How do you simplify (sqrt3 -sqrt6)/(sqrt3 +sqrt 6)?

$- 3 + 2 \sqrt{2}$

Explanation:

We can rationalize the denominator by remembering that:

$\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{a} - \sqrt{b}\right) = a - b$

In this case:

$\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} + \sqrt{6}}$

$\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} + \sqrt{6}} \left(1\right)$

$\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} + \sqrt{6}} \left(\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} - \sqrt{6}}\right)$

${\left(\sqrt{3} - \sqrt{6}\right)}^{2} / \left(3 - 6\right)$

$\frac{3 + 6 - 2 \sqrt{18}}{- 3}$

$\frac{9 - 6 \sqrt{2}}{- 3} = - 3 + 2 \sqrt{2}$