1. **Problem Statement:** Rationalise the denominators of the following expressions and simplify where possible: a) $\frac{1}{1+\sqrt{2}}$ b) $\frac{1}{1-\sqrt{2}}$ c) $\frac{4}{3+\sqrt{5}}$ d) $\frac{10}{2+\sqrt{5}}$ e) $\frac{1}{1+\sqrt{3}}$ f) $\frac{1}{3-\sqrt{7}}$ 2. **Formula and Rule:** To rationalise denominators with surds, multiply numerator and denominator by the conjugate of the denominator. The conjugate of $a+\sqrt{b}$ is $a-\sqrt{b}$ and vice versa. This uses the difference of squares formula: $$ (a+\sqrt{b})(a-\sqrt{b}) = a^2 - (\sqrt{b})^2 = a^2 - b $$ 3. **Step-by-step solutions:** **a)** $$ \frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{1-\sqrt{2}}{(1)^2 - (\sqrt{2})^2} = \frac{1-\sqrt{2}}{1-2} = \frac{1-\sqrt{2}}{-1} = -1 + \sqrt{2} $$ **b)** $$ \frac{1}{1-\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}} = \frac{1+\sqrt{2}}{1 - 2} = \frac{1+\sqrt{2}}{-1} = -1 - \sqrt{2} $$ **c)** $$ \frac{4}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}} = \frac{4(3-\sqrt{5})}{9 - 5} = \frac{12 - 4\sqrt{5}}{4} = \cancel{\frac{12}{4}} - \cancel{\frac{4\sqrt{5}}{4}} = 3 - \sqrt{5} $$ **d)** $$ \frac{10}{2+\sqrt{5}} \times \frac{2-\sqrt{5}}{2-\sqrt{5}} = \frac{10(2-\sqrt{5})}{4 - 5} = \frac{20 - 10\sqrt{5}}{-1} = -20 + 10\sqrt{5} $$ **e)** $$ \frac{1}{1+\sqrt{3}} \times \frac{1-\sqrt{3}}{1-\sqrt{3}} = \frac{1-\sqrt{3}}{1 - 3} = \frac{1-\sqrt{3}}{-2} = -\frac{1}{2} + \frac{\sqrt{3}}{2} $$ **f)** $$ \frac{1}{3-\sqrt{7}} \times \frac{3+\sqrt{7}}{3+\sqrt{7}} = \frac{3+\sqrt{7}}{9 - 7} = \frac{3+\sqrt{7}}{2} = \frac{3}{2} + \frac{\sqrt{7}}{2} $$ 4. **Summary:** Rationalising denominators removes surds from the denominator by multiplying by the conjugate, simplifying the expression to a form easier to work with or interpret.