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question:18 If |x|+x+5 y=2 and |y|-y+x=7, find the value of x+y+2009.
answer:18 Answer: (2012)Suppose y geq 0. From |y|-y+x=7 we have x=7 and from |x|+x+5 y=2 we have y=-frac{12}{5}0(rightarrow leftarrow). So x>0.Hence the two equations become -2 y+x=7 and 2 x+5 y=2 Rightarrow x=frac{13}{3}, y=-frac{4}{3}.
question:Given A=2x^{2}-x+y-3xy and B=x^{2}-2x-y+xy. Find:(1) Simplify A-2B;(2) When x+y=4 and xy=-frac{1}{5}, find the value of A-2B.
answer: Problem:Given A=2x^{2}-x+y-3xy and B=x^{2}-2x-y+xy. Find:(1) Simplify A-2B;(2) When x+y=4 and xy=-frac{1}{5}, find the value of A-2B. Solution:# Part 1: Simplify A-2BWe start with the given expressions for A and B and perform the operation A-2B:[begin{align*}A-2B &= (2x^{2}-x+y-3xy) - 2(x^{2}-2x-y+xy) &= 2x^{2}-x+y-3xy - 2x^{2} + 4x + 2y - 2xy &= -x + 4x + y + 2y - 3xy - 2xy &= 3x + 3y - 5xyend{align*}]Therefore, the simplified form of A-2B is boxed{3x + 3y - 5xy}.# Part 2: Find the value of A-2B when x+y=4 and xy=-frac{1}{5}Given the simplified form of A-2B from Part 1 and the conditions x+y=4 and xy=-frac{1}{5}, we substitute these values into the expression:[begin{align*}A-2B &= 3(x+y) - 5xy &= 3 cdot 4 - 5 cdot left(-frac{1}{5}right) &= 12 + 1 &= 13end{align*}]Thus, under the given conditions, the value of A-2B is boxed{13}.
question:From the seven numbers -5, -3, -1, 0, +2, +4, +6, select any three to form an equation that satisfies the following conditions, with each number being used only once in each equation. (1) The equation that results in the smallest sum of the three numbers is. (2) The equation that results in the largest product of the three numbers is.
answer:(1) The equation that results in the smallest sum is: (-5) + (-3) + (-1). (2) The equation that results in the largest product is: (-5) times (-3) times 6. Therefore, the answers are: (-5) + (-3) + (-1), (-5) times (-3) times 6. Thus, the final answers are boxed{(-5) + (-3) + (-1)} for the smallest sum and boxed{(-5) times (-3) times 6} for the largest product.
question:Consider a hexagon inscribed in a circle of radius R, and satisfying: A B= B C=C D=a, and D E=E F=F A=b. Calculate R in terms of a and b.
answer:It follows from these equalities that the arc C E containing D is equal to one third of the entire circle, so the angle widehat{C D E}=120. Therefore, C E=R sqrt{3} and according to Al Kashi, C E^{2}=C D^{2}+ D E^{2}+C D . D E, which ultimately gives: R^{2}=frac{a^{2}+b^{2}+a b}{3}.
question:The maximum value of the inequality y=x(1-3x) (0 < x < frac{1}{3}) is ( )A: frac{4}{243}B: frac{1}{12}C: frac{1}{64}D: frac{1}{72}
answer:AnalysisThis question examines the application of basic inequalities, paying attention to the judgment of being greater than zero.SolutionSince 0 0, 3x > 0.Thus, y=x(1-3x)= frac{3x(1-3x)}{3} leqslant frac{1}{3}left( frac{3x +1-3x}{2}right)^{2}= frac{1}{12}.Therefore, the answer is boxed{text{B}}.
question:If point P is on the terminal side of an angle of frac{2pi}{3}, and the coordinates of P are (-1, y), then y equals ( )A: sqrt{3}B: -sqrt{3}C: -frac{sqrt{3}}{3}D: frac{sqrt{3}}{3}
answer:Since point P is on the terminal side of an angle of frac{2pi}{3}, and the coordinates of P are (-1, y), we have tan frac{2pi}{3} = -tan frac{pi}{3} = -sqrt{3} = frac{y}{-1}. Therefore, y = sqrt{3}, hence the correct choice is: boxed{A}. This problem is solved by using the definition of trigonometric functions for any angle, to find the value of y. The main focus of this problem is on the definition of trigonometric functions for any angle, making it a basic question.