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Life> 0,1 / 1 Similarly,[link widoczny dla zalogowanych], b + made ≥ 5, /> 0 · second type was multiplied by (n + zoned 6 +) ≥ 25 √. Established by the title n6 ≤ () a throw, was ≥ 4, so the (n +1) (b +) ≥ 25 √ raise one,[link widoczny dla zalogowanych], if and only if n-b ÷ time to take a \Note that the distinctive license law, is based on deep understanding of the conditions dug. Example 7 known function Factory () (∈ R) satisfies the condition: for any real number, there is (a) ≤ (1-2) [plant (1) a f (x2)], (1) and lf (x ) a f (x) l ≤ l a l,[link widoczny dla zalogowanych], (2) where is a constant greater than 0. Real number n. , N, b satisfy f (a.) 10 and b one of a plant (population). (3) (I) proved ≤ 1, and there is no b. ≠ mouth. Such that f (b.) 一 0; (Ⅱ) proved that (6 a ao) ≤ (1 a) (Port a o); (Ⅲ) prove that If (b)] ≤ (1 a) If (a)]. (2004 Jiangsu College Entrance Examination) to analyze the conditions of many candidates face of numerous, do nothing. In fact, no skill demonstrate the problem, as long as the conditions can be fully used. For ease of description, the conditions as (1), (2), (3). There: (I) either take 1,2 ∈ R,[link widoczny dla zalogowanych], 1 ≠ 2,[link widoczny dla zalogowanych], the condition (1), (2) get (1 a 2) ≤ (1-2) [plant (1) a f (x)] ≤ l a 1.1f (x) a f (x) l, then (1 A 2) ≤ l1 a 2l one (1 A 2), so ≤ 1. At this point, for one is also true. With anti-b. ≠ mouth. , So f (b.) 10. By (1), and 0 <(mouth. A bo) ≤ (mouth. A bo) [Factory (mouth.) A f (b.)]. By (3), knowing f (a.) A f (b.) 10, so 0 <(mouth. A bo) ≤ 0, contradiction. That does not exist b. ≠ mouth. Such that f (b.) 10. (Ⅱ) eliminate the use of condition b, then eliminate f (a) can be. * 34 * High School Mathematics Monthly No. 7, 2007 by (2) know that If (b) a f (a)] ≤ (6 ~ a). By (3), (1), known 2f (a) If (b) a / '(mouth)] = a If (b) a plant (population)] ≤ ~ 2 (6 ~ port) z, so If ( b)] ≤ [Factory (port)] a (6 a). Then (3) knowledge (6 a) a [plant (port)], so If (b)] ≤ (1 a) [Factory (port)]. As seen, the proof is repeated to the conditions, and gives a special value. Based on our years of teaching experience, introduced to help students get rid of problem solving \When the right to initiate widespread expectations in. Meanwhile, we hope to teach important to note that in the teaching of basic concepts and methods, to reinforce the foundation, which is the key to improving teaching quality. Fill in the blank on a college entrance examination to explore Hsu, Chi (one in 413,400 Taojiang) 2006 Volume 16, college entrance examination, Fujian, entitled: Figure 1, the link △. B. c. The midpoint of each side to get a new △ Bc, and links △ Bc midpoint of each side are △ AB: c:, so unlimited continues, get a series of triangles: △. B. c. , △ Bc, △. BC, .... This series of triangles tends to a point. Known. (O, O), B. (3, O), C. (2,2), the coordinates of the point is. This will give the title of a clever solution and the three promotion. Lemma 1 Lemma Let G is the focus of AABC, O is any point, then + + a 3. Card slightly. Solving 2 graph 1NA, B, C respectively, △. BoC. The midpoint of each side, so o One (core +), a 1 (O --- ~ o + ox; o ~, One (+),( them. Is an arbitrary point) the sum obtained by the above three type Two + + A + + Odo. Let △ BC's center of gravity for the G (∈ N), obtained by the above formula and Lemma 3o - V, === 3, so G. and G coincide. Similarly, G and G coincide, ..., G ~ and G coincide. So G. and G coincide. and easy to get C9G 。(,),- by the conditions of 3-3C9 So the focus of △ ABC is G 。(,). for each △ Bc contains the point G., and set up by the title of this series of triangles tend to know a point. co so the point will be the point of G., that is, (Lu) .3 Promotion Promotion 1 Figure 2, the respective sides of the AABC's access point, B, c, so that A - e; a, a, = (normal number). and then the edge in △ Bc were taken of each point, B, C, to make a puree, a, a Zhi ~, How V / ~ / eleven thousand one hundred and eleven iv a ~ ~ a ~ one thousand one hundred eleven one hundred and eleven ~ ~ ~ ~ a ~ eleven ~ 1111111 ~ a
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