To rationalize the denominator of the expression \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}\), we can multiply both the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{2} + \sqrt{3}\). This gives us:

\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} * \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} = \frac{(\sqrt{2} + \sqrt{3})^2}{(\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3})}\)

Expanding the numerator and denominator, we get:

\(\frac{(\sqrt{2})^2 + 2\sqrt{2}\sqrt{3} + (\sqrt{3})^2}{(\sqrt{2})^2 - (\sqrt{3})^2} = \frac{2 + 2\sqrt{6} + 3}{2 - 3}\)

Simplifying further, we get:

\(\frac{5 + 2\sqrt{6}}{-1} = -5 - 2\sqrt{6}\)

So, the rationalized form of the expression is \(-5 - 2\sqrt{6}\).