answer:(1) We have f(x) = 2sin(2x + frac{pi}{3}) - sqrt{3}. Since -1 leq sin(2x + frac{pi}{3}) leq 1, we have -2 - sqrt{3} leq 2sin(2x + frac{pi}{3}) - sqrt{3} leq 2 - sqrt{3}, and T = frac{2pi}{2} = pi, which means the range of f(x) is left[-2 - sqrt{3}, 2 - sqrt{3}right] and its smallest positive period is pi. boxed{7 text{ points}} (2) When x in left[0, frac{pi}{6}right], 2x + frac{pi}{3} in left[frac{pi}{3}, frac{2pi}{3}right], thus sin(2x + frac{pi}{3}) in left[frac{sqrt{3}}{2}, 1right], at this time f(x) + sqrt{3} = 2sin(2x + frac{pi}{3}) in left[sqrt{3}, 2right]. From m[f(x) + sqrt{3}] + 2 = 0, we know m neq 0, thus f(x) + sqrt{3} = -frac{2}{m}, which means sqrt{3} leq -frac{2}{m} leq 2, solving this gives -frac{2sqrt{3}}{3} leq m leq -1. Therefore, the range of the real number m is boxed{left[-frac{2sqrt{3}}{3}, -1right]}.

answer:Given: There are (2k+1) planets, each hosting an astronomer observing the nearest planet (all planet pair distances are distinct). **Objective:** Prove that there is at least one planet that is not observed by any astronomer. We'll elaborate on the reference solution. 1. **Consider Planets with Minimum Distance:** Let (A) and (B) be the two planets such that the distance between them, (d(A, B)), is the minimum among all possible distances between any two planets. - The astronomer on planet (A) observes planet (B) because (B) is the nearest planet to (A). - Similarly, the astronomer on planet (B) observes planet (A) because (A) is the nearest planet to (B). 2. **Case Analysis:** We examine two possible situations: **Case 1:** At least one other astronomer observes either planet (A) or (B). In this case, let's say an astronomer on another planet (C) observes planet (A). Then we have: [ 2k+1 text{ (total planets)} - (1 text{ observing } B text{ from } A + 1 text{ observing } A text{ from } B + 1 text{ other observer}) = 2k - 2 ] Planets remaining for (2k-2) observers. The remaining number of observers is (2k - 2). Since each of the remaining (2k) planets is observed by one or fewer astronomers, it follows by the Pigeonhole Principle that at least one of the remaining (2k) planets is not being observed by any astronomer. **Case 2:** No astronomer apart from those on (A) and (B) observes either (A) or (B). In this situation, we can remove the pair (A) and (B) and consider the new system of (2k-1) planets. This new system still retains the crucial property (each planet is watched by an astronomer observing the closest planet). We may now apply the same logic recursively. We either find another pair with the same conditions with fewer planets, or eventually reduce to the base case: - If we ever encounter the situation in Case 1 directly in the recursion, we find an unobserved planet as detailed above in Case 1. - If we reduce the system enough (i.e., repeated application of Case 2 leading to reduction), we end up with (1) planet (when (k=0)), which inherently isn't being observed by any other. By recursively removing pairs of planets (as seen in Case 2) until either an unobserved planet situation arises (Case 1) or reducing to the single planet case, we conclude that in all scenarios: boxed{text{There exists at least one planet that is not observed by any astronomer.}}

answer:The total number of ways to choose any five slips from fifty slips is ( binom{50}{5} ). For ( p ), the number of ways to choose five slips all bearing the same number: - There are 10 numbers, and for each number, there are ( binom{5}{5} = 1 ) way to choose five slips. - Hence, there are ( 10 times 1 = 10 ) favorable outcomes. For ( q ), the number of ways to choose three slips with a number ( a ) and two slips with a number ( b neq a ): - Choose two distinct numbers ( a ) and ( b ) from the 10 numbers: ( binom{10}{2} = 45 ) ways. - For number ( a ), choose three slips: ( binom{5}{3} = 10 ) ways. - For number ( b ), choose two slips: ( binom{5}{2} = 10 ) ways. - Total ways for ( q ) is ( 45 times 10 times 10 = 4500 ). Thus, the probabilities ( p ) and ( q ) are: [ p = frac{10}{binom{50}{5}}, quad q = frac{4500}{binom{50}{5}} ] Therefore, [ frac{q}{p} = frac{4500}{10} = boxed{450} ]

answer:Given the set (A={1,2,3,5}) and (B={2,3,6}), then (A∪B={1,2,3,5,6}). Therefore, the answer is: (boxed{{1,2,3,5,6}}). This problem can be directly solved by using the definition of the union of sets. This question tests the basic operation of sets and the application of the definition of union, and it is a basic question.