answer:Since the universal set U=mathbb{R}, we have A={x| frac {1}{x}≥1}={x|0<x≤1}, Therefore, complement_{U}A={x|x≤0 text{ or } x>1}. Hence, the answer is: boxed{{x|x≤0 text{ or } x>1}}. Given the universal set U=mathbb{R} and the set A={x|0<x≤1}, we derive the answer from the definition of the complement of a set. This problem tests the properties and operations of sets. When solving, it is important to carefully read the problem, meticulously solve it, and pay attention to the proper application of the properties of inequalities.

answer:1. **Probability of all balls in the same bin**: - Probability that any two balls land in the same bin k is (2^{-k})(2^{-k}) = 2^{-2k}. - Extending this to three balls for the same bin: [ sum_{k=1}^{infty} (2^{-k})^3 = sum_{k=1}^{infty} 2^{-3k} = frac{1}{7} ] The formula for the sum of an infinite geometric series is applied here with r = 2^{-3}. 2. **Probability of all balls in different bins**: - Subtract the probability of any same bin from 1: [ 1 - frac{1}{7} = frac{6}{7} ] 3. **Calculating the required probability**: - Since there are three balls and each has an equal chance of landing in any bin independently, the probability of any specific ball (like the red one) being higher than the other two is frac{1}{3} of the probability of all being in different bins: [ frac{1}{3} times frac{6}{7} = frac{2}{7} ] Thus, the probability that the red ball is tossed into a higher-numbered bin than both the green and the blue balls is frac{2{7}}. The final answer is boxed{D}

answer:Given the equation: [ a_{1}^{2} + (2a_{2})^{2} + (3a_{3})^{2} + (4a_{4})^{2} + (5a_{5})^{2} + (6a_{6})^{2} + (7a_{7})^{2} + (8a_{8})^{2} = 204 ] We need to find the sum (a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} + a_{8}). 1. **Lower Bound Analysis**: Observe that the expression on the left-hand side is the sum of squares of different multiples of (a_i). To find a minimum possible value, assume each (a_i = 1): [ a_{1}^{2} + (2 cdot 1)^{2} + (3 cdot 1)^{2} + (4 cdot 1)^{2} + (5 cdot 1)^{2} + (6 cdot 1)^{2} + (7 cdot 1)^{2} + (8 cdot 1)^{2} ] 2. **Calculate Each Term**: Each term becomes: [ 1^2,quad 2^2,quad 3^2,quad 4^2,quad 5^2,quad 6^2,quad 7^2,quad 8^2 ] Therefore: [ 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 ] 3. **Summing Up**: Adding these values together, we get: [ 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = 204 ] This sum matches exactly the value on the right-hand side of the given equation, confirming that: [ a_{1} = a_{2} = a_{3} = a_{4} = a_{5} = a_{6} = a_{7} = a_{8} = 1 ] 4. **Summing (a_i)'s**: Hence, the sum of all (a_i)'s is: [ a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} + a_{8} = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8 ] # Conclusion: [ boxed{8} ]

answer:1. We need to prove that there exists a sequence of positive integers (a_1 < a_2 < ldots < a_{100}) such that for (2 le k le 100), the least common multiple (LCM) of (a_{k-1}) and (a_k) is greater than the LCM of (a_k) and (a_{k+1}). 2. Let's start with the base case where (n = 2). For any two positive integers (a_1) and (a_2) with (a_1 < a_2), there is nothing to prove since there is no (a_{k+1}) to compare with. 3. Now, assume that for some (n geq 2), there exists a sequence (x_1, x_2, ldots, x_n) such that (text{lcm}(x_{k-1}, x_k) > text{lcm}(x_k, x_{k+1})) for (2 le k le n). 4. We need to show that we can extend this sequence to (n+1) terms while maintaining the property. Consider the sequence (dx_1, dx_2, ldots, dx_n) for some positive integer (d). This sequence also satisfies the property because: [ text{lcm}(dx_{k-1}, dx_k) = d cdot text{lcm}(x_{k-1}, x_k) ] and [ text{lcm}(dx_k, dx_{k+1}) = d cdot text{lcm}(x_k, x_{k+1}) ] Since (d) is a common factor, the inequality (text{lcm}(x_{k-1}, x_k) > text{lcm}(x_k, x_{k+1})) is preserved. 5. Now, choose (d) large enough such that: [ d cdot text{lcm}(x_1, x_2) < d x_1 (d x_1 - 1) ] This ensures that the new sequence (dx_1 - 1, dx_1, dx_2, ldots, dx_n) maintains the property. Specifically, we need: [ text{lcm}(dx_1 - 1, dx_1) > text{lcm}(dx_1, dx_2) ] Since (text{lcm}(dx_1 - 1, dx_1) = dx_1 (dx_1 - 1)) and (text{lcm}(dx_1, dx_2) = d cdot text{lcm}(x_1, x_2)), we have: [ dx_1 (dx_1 - 1) > d cdot text{lcm}(x_1, x_2) ] which is true by our choice of (d). 6. By induction, we can construct a sequence of any length (n geq 2) that satisfies the given property. (blacksquare)