Solve `sqrt2+sqrtx=sqrt3`

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

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

Solve (sqrt3-sqrt2)(2-sqrt2)

Differentiate (sqrt2 +sqrtx)/(sqrt2 - sqrtx) w.r.t x

Solve {sqrt(3+2 sqrt2)}^x + {sqrt(3-2 sqrt2)}^x =6 .

y = sqrt(2 -sqrtx)

If x=3+2 sqrt2 , then the value of sqrtx-1/sqrtx is : यदि x=3+2 sqrt2 , तब sqrtx-1/sqrtx का मान :

Solve (sqrt3 +sqrt2)^x + (sqrt3 -sqrt2)^x= 10 .

Solve: (sqrt(2+sqrt(3)))^(x)+(sqrt(2-sqrt(3)))^(x)=2^(x)

If sqrtx+ 1/sqrtx=sqrt 6 , then x^2+1/x^2 is equal to: यदि sqrtx+ 1/sqrtx=sqrt 6 है, तो x^2+1/x^2 का मान क्या होगा ?