1. **Problem statement:** Simplify the expression $7\sqrt{12} + 8\sqrt{147} - 15\sqrt{27}$. 2. **Formula and rules:** To simplify expressions with square roots, factor the radicand into prime factors and extract perfect squares using $\sqrt{a^2 b} = a\sqrt{b}$. 3. **Simplify each term:** - $7\sqrt{12} = 7\sqrt{4 \times 3} = 7 \times 2 \sqrt{3} = 14\sqrt{3}$ - $8\sqrt{147} = 8\sqrt{49 \times 3} = 8 \times 7 \sqrt{3} = 56\sqrt{3}$ - $15\sqrt{27} = 15\sqrt{9 \times 3} = 15 \times 3 \sqrt{3} = 45\sqrt{3}$ 4. **Combine like terms:** $$14\sqrt{3} + 56\sqrt{3} - 45\sqrt{3} = (14 + 56 - 45)\sqrt{3} = 25\sqrt{3}$$ 5. **Final answer:** $25\sqrt{3}$. 1. **Problem statement:** Rationalise the denominators of the following expressions. 2. **Formula and rules:** To rationalise denominators with square roots, multiply numerator and denominator by the conjugate or the root itself to eliminate roots in the denominator. 3. **Part a:** Given $\frac{2}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{2\sqrt{6}}{6}$ (already rationalised). 4. **Part b:** $\frac{\sqrt{3}}{2\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{3} \times \sqrt{6}}{2 \times 6} = \frac{\sqrt{18}}{12} = \frac{3\sqrt{2}}{12} = \frac{\cancel{3}\sqrt{2}}{\cancel{12}4} = \frac{\sqrt{2}}{4}$ 5. **Part c:** $\frac{2}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{2(\sqrt{5} + 2)}{(\sqrt{5})^2 - 2^2} = \frac{2\sqrt{5} + 4}{5 - 4} = 2\sqrt{5} + 4$ 6. **Part d:** $\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = \frac{\cancel{2}(2 - \sqrt{3})}{\cancel{2}1} = 2 - \sqrt{3}$ 7. **Final answers:** - a) $\frac{2\sqrt{6}}{6}$ - b) $\frac{\sqrt{2}}{4}$ - c) $2\sqrt{5} + 4$ - d) $2 - \sqrt{3}$