answer:(1) Let event A denote that the equation x² + 2ax + b² = 0 has real roots. When a ≥ 0 and b ≥ 0, the necessary and sufficient condition for the equation to have real roots is a ≥ b.There are a total of 36 basic events, which are:(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4),(2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2),(4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)where the first number represents the value of a and the second number represents the value of b. Event A contains 21 of these basic events. Therefore, the probability of event A occurring is P(A) = frac{7}{12}.(2) The region formed by all possible outcomes of the experiment is {(a, b) | 0 ≤ a ≤ 5, 2 ≤ b ≤ 4}. The region forming event A is {(a, b) | 0 ≤ a ≤ 5, 2 ≤ b ≤ 4, a ≥ b}. The probability is the ratio of the areas of these two regions.Hence, the required probability is P(A) = boxed{frac{2}{5}}.

answer:2. f(x)=x^{2} is a convex function on (-infty,+infty), so frac{1}{n}left[fleft(a_{1}right)+cdots+fleft(a_{n}right)right] geqslant fleft(frac{a_{1}+cdots+a_{n}}{n}right). Therefore, a_{1}^{2}+a_{2}^{2}+cdots+a_{n}^{2} geqslant n cdotleft(frac{k}{n}right)^{2}=frac{k^{2}}{n}.

answer:1. 283.From 12^{3}=1728<2011<13^{3}=2197, we know the minimum value of x is2011-1728=283 text {. }

answer:6. For the product a b c to be a multiple of 12 , the integers a, b and c must include between them at least 1 factor of 3 and 2 factors of 2 .Among the integers in the list, there are five categories with respect to divisibility by 2 and 3 :- (A): 3 are not divisible by 2 or 3 (these are 1,5,37 )- (B): 3 are divisible by 2 but not by 4 or 3 (these are 22,46,50 )- (C): 1 is divisible by 4 but not by 3 (this is 8 )- (D): 4 are divisible by 3 but not by 2 (these are 21,27,33,39 )- (E): 1 is divisible by 2 and by 3 (this is 30 )We work through two possibilities: either 30 is chosen or 30 is not chosen. We do not need to worry about whether 30 is a, b or c, since the three chosen integers can be arranged in increasing order after they are chosen.Case 1: 30 is chosenIn this case, at least one of the remaining integers is even, since 30 includes only 1 factor of 2 .If both of the remaining integers are even, then 2 of the 4 integers from (B) and (C) are chosen. There are 6 ways to do this: 22 and 46 ; 22 and 50 ; 22 and 8 ; 46 and 50 ; 46 and 8 ; 50 and 8 . Therefore, there are 6 ways to choose in this sub-case.If only one of the remaining integers is even, this even integer can come from either (B) or (C). There are 4 such integers.The third integer chosen must be odd, and so is from (A) or (D). There is a total of 7 integers in these.Thus, in this sub-case there are 4 times 7=28 ways to choose since each of the 4 even integers can be paired with each of the 7 odd integers.Case 2: 30 is not chosenIn this case, one or two of the three integers chosen must be from (D), since a b c must include at least 1 factor of 3 . (If all three were from (D), there would be no factors of 2.)If two of the integers are from (D), there are 6 ways of choosing these 2 integers (21 and 27;21 and 33 ; 21 and 39 ; 27 and 33 ; 27 and 39 ; 33 and 39 ).The third integer chosen must then be a multiple of 4 in order for a b c to have at least 2 factors of 2 . Thus, the third integer is 8 .Thus, in this sub-case there are 6 ways to choose.If one of the integers is from (D), there are 4 ways of choosing that integer.The remaining two integers chosen then need to include a multiple of 4 and another even integer, or a multiple of 4 and an odd integer not divisible by 3 , or two even integers that arenot multiples of 4. (Can you see why these are all of the possible cases?)If these integers are a multiple of 4 and another even integer, the multiple of 4 must be 8 and the even integer comes from the 3 integers in (B).Thus, there are 4 times 3=12 ways to choose in this sub-case, since each of the 4 integers from (D) can be paired with each of the 3 integers from (B).If the two integers are a multiple of 4 and an odd integer not divisible by 3 , the multiple of 4 must be 8 and the odd integer comes from the 3 integers in (A).Thus, there are 4 times 3=12 ways to choose in this sub-case.If the two integers are even and not multiples of 4 , we choose 2 of the 3 integers from (B) and there are 3 ways to do this, as we saw earlier.Thus, there are 4 times 3=12 ways to choose in this sub-case.In total, there are 6+28+6+12+12+12=76 ways to choose the three integers.ANSWER: 76

answer:(1) When x = 40, the car travels from city A to city B for frac{100}{40} = 2.5 hours. It consumesleft( frac{1}{128000} times 40^3 - frac{3}{80} times 40 + 8 right) times 2.5 = 17.5 text{ liters.}Answer: When the car travels at a constant speed of 40 kilometers per hour, it consumes 17.5 liters of fuel from city A to city B.(2) When the speed is x kilometers per hour, the car travels from city A to city B for frac{100}{x} hours. Let the fuel consumption be h(x) liters. According to the problem, we haveh(x) = left( frac{1}{128000}x^3 - frac{3}{80}x + 8 right) cdot frac{100}{x} = frac{1}{1280}x^2 + frac{800}{x} - frac{15}{4} (0 0, h(x) is an increasing function.Therefore, when x = 80, h(x) reaches its minimum value h(80) = 11.25.Since h(x) has only one extreme value on (0, 120], this is the minimum value.Answer: When the car travels at a constant speed of 80 kilometers per hour, it consumes the least fuel from city A to city B, with a minimum fuel consumption of boxed{11.25} liters.

answer:Analysis: This problem is related to the calculation of interest on deposits and taxation. It is a taxation problem, and the key is to use the relationship: taxable amount × tax rate = personal income tax paid, to find the amount of personal income tax that should be paid.According to the problem, we first calculate the amount exceeding 3,500 yuan, and then solve the problem by understanding the meaning of multiplication.Let's denote the taxable income exceeding 3,500 yuan as x. The tax rate for this portion is 3%, or 0.03. The total income before tax can be represented as 3500 + x, and the income after tax is this total income minus the tax paid on x, which is 3500 + x - 0.03x. According to the problem, this amount after tax is 4,761 yuan. Therefore, we have:[3500 + x - 0.03x = 4761]Solving this equation for x gives us:[x - 0.03x = 4761 - 3500][0.97x = 1261][x = frac{1261}{0.97}][x = 1300]Therefore, the taxable income exceeding 3,500 yuan is 1,300 yuan, and the tax paid on this amount at a rate of 3% is:[1300 times 0.03 = 39]Thus, Mr. Zhang paid boxed{39} yuan in personal income tax last month.