1. **Stating the problem:** Simplify and evaluate the expressions for $X$, $Y$, $A$, $B$, and $C$, and solve the geometric problem involving triangle $ABC$ with given segment ratios and parallel lines. 2. **Expression for $X$:** $$X = -\sqrt{3} - \left[-2 - \sqrt{3}(\sqrt{2} - 1)\right] - (2 - \sqrt{2})(1 - \sqrt{3})$$ - First, simplify inside the brackets: $$-2 - \sqrt{3}(\sqrt{2} - 1) = -2 - \sqrt{3}\sqrt{2} + \sqrt{3} = -2 - \sqrt{6} + \sqrt{3}$$ - So, $$X = -\sqrt{3} - \left(-2 - \sqrt{6} + \sqrt{3}\right) - (2 - \sqrt{2})(1 - \sqrt{3})$$ - Distribute the minus sign: $$X = -\sqrt{3} + 2 + \sqrt{6} - \sqrt{3} - (2 - \sqrt{2})(1 - \sqrt{3})$$ - Simplify the first terms: $$-\sqrt{3} - \sqrt{3} = -2\sqrt{3}$$ - Now expand the last product: $$(2 - \sqrt{2})(1 - \sqrt{3}) = 2 \cdot 1 - 2 \sqrt{3} - \sqrt{2} \cdot 1 + \sqrt{2} \sqrt{3} = 2 - 2\sqrt{3} - \sqrt{2} + \sqrt{6}$$ - Substitute back: $$X = 2 + \sqrt{6} - 2\sqrt{3} - (2 - 2\sqrt{3} - \sqrt{2} + \sqrt{6})$$ - Distribute the minus: $$X = 2 + \sqrt{6} - 2\sqrt{3} - 2 + 2\sqrt{3} + \sqrt{2} - \sqrt{6}$$ - Combine like terms: $$2 - 2 = 0, \sqrt{6} - \sqrt{6} = 0, -2\sqrt{3} + 2\sqrt{3} = 0$$ - Remaining term: $$X = \sqrt{2}$$ 3. **Expression for $Y$:** $$Y = 1 + \frac{9}{1 + \frac{1}{2} - \frac{\frac{1}{3}}{\frac{5}{4}} + \frac{5}{6}}$$ - Simplify the complex fraction: $$\frac{\frac{1}{3}}{\frac{5}{4}} = \frac{1}{3} \times \frac{4}{5} = \frac{4}{15}$$ - Substitute back: $$Y = 1 + \frac{9}{1 + \frac{1}{2} - \frac{4}{15} + \frac{5}{6}}$$ - Find common denominator for the denominator terms (denominator of denominator): 30 - Convert each term: $$1 = \frac{30}{30}, \frac{1}{2} = \frac{15}{30}, \frac{4}{15} = \frac{8}{30}, \frac{5}{6} = \frac{25}{30}$$ - Sum inside denominator: $$\frac{30}{30} + \frac{15}{30} - \frac{8}{30} + \frac{25}{30} = \frac{30 + 15 - 8 + 25}{30} = \frac{62}{30} = \frac{31}{15}$$ - So, $$Y = 1 + \frac{9}{\frac{31}{15}} = 1 + 9 \times \frac{15}{31} = 1 + \frac{135}{31} = \frac{31}{31} + \frac{135}{31} = \frac{166}{31}$$ 4. **Expression for $A$ (first):** $$A = \frac{2\sqrt{2} + 2}{3 - \sqrt{3}} + \frac{1}{\sqrt{6} - 3}$$ - Rationalize denominators: - For first term multiply numerator and denominator by conjugate $3 + \sqrt{3}$: $$\frac{2\sqrt{2} + 2}{3 - \sqrt{3}} \times \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{(2\sqrt{2} + 2)(3 + \sqrt{3})}{9 - 3} = \frac{(2\sqrt{2} + 2)(3 + \sqrt{3})}{6}$$ - Expand numerator: $$2\sqrt{2} \times 3 = 6\sqrt{2}, \quad 2\sqrt{2} \times \sqrt{3} = 2\sqrt{6}, \quad 2 \times 3 = 6, \quad 2 \times \sqrt{3} = 2\sqrt{3}$$ - Sum numerator: $$6\sqrt{2} + 2\sqrt{6} + 6 + 2\sqrt{3}$$ - So first term: $$\frac{6 + 6\sqrt{2} + 2\sqrt{6} + 2\sqrt{3}}{6} = 1 + \sqrt{2} + \frac{\sqrt{6}}{3} + \frac{\sqrt{3}}{3}$$ - For second term: $$\frac{1}{\sqrt{6} - 3} \times \frac{\sqrt{6} + 3}{\sqrt{6} + 3} = \frac{\sqrt{6} + 3}{6 - 9} = \frac{\sqrt{6} + 3}{-3} = -\frac{\sqrt{6}}{3} - 1$$ - Add both terms: $$\left(1 + \sqrt{2} + \frac{\sqrt{6}}{3} + \frac{\sqrt{3}}{3}\right) + \left(-\frac{\sqrt{6}}{3} - 1\right) = \sqrt{2} + \frac{\sqrt{3}}{3}$$ 5. **Expression for $B$ (first):** $$B = \frac{\sqrt{288} - \sqrt{243} + \sqrt{108} - \sqrt{18}}{15}$$ - Simplify radicals: $$\sqrt{288} = \sqrt{144 \times 2} = 12\sqrt{2}$$ $$\sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3}$$ $$\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}$$ $$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$$ - Substitute: $$B = \frac{12\sqrt{2} - 9\sqrt{3} + 6\sqrt{3} - 3\sqrt{2}}{15} = \frac{(12\sqrt{2} - 3\sqrt{2}) + (-9\sqrt{3} + 6\sqrt{3})}{15} = \frac{9\sqrt{2} - 3\sqrt{3}}{15}$$ - Simplify fraction: $$B = \frac{3(3\sqrt{2} - \sqrt{3})}{15} = \frac{3\sqrt{2} - \sqrt{3}}{5}$$ 6. **Expression for $C$:** $$C = \frac{\sqrt{43} - 3\sqrt{2} - 5\sqrt{5}}{5}$$ - This is already simplified. 7. **Expression for $A$ (second):** $$A = \frac{3\sqrt{2} + \sqrt{3}}{3}$$ - Already simplified. 8. **Expression for $B$ (second):** $$B = \frac{3\sqrt{2} - \sqrt{3}}{5}$$ - Already simplified. 9. **Geometric problem:** - Given triangle $ABC$ with $AC = 6$, $BC = 5$. - Point $M$ on $AB$ such that $AM = \frac{2}{5} AB$. - Line through $M$ parallel to $BC$ intersects $AC$ at $N$. - Given $AN = 2.4$, find $NC$. - Since $MN \parallel BC$, triangles $AMN$ and $ABC$ are similar. - Ratio of similarity: $$\frac{AM}{AB} = \frac{AN}{AC} = \frac{2}{5}$$ - Given $AN = 2.4$, check ratio: $$\frac{AN}{AC} = \frac{2.4}{6} = 0.4 = \frac{2}{5}$$ - So, $$NC = AC - AN = 6 - 2.4 = 3.6$$ 10. **Additional points:** - Point $E$ on $BC$ with $BE = 2$. - Point $F$ on $AB$ with $AF = \frac{3}{2} AB$ (which is longer than $AB$, so $F$ lies beyond $B$). - Point $K$ on $AC$ with $AK = 1.6$. - Given ratios: $$\frac{AK}{AN} = \frac{AB}{AF}$$ $$\frac{AF}{AM} = \frac{AC}{AK}$$ - Using $AM = \frac{2}{5} AB$, $AF = \frac{3}{2} AB$, $AN = 2.4$, $AK = 1.6$, $AC = 6$. - Check first ratio: $$\frac{1.6}{2.4} = \frac{AB}{\frac{3}{2} AB} = \frac{AB}{1.5 AB} = \frac{1}{1.5} = \frac{2}{3}$$ - Calculate left side: $$\frac{1.6}{2.4} = \frac{2}{3}$$ - Ratio holds. - Check second ratio: $$\frac{\frac{3}{2} AB}{\frac{2}{5} AB} = \frac{6}{1.6}$$ - Simplify left side: $$\frac{1.5 AB}{0.4 AB} = \frac{1.5}{0.4} = 3.75$$ - Right side: $$\frac{6}{1.6} = 3.75$$ - Ratio holds. 11. **Parallel lines:** - Lines $CF$ and $KM$ are parallel, consistent with the given ratios and points. **Final answers:** $$X = \sqrt{2}$$ $$Y = \frac{166}{31}$$ $$A = \sqrt{2} + \frac{\sqrt{3}}{3}$$ $$B = \frac{3\sqrt{2} - \sqrt{3}}{5}$$ $$C = \frac{\sqrt{43} - 3\sqrt{2} - 5\sqrt{5}}{5}$$ $$NC = 3.6$$