`2x^2 + 3 sqrt3x + 3`

3 sqrt(3x)^3 - 2 sqrt2y^3 = (sqrt3x - sqrt2y)(Ax^2 - Bxy + Cy^2) , then the value of (A^2 - B^2 + C^2) is: 3 sqrt(3x)^3 - 2 sqrt2y^3 = (sqrt3x - sqrt2y)(Ax^2 - Bxy + Cy^2) , है तो (A^2 - B^2 + C^2) का मान हैः

Solve for x : sqrt3x^2 - 2sqrt2x - 2sqrt3 = 0

If ( 3x- 4)/( sqrt(3x)+2) = 2+ ( sqrt(3x)- 2)/(2) , then find the value of x

If (sqrt(x + 2) + sqrt(x-3))/(sqrt(x+2) - sqrt(x - 3)) = 5 , then the value of x is

If 3 sqrt3 x^3-2sqrt2 y^3=(sqrt3x- sqrt2y) (Ax^2-Bxy+Cy^2) , then the value of (A^2-B^2+C^2) is : यदि 3 sqrt3 x^3-2sqrt2 y^3=(sqrt3x- sqrt2y) (Ax^2-Bxy+Cy^2) है, तो (A^2-B^2+C^2) का मान क्या होगा ?

Differentiate the following w.r.t. x: frac{(x^2 +2x + 2)^(3/2)} {(sqrt x + 3 )^3 (cos x)^x}

If the represented by the equation 3y^2-x^2+2sqrt(3)x-3=0 are rotated about the point (sqrt(3),0) through an angle of 15^0 , on in clockwise direction and the other in anticlockwise direction, so that they become perpendicular, then the equation of the pair of lines in the new position is (1) y^2-x^2+2sqrt(3)x+3=0 (2) y^2-x^2+2sqrt(3)x-3=0 (3) y^2-x^2-2sqrt(3)x+3=0 (4) y^2-x^2+3=0

If the represented by the equation 3y^2-x^2+2sqrt(3)x-3=0 are rotated about the point (sqrt(3),0) through an angle of 15^0 , on in clockwise direction and the other in anticlockwise direction, so that they become perpendicular, then the equation of the pair of lines in the new position is (1) y^2-x^2+2sqrt(3)x+3=0 (2) y^2-x^2+2sqrt(3)x-3=0 (3) y^2-x^2-2sqrt(3)x+3=0 (4) y^2-x^2+3=0