqrt2 + sqrt3 + sqrt4 + sqrt6 = (sqrt3 + sqrt2) (sqrt2 +1 ).

2/sqrt6 xx 3/sqrt6 + 2/6

(i) If x = (6ab)/(a + b) , find the value of : (x + 3a)/(x - 3a) + (x + 3b)/(x - 3b) . (ii) a = (4sqrt(6))/(sqrt(2) + sqrt(3)) , find the value of : (a + 2sqrt(2))/(a - 2sqrt(2)) + (a + 2sqrt(3))/(a - 2sqrt(3)) .

Let k be a real number such that the inequality sqrt(x-3) +sqrt(6 -x) ge k has a solution then the maximum value of k is sqrt3 (2) sqrt6 -sqrt3 (3) sqrt6 (4) sqrt6 +sqrt3

Using binomial theorem, write the value of (a+b)^(6) + (a-b)^(6) and hence find the value of ( sqrt(3) + sqrt(2) )^(6) + ( sqrt(3) - sqrt(2) )^(6) .