\[ \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \Rightarrow \frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \] \[ \Rightarrow \frac{x - (a + b + x)}{x(a + b + x)} = \frac{a + b}{ab} \Rightarrow \frac{x - a - b - x}{x(a + b + x)} = \frac{a + b}{ab} \Rightarrow \frac{-(a + b)}{ax + bx + x^2} = \frac{a + b}{ab} \] \[ \Rightarrow \frac{-1}{x^2 + ax + bx} = \frac{1}{ab} \Rightarrow x^2 + ax + bx = -ab \Rightarrow x^2 + ax + bx + ab = 0 \Rightarrow x(x + a) + b(x + a) = 0 \Rightarrow (x + a)(x + b) = 0 \] That means either \(x + a = 0 \Rightarrow x = -a\), or \(x + b = 0 \Rightarrow x = -b\) Therefore, the solutions to the equation \[ \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \] are \(x = -a\) and \(x = -b\).