answer:**Answer**: When a > b, then (a-b)^2 + (b-a)|a-b| = (a-b)^2 + (b-a)(a-b) = 0, which contradicts ab neq 0, so this case is excluded. When a < b, then (a-b)^2 + (b-a)|a-b| = ab Rightarrow 2(a-b)^2 = ab Rightarrow (2a-b)(a-2b) = 0, Therefore, 2a = b or a = 2b, - When b = 2a and a < b, then b - a = a > 0, which means b > a > 0, the possible relationships are ab > 0 or a + b > 0; - When a = 2b and a < b, then a - b = b < 0, which means a < b < 0, the possible relationship is a + b < 0; Thus, the relationship that definitely does not hold is ab < 0. Therefore, the correct choice is boxed{A}.

answer:1. **Understanding the problem**: We need to find the maximum possible value of ( n ) such that the product ((a_1 - a_2)(a_2 - a_3) cdots (a_{17} - a_1) = n^{17}), where (a_1, a_2, ldots, a_{17}) is a permutation of (1, 2, ldots, 17). 2. **Initial observation**: Since (a_1, a_2, ldots, a_{17}) is a permutation of (1, 2, ldots, 17), each (a_i) is distinct. Therefore, each difference (a_i - a_{i+1}) is non-zero. 3. **Absolute differences**: Define (left| a_i - a_{i+1} right| = x_i - y_i), where (x_i = max{a_i, a_{i+1}}) and (y_i = min{a_i, a_{i+1}}). Each (a_k) can appear at most twice among the numbers (x_1, x_2, ldots, x_{17}) and at most twice among the numbers (y_1, y_2, ldots, y_{17}). 4. **Sum of differences**: The sum of the maximum values minus the sum of the minimum values is: [ sum x_i - sum y_i leq left[ 2(17 + 16 + cdots + 10) + 9 right] - left[ 2(1 + 2 + cdots + 8) + 9 right] ] Calculating the sums: [ 17 + 16 + cdots + 10 = sum_{k=10}^{17} k = frac{8(17 + 10)}{2} = 108 ] [ 1 + 2 + cdots + 8 = sum_{k=1}^{8} k = frac{8(8 + 1)}{2} = 36 ] Therefore: [ sum x_i - sum y_i leq 2 cdot 108 + 9 - 2 cdot 36 - 9 = 216 + 9 - 72 - 9 = 144 ] 5. **Inequality of means**: Using the inequality of means, we have: [ n leq left| n right| = sqrt[17]{prod_{i=1}^{17} left| a_i - a_{i+1} right|} leq frac{sum x_i - sum y_i}{17} leq frac{144}{17} approx 8.47 ] Therefore, (n leq 8). 6. **Checking (n = 8)**: If (n = 8), then: [ (a_1 - a_2)(a_2 - a_3) cdots (a_{17} - a_1) = 8^{17} ] All 17 factors cannot be multiples of 8, otherwise all 17 (a_1, a_2, ldots, a_{17}) numbers would have the same parity, which is impossible. Therefore, at least one of the 17 factors is a multiple of 16. This is only possible if the numbers in that factor are 1 and 17. Of the remaining 16 factors, only 9 can be multiples of 8: [ (17, 9), (16, 8), (15, 7), (14, 6), (13, 5), (12, 4), (11, 3), (10, 2), (9, 1) ] Therefore: [ 51 = 3 cdot 17 = v_2 left( prod_{i=1}^{17} (a_i - a_{i+1}) right) leq 4 + 9 cdot 3 + 7 cdot 2 = 45 ] which is false. This means that (n leq 6). 7. **Example for (n = 6)**: The example for (n = 6) is: [ (17 - 8)(8 - 16)(16 - 7)(7 - 15)(15 - 6)(6 - 14)(14 - 5)(5 - 13)(13 - 4)(4 - 2)(2 - 11)(11 - 10)(10 - 1)(1 - 3)(3 - 12)(12 - 9)(9 - 17) = 6^{17} ] The final answer is ( boxed{ 6 } ).

answer:Given that the average score is 124, we can find the value of a by using the formula for the average score: frac{121 + 127 + 123 + a + 125}{5} = 124 Solving for a, we get: 496 + a = 620 a = 124 Now, to find the variance, we first calculate the sum of the squared differences from the mean: (121-124)^2 + (127-124)^2 + (123-124)^2 + (124-124)^2 + (125-124)^2 = 9 + 9 + 1 + 0 + 1 = 20 The variance is the average of these squared differences: frac{20}{5} = 4 Therefore, the variance of this set of data is boxed{4}.

answer:Solution: (I) f'(x)=x+2e, g′(x)= frac {3e^{2}}{x}. Let the common point be (x_{0},y_{0}), then begin{cases} frac {1}{2}x_{0}^{2}+2ex_{0}=3e^{2}ln x_{0}+b x_{0}+2e= frac {3e^{2}}{x_{0}}(*)end{cases}, From (*), we get x_{0}=e or x_{0}=-3e (discard this), substituting back into the first equation, we solve to get b=- frac {e^{2}}{2}. (II) Let F(x)=f(x)-g(x)+b= frac {1}{2}x^{2}+2ax-3a^{2}ln x(x > 0), F′(x)=x+2a- frac {3a^{2}}{x}=x+2a- frac {3a^{2}}{x}= frac {(x-a)(x+3a)}{x}, Since x > 0 and a > 0, F(x) is decreasing on (0,a) and increasing on (a,+infty), Therefore, F(x) reaches its minimum value at x=a, then F(a)= frac {1}{2}a^{2}+2a^{2}-3a^{2}ln ageqslant 0, Solving this, we get 0 < aleqslant e^{ frac {5}{6}}. Therefore, the range of the real number a is boxed{(0,e^{ frac {5}{6}}]}.