answer:A Briefly, the solution involves converting 189 into a ternary (base-3) number. The process of conversion will show that the last digit of 189 in base-3 is 0. Therefore, the correct answer is boxed{A}.

answer:- **Partition the Square**: Divide the square of side length 2 into four squares, each square having a side length of 1. - **Apply Pigeonhole Principle**: With six points and only four smaller squares, at least one square must contain at least two of the points (by the Pigeonhole Principle). - **Calculate the Maximum Distance in the 1x1 Square**: The furthest distance between any two points within a square of side 1 is the length of the diagonal. Therefore, the maximum possible distance between two points in the same 1x1 square is calculated by the Pythagorean theorem: ( sqrt{1^2 + 1^2} = sqrt{2} ). - **Conclusion**: Thus, b is at most sqrt{2}. [ boxed{b = sqrt{2}} ]

answer:To analyze each option systematically: **Option A:** Given x > 1, we can say x - 1 > 0. We examine the expression x + frac{1}{x-1}. We rewrite it as: [ x + frac{1}{x-1} = (x-1) + frac{1}{x-1} + 1 ] Applying AM-GM inequality, since x-1 > 0, we get: [ (x-1) + frac{1}{x-1} geq 2sqrt{(x-1)cdotfrac{1}{x-1}} ] Simplifying, we find: [ 2sqrt{(x-1)cdotfrac{1}{x-1}} = 2 ] Therefore: [ x + frac{1}{x-1} geq 2 + 1 = 3 ] Equality holds when x = 2, thus proving option A is boxed{text{correct}}. **Option B:** We analyze the expression frac{x^2+5}{sqrt{x^2+4}}. Rewriting it, we get: [ frac{x^2+5}{sqrt{x^2+4}} = frac{x^2+4+1}{sqrt{x^2+4}} = sqrt{x^2+4} + frac{1}{sqrt{x^2+4}} ] Applying AM-GM inequality, we have: [ sqrt{x^2+4} + frac{1}{sqrt{x^2+4}} geq 2 ] Equality would hold if x^2 + 4 = 1, which implies x^2 = -3. This is not possible, so the minimum value being 2 is not achievable. Thus, option B is boxed{text{incorrect}}. **Option C:** Given 0 < x < 10, we examine sqrt{x(10-x)}. Using AM-GM inequality, we have: [ sqrt{x(10-x)} leq frac{x + (10-x)}{2} = 5 ] Equality holds when x = 10 - x, which simplifies to x = 5. Therefore, option C is boxed{text{correct}}. **Option D:** Given the solution set of ax^2 + bx + c > 0 is {x | 2 < x < 3}, we deduce a < 0. From the conditions, we have: [ begin{cases} -frac{b}{a} = 2 + 3 frac{c}{a} = 2 times 3 end{cases} ] This leads to b = -5a and c = 6a. Thus: [ a - b + c = a + 5a + 6a = 12a ] Since a < 0, we conclude 12a < 0, making option D boxed{text{incorrect}}. **Final Answer:** The correct options are boxed{text{A and C}}.

answer:(1) Let the common ratio of the geometric sequence {a_n} be q. From the given information, we have a_1+a_2+a_3=frac{4}{q}+4+4q=14. Solving for q, we get q=frac{1}{2} or q=2. Since {a_n} is an increasing geometric sequence and a_2=4, we have q=2. Therefore, the general term formula of the sequence is a_n=2^n. (2) Proof: Suppose three terms a_m, a_n, and a_p in the sequence {a_n} with m < n < p form an arithmetic sequence. Then, by the definition of an arithmetic sequence, we have 2a_n=a_m+a_p. Substituting the general term formula, we get 2^n=2^{m-1}+2^{p-1}. Dividing both sides by 2^{m-1}, we get 2^{n-m+1}=1+2^{p-m} ldots (*). Since m < n < p, we have p-m geq 2 and n-m geq 1. Therefore, the left side of (*) is an even number, while the right side is an odd number, which is a contradiction. Hence, any three terms in the sequence {a_n} cannot form an arithmetic sequence. Therefore, the final answers are: (1) boxed{a_n=2^n} (2) boxed{text{Any three terms in the sequence } {a_n} text{ cannot form an arithmetic sequence}}