The value of $$x$$ which satisfies $$\frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{x+a+b} + \frac{1}{x}$$ is

$$\dfrac{1}{x+a} + \dfrac{1}{x+b} = \dfrac{1}{x+a+b} + \frac{1}{x}$$

$$\dfrac{2x+a+b}{(x+a)(x+b)}= \dfrac{2x+a+b}{x(x+a+b)} $$

So, for the equality to hold, we get $$2$$ cases here

CASE 1: $$2x+a+b=0$$

So, $$x=\dfrac{-(a+b)}{2}$$

CASE 2: $$2x+a+b\neq0$$

$$\dfrac{2x+a+b}{(x+a)(x+b)}= \dfrac{2x+a+b}{x(x+a+b)} $$

So, $$2x+a+b$$ can be cancelled out from both sides of the equality, as it's a non-zero number

$$\dfrac{1}{(x+a)(x+b)}= \dfrac{1}{x(x+a+b)} $$

Cross-multiplying, we get:

$$x(x+a+b)=(x+a)(x+b)$$

or, $$x^2+ax+bx=x^2+ax+bx+ab$$

$$ab=0$$

So, this means either $$a$$ or $$b$$ or both are zero.

That means the equality will be satisfied here at some particular value of $$a$$ and $$b$$ and not every value of $$a$$ and $$b$$

So, for any value of $$a$$ and $$b$$, the equality holds only if $$x=\dfrac{-(a+b)}{2}$$

Hence, Option D is correct.