(a) Rationalise the denominator and leave the answer in simplest surd form: \( \begin{array}{llll}\text { (1) } \frac{3}{\sqrt{6}} & \text { (2) } \frac{2}{5 \sqrt{2}} & \text { (3) } \frac{3}{\sqrt[3]{9}} & \text { (4) } \frac{2}{4+\sqrt{2}}\end{array} \)

Let's rationalize each denominator and simplify the expressions to their simplest surd form. ### (1) \(\frac{3}{\sqrt{6}}\) **Rationalization:** Multiply the numerator and the denominator by \(\sqrt{6}\): \[ \frac{3}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2} \] **Simplified Form:** \[ \frac{\sqrt{6}}{2} \] --- ### (2) \(\frac{2}{5\sqrt{2}}\) **Rationalization:** Multiply the numerator and the denominator by \(\sqrt{2}\): \[ \frac{2}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{10} = \frac{\sqrt{2}}{5} \] **Simplified Form:** \[ \frac{\sqrt{2}}{5} \] --- ### (3) \(\frac{3}{\sqrt[3]{9}}\) **Rationalization:** To rationalize a cube root in the denominator, multiply the numerator and the denominator by \(\sqrt[3]{81}\) (since \(9 = 3^2\), and \(81 = 3^4\)): \[ \frac{3}{\sqrt[3]{9}} \times \frac{\sqrt[3]{81}}{\sqrt[3]{81}} = \frac{3\sqrt[3]{81}}{9} \] Simplify \(\sqrt[3]{81}\): \[ \sqrt[3]{81} = \sqrt[3]{3^4} = 3^{1 + \frac{1}{3}} = 3\sqrt[3]{3} \] Substitute back: \[ \frac{3 \times 3\sqrt[3]{3}}{9} = \frac{9\sqrt[3]{3}}{9} = \sqrt[3]{3} \] **Simplified Form:** \[ \sqrt[3]{3} \] --- ### (4) \(\frac{2}{4 + \sqrt{2}}\) **Rationalization:** Multiply the numerator and the denominator by the conjugate of the denominator, which is \(4 - \sqrt{2}\): \[ \frac{2}{4 + \sqrt{2}} \times \frac{4 - \sqrt{2}}{4 - \sqrt{2}} = \frac{2(4 - \sqrt{2})}{(4)^2 - (\sqrt{2})^2} = \frac{8 - 2\sqrt{2}}{16 - 2} = \frac{8 - 2\sqrt{2}}{14} \] Simplify by dividing numerator and denominator by 2: \[ \frac{4 - \sqrt{2}}{7} \] **Simplified Form:** \[ \frac{4 - \sqrt{2}}{7} \] --- ### **Final Answers:** 1. \(\dfrac{\sqrt{6}}{2}\) 2. \(\dfrac{\sqrt{2}}{5}\) 3. \(\sqrt[3]{3}\) 4. \(\dfrac{4 - \sqrt{2}}{7}\)