# How do you rationalize (sqrt3-sqrt2)/ ( sqrt3+sqrt2)?

You can multiply and divide by: $\sqrt{3} - \sqrt{2}$ to get:
$\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} =$
$= {\left(\sqrt{3} - \sqrt{2}\right)}^{2} / \left(3 - 2\right) = 3 - 2 \sqrt{6} + 2 = 5 - 2 \sqrt{6}$