To rationalize the denominator of the given expression \(\frac{\sqrt{1 + a} - \sqrt{a}}{1 + a + \sqrt{a}}\), we can multiply both the numerator and denominator by the conjugate of the denominator, which is \(1 + a - \sqrt{a}\). This gives us:

\[\frac{\sqrt{1 + a} - \sqrt{a}}{1 + a + \sqrt{a}} \times \frac{1 + a - \sqrt{a}}{1 + a - \sqrt{a}} = \frac{\left(\sqrt{1 + a} - \sqrt{a}\right)\left(1 + a - \sqrt{a}\right)}{\left(1 + a + \sqrt{a}\right)\left(1 + a - \sqrt{a}\right)}\]

Expanding the numerator and denominator, we get:

\[\frac{\left(\sqrt{1 + a} - \sqrt{a}\right)\left(1 + a - \sqrt{a}\right)}{\left(1 + a\right)^2 - \left(\sqrt{a}\right)^2} = \frac{\left(\sqrt{1 + a} - \sqrt{a}\right)\left(1 + a - \sqrt{a}\right)}{(1 + 2a) - a} = \frac{\left(\sqrt{1 + a} - \sqrt{a}\right)\left(1 + a - \sqrt{a}\right)}{(1 + a)}\]

Simplifying further, we get:

\[\frac{\left(\sqrt{1 + a} - \sqrt{a}\right)\left(1 + a - \sqrt{a}\right)}{(1 + a)} = 1 + 2a - 2\sqrt{(1+a)a}\]

So, the correct answer is 1 + 2a - 2\(\sqrt{(1+a)a}\)