answer:Solution: Given log_{frac{1}{2}}(2x-1) < log_{frac{1}{2}}(-x+5), we get begin{cases} 2x-1>0 -x+5>0 2x-1>-x+5end{cases}, solving this gives 2<x<5. Therefore, the solution set of the inequality log_{frac{1}{2}}(2x-1) < log_{frac{1}{2}}(-x+5) is (2,5). Hence, the answer is: boxed{(2,5)}. This problem is solved by converting the logarithmic inequality into a system of linear inequalities, utilizing the monotonicity of the logarithmic function. This question examines the method of solving logarithmic inequalities and the monotonicity of logarithmic functions, making it a basic question.

answer:The function f(x) = sin frac{x}{4} + sin frac{x}{7} achieves its maximum value when both sine functions equal 1. That is when [ sin frac{x}{4} = 1 quad text{and} quad sin frac{x}{7} = 1. ] This implies that [ frac{x}{4} = 90^circ + 360^circ k quad text{and} quad frac{x}{7} = 90^circ + 360^circ j ] for integers k and j. Rewriting these, [ x = 360^circ k + 90^circ = 280^circ j + 70^circ. ] Equating and simplifying, [ 360k + 90 = 280j + 70. ] [ 360k = 280j - 20. ] [ 36k = 28j - 2. ] [ 18k = 14j - 1. ] This simplifies to [ 18k - 14j = -1, ] a linear Diophantine equation. Solving for j and k such that k and j are integers, we find the smallest solution that fits: [ k = 7, quad j = 9. ] Thus, [ x = 360^circ times 7 + 90^circ = boxed{2610^circ}. ]

answer:The problem asked us to count the number of distinct sequences formed by arranging 6 positive integers, a, b, c, d, e, f in such a way that there is a specific constraint provided. The constraint ensures that if any integer is greater than 1, then the corresponding integer that is one less must appear to its left. We define b_n as the number of valid sequences of length n that satisfy the given constraint. 1. **Base Case Calculation**: - For n = 1: There is only 1 sequence possible: {1}. [ b_1 = 1 ] 2. **Calculation for n = 2**: - The sequences are composed of two integers starting with 1 (a=1): 1. {a=1, b=1} results in the largest number being 1. 2. {a=1, b=2} results in the largest number being 2. Therefore, we can write the contribution as: [ b_2 = 1_{text{max 1}} + 1_{text{max 2}} = 1 + 1 = 2 ] 3. **General Recurrence Relationship**: - For constructing b_n, we consider each possible maximum integer from 1 up to n for the sequence. - The sequence construction step can be described as appending the new maximum value to all possible sequences of length n-1. - Hence, for each k (k leq n), b_k ways are extended further considering valid placements of the new maximum value using previous b_k computations. 4. **Calculation for b_3**: - To find the sequences: 1. Sequence max is 1: Only {1,1,1}. 2. Sequence max is 2: Valid sequences are: - To ensure 2 appears, 1_1 or 1_1, 2. 3. Sequence max is 3: Valid sequence formation from sequence max 2. - Leading to valid sequence 1_1,2 and substitutes of 1,2_1 patterns. Count: [ b_3 = 1_{text{max 1}} + (1 + 1 times 2)_{text{max 2}} + 1_{text{max 3}} = 1 + 3 + 1 = 5 ] 5. **Continue this method to compute b_4 and b_5 until desired b_6**. We build Table incrementally: [ begin{aligned} b_1 & = 1 b_2 & = 1 + 1 = 2 b_3 & = 1 + (1 + 1 times 2) + 1 = 1 +3 + 1 = 5 b_4 & = 1 + (1 + 3 times 2) + (3 + 2 times 3) + 1 = 1 + 7 + 9 + 1 = 18 b_5 & = 1 + (1 + 7 times 2) + (7 + 6 times 3) + (6 + 1 times 4) + 1 = 1 + 15 + 25 + 10 + 1= 52 b_6 & = 1 + (1 + 15 times 2) + (15 + 25 times 3) + (25 + 10 times 4) + (10 + 1 times 5)+ 1 & = 1 + 31 + 90 + 65 + 15 + 1 = 203 end{aligned} ] Therefore, the number of valid sequences of length 6 is: [ Conclusion: boxed{203} ]

answer:First, factor and simplify the expression (p^2 - q^2 + 2p - 2q): [ begin{align*} p^2 - q^2 + 2p - 2q &= (p^2 - q^2) + 2(p - q) &= (p+q)(p-q) + 2(p-q) &= (p-q)(p+q+2). end{align*} ] Substituting (p = 2a+1) and (q = 2b+1): [ begin{align*} (p-q)(p+q+2) &= (2a-2b)(2a+2b+4) &= 4(a-b)(a+b+2). end{align*} ] Analyze divisibility: - If (a) and (b) have the same parity (both even or both odd), (a-b) is even. - (a+b+2) is always even since (a) and (b) are integers. Thus, the expression (4(a-b)(a+b+2)) is divisible by (8) in all cases, as (4) multiplies an even number times another even number. Conclusion: The largest integer dividing all expressions of the form (p^2 - q^2 + 2p - 2q) for odd (p) and (q) is (boxed{8}).