$$\sqrt{a}\times\sqrt{b} = \sqrt{ab} \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \qquad \sqrt{a^2 b} = a\sqrt{b}$$ $$(\sqrt{a})^2 = a \qquad (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$$

Rationalising the denominator

$$\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a} \qquad \frac{1}{a + \sqrt{b}} = \frac{a - \sqrt{b}}{a^2 - b} \qquad \frac{1}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{a} + \sqrt{b}}{a - b}$$ Multiply top and bottom by the conjugate of the denominator.