`A=1/(sqrt(5)-sqrt(3)), B=1/(sqrt(5)+sqrt(3))=t h e n_(AxxB)`

1/(sqrt5 +sqrt3 +2) + 1/(-sqrt5 +sqrt3 +2 ) =

Sum of 1/(sqrt(2)+sqrt(5))+1/(sqrt(5)+sqrt(8))+1/(sqrt(8)+sqrt(11))+1/(sqrt(11)+sqrt(14))+..to n terms= (A) n/(sqrt(3n+2)-sqrt(2)) (B) 1/3 (sqrt(2)-sqrt(3n+2) (C) n/(sqrt(3n+2)+sqrt(2)) (D) none of these

Sum of 1/(sqrt(2)+sqrt(5))+1/(sqrt(5)+sqrt(8))+1/(sqrt(8)+sqrt(11))+1/(sqrt(11)+sqrt(14))+..to n terms= (A) n/(sqrt(3n+2)-sqrt(2)) (B) 1/3 (sqrt(2)-sqrt(3n+2) (C) n/(sqrt(3n+2)+sqrt(2)) (D) none of these

Sum of 1/(sqrt(2)+sqrt(5))+1/(sqrt(5)+sqrt(8))+1/(sqrt(8)+sqrt(11))+1/(sqrt(11)+sqrt(14))+..to n terms= (A) n/(sqrt(3n+2)-sqrt(2)) (B) 1/3 (sqrt(2)-sqrt(3n+2) (C) n/(sqrt(3n+2)+sqrt(2)) (D) none of these

The sum up to n terms of the series 1/(sqrt(1) + sqrt(3)) + 1/(sqrt(3) + sqrt(5)) + 1/(sqrt(5) + sqrt(7)) +… is:

1/(sqrt(2)+sqrt(3)+sqrt(5))+1/(sqrt(2)+sqrt(3)-sqrt(5))

The value of {1/((sqrt(6) - sqrt(5))) + 1/((sqrt(5) + sqrt(4))) + 1/((sqrt(4) + sqrt(3))) - 1/((sqrt(3) - sqrt(2))) + 1/((sqrt(2) - 1))} is :

If both a and b are rational numbers,find the values of a and b in each of the following equalities :(sqrt(3)-1)/(sqrt(3)+1)=a+b sqrt(3)( ii) (3+sqrt(7))/(3-sqrt(7))=a+b sqrt(7)(5+2sqrt(3))/(7+4sqrt(3))=a+b sqrt(3)( iv) (5+sqrt(3))/(7-sqrt(3))=47a+sqrt(3)b(sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))=a+b sqrt(15) (iv) (sqrt(2)+sqrt(3))/(3sqrt(2)-2sqrt(3))=1-b sqrt(3)

(1)/(2sqrt(5)-sqrt(3))-(2sqrt(5)+sqrt(3))/(2sqrt(5)+sqrt(3)) =

1/(sqrt(3)+sqrt(2))-2/(sqrt(5)-sqrt(3))-3/(sqrt(2)-sqrt(5))