answer:**Step 1: Understanding the Problem** This problem requires knowledge of binomial expansion and finding specific coefficients within the expansion. To solve it, we need to find the coefficients of the terms containing x^{2} and x^{3} in the expansion of (1-2x)^{4}. **Step 2: Binomial Expansion** The binomial theorem states that: (a+b)^n = sum_{k=0}^{n} {n choose k} a^{n-k} b^k Using this formula, we can expand (1-2x)^{4} as follows: (1-2x)^{4} = {4 choose 0} (1)^{4} (-2x)^{0} + {4 choose 1} (1)^{3} (-2x)^{1} + {4 choose 2} (1)^{2} (-2x)^{2} + {4 choose 3} (1)^{1} (-2x)^{3} + {4 choose 4} (1)^{0} (-2x)^{4} **Step 3: Finding the Coefficients** We are interested in the coefficients of the terms containing x^{2} and x^{3}. Thus, we focus on the following terms: begin{align*} text{Term with } x^{2}: & quad {4 choose 2} (1)^{2} (-2x)^{2} = 6 cdot 4x^{2} = 24x^{2} text{Term with } x^{3}: & quad {4 choose 3} (1)^{1} (-2x)^{3} = 4 cdot (-8x^{3}) = -32x^{3} end{align*} **Step 4: Calculating the Desired Coefficient** Now, we multiply these terms by left(2-dfrac{1}{x}right) and extract the x^{2} terms: begin{align*} 2 cdot 24x^{2} & = 48x^{2} -frac{1}{x} cdot (-32x^{3}) & = 32x^{2} end{align*} Adding these two coefficients, we get 48+32 = boxed{80}.

answer:1. Given the points and lines, let's first establish that ( C I ) is the angle bisector of (angle P C Q): [ P I = I Q ] This implies that point ( I ) lies at the midpoint of the arc ( P Q ) on the circumcircle of triangle ( A I C ). 2. Define ( T ) as the intersection point of lines ( A P ) and ( C Q ). 3. Apply the Thales Theorem to the situation. The theorem suggests that the sought parallelism ( L S parallel P Q ) is equivalent to the proportionality condition: [ frac{T L}{T P} = frac{T S}{T Q} ] 4. Using the properties of the angle bisector and triangle similarity: - For triangle ( T P C ): [ frac{T L}{T P} = frac{T C}{P C} ] - For triangle ( T Q A ): [ frac{T S}{T Q} = frac{T A}{Q A} ] 5. Because triangles ( T P C ) and ( T Q A ) are similar by the Angle-Angle (AA) criterion (their corresponding angles are equal), the following relation holds: [ frac{T C}{P C} = frac{T A}{Q A} ] 6. Now, using these established proportions: - From the similarity ratios, we substitute back into the initial proportion: [ frac{T L}{T P} = frac{T C}{P C} quad text{and} quad frac{T S}{T Q} = frac{T A}{Q A} ] - Since ( frac{T C}{P C} = frac{T A}{Q A} ), this implies: [ frac{T L}{T P} = frac{T S}{T Q} ] which directly leads to the parallelism condition: [ L S parallel P Q ] # Conclusion: [ boxed{L S parallel P Q} ]

answer:1. **Identify the problem and given conditions:** - We have a regular hexagon (ABCDEF) inscribed in a circle. - Point (P) is on the shorter arc (EF) of the circumcircle. - We need to prove that the value of (frac{AP + BP + CP + DP}{EP + FP}) is constant and find its value. 2. **Apply Ptolemy's theorem:** - Ptolemy's theorem states that for a cyclic quadrilateral (ABCD), the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals: [ AC cdot BD = AB cdot CD + AD cdot BC ] - We will use this theorem to find the lengths of (AP), (BP), (CP), and (DP) in terms of (EP) and (FP). 3. **Express (AP), (BP), (CP), and (DP) using Ptolemy's theorem:** - For (AP): [ AP = EP + sqrt{3} FP ] - For (BP): [ BP = sqrt{3} EP + 2 FP ] - For (CP): [ CP = 2 EP + sqrt{3} FP ] - For (DP): [ DP = sqrt{3} EP + FP ] 4. **Sum the expressions for (AP), (BP), (CP), and (DP):** [ AP + BP + CP + DP = (EP + sqrt{3} FP) + (sqrt{3} EP + 2 FP) + (2 EP + sqrt{3} FP) + (sqrt{3} EP + FP) ] Simplify the sum: [ AP + BP + CP + DP = EP + sqrt{3} FP + sqrt{3} EP + 2 FP + 2 EP + sqrt{3} FP + sqrt{3} EP + FP ] Combine like terms: [ AP + BP + CP + DP = (EP + 2 EP + sqrt{3} EP + sqrt{3} EP) + (sqrt{3} FP + sqrt{3} FP + 2 FP + FP) ] [ AP + BP + CP + DP = (3 + 2sqrt{3}) EP + (3 + sqrt{3}) FP ] 5. **Factor out (EP + FP):** [ AP + BP + CP + DP = (3 + sqrt{3})(EP + FP) ] 6. **Form the ratio and simplify:** [ frac{AP + BP + CP + DP}{EP + FP} = frac{(3 + sqrt{3})(EP + FP)}{EP + FP} = 3 + sqrt{3} ] Conclusion: [ boxed{3 + sqrt{3}} ]

answer:- Consider intervals for x based on lfloor x rfloor: - When 1 le x < 2, lfloor x rfloor = 1, and thus x - 1 = frac{2}{1}=2, but this does not belong to the interval as x would be 3. So, there's no solution in this range. - When 2 le x < 3, lfloor x rfloor = 2, thus x - 2 = frac{2}{2} = 1. Therefore, x = 3, which does fall within the interval. - When 3 le x < 4, lfloor x rfloor = 3, thus x - 3 = frac{2}{3}. Therefore, x = 3frac{2}{3}. - When 4 le x < 5, lfloor x rfloor = 4, thus x - 4 = frac{2}{4} = frac{1}{2}. Therefore, x = 4frac{1}{2}. - Sum of the three smallest solutions: 3 + 3frac{2}{3} + 4frac{1}{2} = 3 + 3frac{2}{3} + 4frac{1}{2} = 11frac{1}{6}. Conclusion: The sum of the three smallest positive solutions to the equation is boxed{11frac{1}{6}}.