To simplify the expression \(\frac{\sqrt{2}}{\sqrt{2} + \sqrt{3} - \sqrt{5}} - \frac{1}{2}\) with a rational denominator, we first need to work on the term \(\frac{\sqrt{2}}{\sqrt{2} + \sqrt{3} - \sqrt{5}}\).
We can rationalize the denominator of \(\frac{\sqrt{2}}{\sqrt{2} + \sqrt{3} - \sqrt{5}}\) by multiplying the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{2} + \sqrt{3} + \sqrt{5}\):
\[ \frac{\sqrt{2}(\sqrt{2} + \sqrt{3} + \sqrt{5})}{(\sqrt{2} + \sqrt{3} - \sqrt{5})(\sqrt{2} + \sqrt{3} + \sqrt{5})} \]
Now, let's calculate the denominator:
\[ (\sqrt{2} + \sqrt{3})^2 - (\sqrt{5})^2 = (\sqrt{2})^2 + 2\sqrt{2}\sqrt{3} + (\sqrt{3})^2 - (\sqrt{5})^2 \]
Calculating the squares:
\[ = 2 + 3 + 2\sqrt{6} - 5 = 0 + 2\sqrt{6} = 2\sqrt{6} \]
Thus, the denominator becomes \(2\sqrt{6}\).
Now substituting back:
\[ \frac{\sqrt{2}(\sqrt{2} + \sqrt{3} + \sqrt{5})}{2\sqrt{6}} = \frac{(\sqrt{2})^2 + \sqrt{2}\sqrt{3} + \sqrt{2}\sqrt{5}}{2\sqrt{6}} = \frac{2 + \sqrt{6} + \sqrt{10}}{2\sqrt{6}} \]
Now, we have the expression:
\[ \frac{2 + \sqrt{6} + \sqrt{10}}{2\sqrt{6}} - \frac{1}{2} \]
To combine the two fractions, we can express \(-\frac{1}{2}\) with a denominator of \(2\sqrt{6}\):
\[ -\frac{1}{2} = -\frac{\sqrt{6}}{2\sqrt{6}} \]
So we can rewrite the entire expression:
\[ \frac{2 + \sqrt{6} + \sqrt{10}}{2\sqrt{6}} - \frac{\sqrt{6}}{2\sqrt{6}} = \frac{2 + \sqrt{6} + \sqrt{10} - \sqrt{6}}{2\sqrt{6}} = \frac{2 + \sqrt{10}}{2\sqrt{6}} \]
Now, simplify the expression:
\[ = \frac{2 + \sqrt{10}}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{(2 + \sqrt{10})\sqrt{6}}{12} = \frac{2\sqrt{6} + \sqrt{60}}{12} \]
Thus, the final expression with a rational denominator is:
\[ \frac{2\sqrt{6} + \sqrt{60}}{12} \]