1. Simplify the expressions: (i) Simplify $\frac{\sqrt{3} + \sqrt{12} + \sqrt{108} - \sqrt{15}}{\sqrt{6} - \sqrt{96} + \sqrt{105}}$. - First, simplify each surd: $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ $\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}$ $\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}$ - Substitute back: Numerator: $\sqrt{3} + 2\sqrt{3} + 6\sqrt{3} - \sqrt{15} = (1 + 2 + 6)\sqrt{3} - \sqrt{15} = 9\sqrt{3} - \sqrt{15}$ Denominator: $\sqrt{6} - 4\sqrt{6} + \sqrt{105} = (1 - 4)\sqrt{6} + \sqrt{105} = -3\sqrt{6} + \sqrt{105}$ - So the expression is $\frac{9\sqrt{3} - \sqrt{15}}{-3\sqrt{6} + \sqrt{105}}$. - Factor numerator and denominator if possible: Note $\sqrt{15} = \sqrt{3 \times 5}$ and $\sqrt{105} = \sqrt{3 \times 35}$. - No common factors to simplify further, so this is the simplified form. (ii) Simplify $\frac{\sqrt{8} + 5\sqrt{32} - 4\sqrt{80}}{\sqrt{18}}$. - Simplify each surd: $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$ $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$ $\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$ $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ - Substitute back: Numerator: $2\sqrt{2} + 5 \times 4\sqrt{2} - 4 \times 4\sqrt{5} = 2\sqrt{2} + 20\sqrt{2} - 16\sqrt{5} = 22\sqrt{2} - 16\sqrt{5}$ - Expression becomes $\frac{22\sqrt{2} - 16\sqrt{5}}{3\sqrt{2}}$. - Split the fraction: $\frac{22\sqrt{2}}{3\sqrt{2}} - \frac{16\sqrt{5}}{3\sqrt{2}} = \frac{22}{3} - \frac{16\sqrt{5}}{3\sqrt{2}}$. - Rationalize the second term: $\frac{16\sqrt{5}}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{16\sqrt{10}}{3 \times 2} = \frac{16\sqrt{10}}{6} = \frac{8\sqrt{10}}{3}$. - Final simplified form: $\frac{22}{3} - \frac{8\sqrt{10}}{3}$. 2. Expand and simplify: (i) Expand $(\sqrt{6} - 2)^2$. - Use formula $(a - b)^2 = a^2 - 2ab + b^2$. - $a = \sqrt{6}$, $b = 2$. - Calculate: $a^2 = (\sqrt{6})^2 = 6$ $-2ab = -2 \times \sqrt{6} \times 2 = -4\sqrt{6}$ $b^2 = 2^2 = 4$ - Sum: $6 - 4\sqrt{6} + 4 = 10 - 4\sqrt{6}$. (ii) Expand $(4 + \sqrt{7})(4 - \sqrt{7})$. - Use formula $(a + b)(a - b) = a^2 - b^2$. - $a = 4$, $b = \sqrt{7}$. - Calculate: $a^2 = 16$ $b^2 = 7$ - Result: $16 - 7 = 9$. 3. Find the positive square root of the expressions: (i) Find positive square root of $28 - 10\sqrt{3}$. - Assume $\sqrt{28 - 10\sqrt{3}} = \sqrt{a} - \sqrt{b}$ with $a > b > 0$. - Square both sides: $28 - 10\sqrt{3} = a + b - 2\sqrt{ab}$. - Equate rational and irrational parts: $a + b = 28$ $2\sqrt{ab} = 10\sqrt{3} \implies \sqrt{ab} = 5\sqrt{3} \implies ab = 25 \times 3 = 75$ - Solve system: $a + b = 28$ $ab = 75$ - $a$ and $b$ are roots of $x^2 - 28x + 75 = 0$. - Solve quadratic: $x = \frac{28 \pm \sqrt{28^2 - 4 \times 75}}{2} = \frac{28 \pm \sqrt{784 - 300}}{2} = \frac{28 \pm \sqrt{484}}{2} = \frac{28 \pm 22}{2}$ - Roots: $x = 25$ or $x = 3$. - So $a = 25$, $b = 3$. - Therefore, $\sqrt{28 - 10\sqrt{3}} = \sqrt{25} - \sqrt{3} = 5 - \sqrt{3}$. (ii) Find positive square root of $24 - 16\sqrt{2}$. - Assume $\sqrt{24 - 16\sqrt{2}} = \sqrt{a} - \sqrt{b}$ with $a > b > 0$. - Square both sides: $24 - 16\sqrt{2} = a + b - 2\sqrt{ab}$. - Equate parts: $a + b = 24$ $2\sqrt{ab} = 16\sqrt{2} \implies \sqrt{ab} = 8\sqrt{2} \implies ab = 64 \times 2 = 128$ - Solve system: $a + b = 24$ $ab = 128$ - $a$ and $b$ are roots of $x^2 - 24x + 128 = 0$. - Solve quadratic: $x = \frac{24 \pm \sqrt{24^2 - 4 \times 128}}{2} = \frac{24 \pm \sqrt{576 - 512}}{2} = \frac{24 \pm \sqrt{64}}{2} = \frac{24 \pm 8}{2}$ - Roots: $x = 16$ or $x = 8$. - So $a = 16$, $b = 8$. - Therefore, $\sqrt{24 - 16\sqrt{2}} = \sqrt{16} - \sqrt{8} = 4 - 2\sqrt{2}$. Final answers: 1.(i) $\frac{9\sqrt{3} - \sqrt{15}}{-3\sqrt{6} + \sqrt{105}}$ 1.(ii) $\frac{22}{3} - \frac{8\sqrt{10}}{3}$ 2.(i) $10 - 4\sqrt{6}$ 2.(ii) $9$ 3.(i) $5 - \sqrt{3}$ 3.(ii) $4 - 2\sqrt{2}$