Answer:
x ∈ {-a, -b}
Explanation:
1/(a+b+x) = 1/a +1/b +1/x . . . . given
abx = bx(a+b+x) +ax(a+b+x) +ab(a+b+x) . . . . multiply by abx(a+b+x)
(a+b)x^2 +(a+b)^2x +ab(a+b) = 0 . . . . . subtract abx
x^2 + (a+b)x +ab = 0 . . . . . divide by (a+b)
This is a quadratic equation in x. It will have two solutions, as given by the quadratic formula.
x = (-(a+b) ±√((a+b)^2 -4(1)(ab))/(2(1)) = (-(a+b) ± |a -b|)/2
Without loss of generality, we can assume a ≥ b, so |a -b| ≥ 0. Then ...
x = (-a -b -a +b)/2 = -a
x = (-a -b +a -b)/2 = -b
There are two solutions: x ∈ {-a, -b}.