` 1/(x+a) + 1/(x+b) = 1/(x+a+b) + 1/x `

Solve for x : (1)/(a + b + x) = (1)/(a) + (1)/(b) + (1)/(x) , a ne b ne 0 , x ne 0 , x ne -(a + b)

Solve : 1/x - 1/(x+b)= 1/a -1/(a+b)[ x ne 0, -b]

(3x + 1)/(( x- 1) (x - 3)) = (A)/(x - 1) + (B)/(x - 3) , then sin^(-1) (A)/(B) =

If y = (1)/(1 + x^(a - b) + x^(c - b)) + (1)/(1 + x^(b-c) + x^(a - c)) + (1)/(1 + x^(b - a) + x^(c - a)) then find (dy)/(dx) at e^(a^(b^(c )))

If y = (1)/(1 + x^(a - b) + x^(c - b)) + (1)/(1 + x^(b-c) + x^(a - c)) + (1)/(1 + x^(b - a) + x^(c - a)) then find (dy)/(dx) at e^(a^(b^(c )))

If x + (1)/(x) = a and x - (1)/(x) = b , then a^(2) - b^(2) = "_______" .

If |(1,1,1),(a,b,c),(a^(3),b^(3),c^(3))| = (a - b) (b - c) (c - a) (a + b + c) , where a,b,c are all different, then the determinant |(1,1,1),((x-a)^(2),(x-b)^(2),(x-c)^(2)),((x-b)(x-c),(x-c)(x-a),(x-a)(x-b))| vanishes when a)a + b + c = 0 b) x = (1)/(3) (a + b + c) c) x = (1)/(2) (a + b + c) d) x = a + b + c