Appearance
question:13.016. The numerators of three fractions are proportional to the numbers 1, 2, and 3, while the reciprocals of the corresponding denominators are proportional to the numbers 1, 1 / 3, and 0.2. Find these fractions, given that their arithmetic mean is 136 / 315.
answer:Solution.Let's compile the following table:| Fraction | Numerator | Denominator || :--- | :---: | :---: || First | x | 1 / y || Second | 2 x | 3 / y || Third | 3 x | 1 / 0.2 y |First fraction -x y; second -frac{2}{3} x y; third -0.6 x y. The arithmetic mean of the fractions frac{x y+frac{2}{3} x y+0.6 x y}{3}=frac{136}{315}, from which x y=frac{136}{238}=frac{4}{7}. Therefore, the first fraction is -frac{4}{7}; the second is -frac{8}{21}; the third is -frac{12}{35}.Answer: frac{4}{7} ; frac{8}{21} ; frac{12}{35}.
question:10. Factorize 2 x^{3}+7 x^{2}-9, one of the factors is (2 x+b), then b=
answer:3
question:Example 5 Factorization:(x-1)(x-2)(x-3)(x-4)-120 text {. }(1991, Jilin Province Participates in the National Junior High School Mathematics League Preliminary Selection Competition)
answer:Solution: Let f(x)=(x-1)(x-2)(x-3)(x-4)-120 text {. }Notice thatbegin{array}{l}f(-1)=(-2) times(-3) times(-4) times(-5)-120=0, f(6)=5 times 4 times 3 times 2-120=0,end{array}Thus, we know that f(x) has factors x+1 and x-6.Expanding and performing division, we getbegin{array}{l}f(x)=x^{4}-10 x^{3}+35 x^{2}-50 x-96 =(x+1)(x-6)left(x^{2}-5 x+16right) .end{array}
question:15. Express 2006 as the sum of 5 positive integers x_{1}, x_{2}, x_{3}, x_{4}, x_{5}. Let S=sum_{1 leqslant i<j leqslant 5} x_{i} x_{j}, find:(1) For what values of x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, does S achieve its maximum value;
answer:15. (1) First, the set of such S values is bounded, so there must exist a maximum and a minimum. If x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=2006, and S=sum_{1 leqslant i < j leqslant 5} x_{i} x_{j},Rewrite S asS=x_{1} x_{2}+left(x_{1}+x_{2}right)left(x_{3}+x_{4}+x_{5}right)+x_{3} x_{4}+x_{3} x_{5}+x_{4} x_{5}At the same time, we haveS^{prime}=x^{prime}{ }_{1} x^{prime}{ }_{2}+left(x_{1}^{prime}+x^{prime}{ }_{2}right)left(x_{3}+x_{4}+x_{5}right)+x_{3} x_{4}+x_{3} x_{5}+x_{4} x_{5}Thus, S^{prime}-S=x^{prime}{ }_{1} x^{prime}{ }_{2}-x_{1} x_{2}>0, which contradicts the assumption that S achieves its maximum value at x_{1}, x_{2}, x_{3}, x_{4}, x_{5}. Therefore, it must be that left|x_{i}-x_{j}right| leqslant 1(1 leqslant i, j leqslant 5). Hence, the maximum value is achieved when x_{1}=402, x_{2}=x_{3}=x_{4}= x_{5}=401.
question:On a wire, 9 swallows are sitting at regular intervals from each other. The distance between the first and the last swallow is 720 mathrm{~cm}.- What is the distance between neighboring swallows?- How many swallows would be sitting on the wire if three more swallows sat between each pair of the already sitting swallows?(M. Krejčová)
answer:Among nine swallows, there are only 8 gaps.1. The distance between neighboring swallows is 720: 8=90(mathrm{~cm}).2. If 3 new swallows perched in each of the eight gaps, a total of 8 cdot 3=24 swallows would perch. There would then be 9+24=33 swallows on the wire.Evaluation. 1 point for the observation about the number of gaps; 2 points for the first part; 3 points for the second part of the problem.
question:Let the function f(x)=x^{2}-5x+6, where xin[0,5]. If a real number x_{0} is randomly selected from the interval [0,5], then the probability that the selected real number x_{0} satisfies f(x_{0})leqslant 0 is ( ).A: 0.2B: 0.3C: 0.4D: 0.5
answer:AnalysisThis question mainly examines the geometric probability model and the solution method for quadratic inequalities. In the examination of probability problems, probability is just a carrier, and other content occupies a larger proportion. This is a basic question.SolutionGiven the problem, this is a geometric probability model, where the value of probability corresponds to the ratio of lengths.From f(x_{0})leqslant 0, we get x^{2}-5x+6leqslant 0,Solving this yields: 2leqslant xleqslant 3,therefore P= dfrac{1}{5} =0.2,Therefore, the correct choice is boxed{A}.