" प्रश्न 7."x sqrt(x+2)

State the coefficient of x^(2) in each of the following polynomials : sqrt(2)x^(2) + 7x - 4sqrt(2)

If x=(1)/(2)(sqrt(7)+(1)/(sqrt(7))) ,then , log_(27)((sqrt(x^(2)-1))/(x-sqrt(x^(2)-1))) is equal to

If sqrt(3x - 7) + sqrt(3x + 7) = 4 + sqrt(2) then value of x + 1/x is

int sqrt (x^2-8 x+7) d x is equal to A) 1/2(x-4) sqrt(x^2-8 x+7)+9 log |x-4+sqrt(x^2-8 x+7)|+C B) 1/2(x-4) sqrt(x^2-8 x+7)+9 log | x+4+sqrt(x^2-8 x+7)+C C) 1/2(x-4) sqrt(x^2-8 x+7)+3 sqrt(2) log | x-4+sqrt(x^2-8 x+7)+C D) 1/2(x-4) sqrt(x^2-8 x+7)-9/2 log |x-4+sqrt(x^2-8 x+7)|+C

Simplify : (x+sqrt(x^(2)-1))^(7) + (x-sqrt(x^(2)-1))^(7)

Complete the following activity to solve the quadratic equation sqrt(3) x^(2) +4x- 7 sqrt(3) = 0 by factorisation method : sqrt(3) x^(2) +4x- 7 sqrt(3) = 0 :. sqrt(3) x^(2) + square - 3x - 7 sqrt(3) = 0 :. x( sqrt(3) x + ) - sqrt(3) ( sqrt(3) x + 7) = 0 :. (" ") (x-sqrt(3))= 0 :. sqrt(3) x + 7 = 0 or square = 0 :. x = ( -7)/( sqrt(3)) or x = square :. (-7)/(sqrt(3)) and sqrt(3) are the roots of the equation.

Solve for 'x' : sqrt(6x+7)-(2x-7)=0 .

Solve sqrt(3x^2 -7x -30) - sqrt(2 x^2 -7x-5) = x-5