Solve `sqrtx-1/sqrt5=sqrt5`

If sqrtx- 1/sqrtx= sqrt 5 , then x^2+1/x^2 is equal to: यदि sqrtx- 1/sqrtx= sqrt 5 है, तो x^2+1/x^2 का मान क्या होगा ?

Solve sqrt(2)sqrt(3)sqrt(5)

If x=(5+sqrt6) then sqrtx+1/sqrtx=sqrt3

(sqrt2 + 1)/sqrt5

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

The value of { 1/(sqrt6 - sqrt5) - 1/(sqrt5 - sqrt4) + 1/(sqrt4 - sqrt3) - 1/(sqrt3 - sqrt2) + 1/(sqrt2 - 1)} is :

Solve ((8+sqrt5)/(8-sqrt5))+((8-sqrt5)/(8+sqrt5))

(i) int_0^a sqrtx/(sqrtx+sqrt(a-x)) dx (ii) int_0^5 sqrtx/(sqrtx+sqrt(5-x)) dx (iii) int_0^a (dx)/(x+sqrt(a^2+x^2))

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

The value of 1/(sqrt5-sqrt(5-sqrt(24)))+1/(sqrt5-sqrt(5+sqrt(24))