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提高初中的生数学解题能力.doc

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数学教学的一个重要任务是提高学生的解题能力。实际上,老师们在教解题上花费了大量的时间,但是效果总是不那么如意。因此,怎样培养学生的解题能力是一个不可忽视的话题。老师们都知道要培养学生的解题能力,然而许多教师老是在激发学生的解题兴趣上做文章,这无可厚非。不过,不论有无兴趣,作为学生,总归要养成一定的解题能力。因此,笔者认为,以下三点,应该才是培养学生解题能力的重要手段。 An important task of mathematics teaching is to improve the students' ability of problem solving. In fact, teachers spend a lot of time on teaching problem solving, but the effect is always not so satisfied. Therefore, how to develop the students' ability of problem solving is a topic that can not be ignored. Teachers know that to develop the students' ability of problem solving, however, many teachers always in arousing students' interest in the problem solving, this understandable. However, if any interest, as a student, it will form a certain ability to solve problems. Therefore, the author thinks that, the following three points, should is the important means of training students' problem solving ability.   一、展示教师自身的解题过程 A, show the teacher's own problem solving process   学生形成解题能力是从模仿解题开始的。模仿到一定程度后,方有自己的解题感觉。因此,教师的示范作用就显得非常重要。然而,在学生眼里,教师解题总是有如神来之笔,一下笔,题目的方法就出来了。学生对此除了羡慕外,剩下的感觉就是突兀。 Students form begins with imitate problem solving problem solving ability. Imitate to a certain extent, feel their problem solving. Therefore, the teacher exemplary role is very important. However, in the eyes of students, teachers' problem solving always like a pen, a pen, topic methods came out. Students are in addition to envy, the rest of the feeling is abrupt. 正如波利亚所说的,教师要告诉学生“是怎样想到这个解法的?”“是什么促使你这样想、这样做的?”也就是说,教师要向学生展示自己的解题过程,这样学生才有模仿的可能。 As BoLiYa puts it, the teacher wants to tell students \"is how to think of the solution?\" \"What made you think like this, do that?\" That is to say, teachers should show students own the problem solving process, so that students there is a possibility of imitation.   19世纪末,德国有位天才的数学家叫闵可夫斯基,他曾是爱因斯坦的老师。一天,闵可夫斯基刚走进教室,一名学生就递给他一张纸条,上面写着:“如果把地图上有共同边界的国家涂成不同颜色,那么只需要四种颜色就足够了,您能解释其中的道理吗?” The end of the 19th century, Germany a genius mathematician called minkowski, he was Einstein's teacher. One day, minkowski had just walked into the classroom, a student just handed him a piece of paper, it read: \"if there is a common boundary of the country on the map painted different colors, so only need 4 kinds of color is enough, can you explain the truth?\"   闵可夫斯基微微一笑,对学生们说:“这个问题叫四色问题,是一个著名的数学难题。其实,它之所以一直没有得到解决,仅仅是由于没有第一流的数学家来解决它。” Minkowski smiled and said to the students: \"the problem is called four color problem, is a famous mathematical problems. Actually, the reason it has not been solve, simply because there is no first-class mathematicians to solve it.\"   为证明纸条上写的不是一道大餐,而是小菜一碟,闵可夫斯基决定“当堂掌勺”,问题就会变成定理…… To prove note top write of is not a big meal, but a piece of cake, minkowski decided \"classroom cooking\", the problem will become a theorem...   下课铃响了,可“菜”还是生的。一连好几天,他都挂了黑板。后来有一天,闵可夫斯基走进教室时,忽然雷声大作,他借此自嘲道:“哎,上帝在责备我狂妄自大呢,我解决不了这个问题。” When the bell rang, \"dishes\" or raw. He hung up the blackboard for several days. And then one day, when the minkowski came into the classroom, suddenly write you, he adds: \"ah, god hubris in the blame me, I won't solve the problem.\"   尽管闵可夫斯基没有解答出问题,然而,他的解题方式和思维,在学生头脑中留存了一辈子。 Although minkowski didn't solve the problem, however, his way of problem solving and thinking, in the student mind retained all his life.   对于现今的教师来说,尽管无法做到闵可夫斯基那样,但我们可以通过一连串的问句进行质疑启发,向学生展示解题的思维活动过程。  例1 若函数y=f(x)(x∈R)的图像关于直线x=a与x=b(a≠b)都对称,求证:y=f(x)是周期函数且2(b-a)是它的一个周期。   教师可设置一组问句启发引导:   ①本题的条件是什么?条件中的关键词语是什么? ②本题的结论是什么?问题解决的落脚点在哪里?落脚点有何表达形式? ③条件可进行怎样的转化?转化后的条件与结论有怎样的距离?缩短距离的方法是什么?(探索解题思路)   至此,本题的解题思路基本畅通,但思维活动没有结束,否则,真可谓“进宝山而空返”了。待学生整理完解题过程后,教师可继续设问:  ④本题使用了哪些知识点及思想方法? ⑤由本题可产生哪些猜想?怎样证明猜想?  同学们兴趣盎然,积极讨论,最后师生共同归纳出相并列的两命题:   (1)函数y=f(x)(x∈R)的图像关于点(a,c)和(b,c)(b≠a)成中心对称,则y=f(x)为周期函数,且2(b-a)是它的一个周期;   (2)若函数y=f(x)(x∈R)的图像关于直线x=b对称及关于点(a,c)(b≠a)成中心对称,则f(x)是周期函数且4(a-b)是它的一个周期。   通过对本题的探索研究,学生不但对函数的对称性、周期性有了深刻的理解,而且通过解决问题的思维过程的展示,有效地训练了思维能力,提高了学生发现问题、分析问题并解决问题的能力。   二、帮助学生反思解题的结论   一类数学问题,其解法往往是有规律可循的。要想减轻学生负担,让学生从题海中解脱出来,必须教会学生从解题中及时归纳总结基本的解题规律,以达到举一反三、触类旁通的目的。教学中,教师应经常启发、引导学生在解题后反思这类数学问题的基本解题规律,对方法进行归类。这对提高解题能力尤其重要。 A mathematical problem, its solution is often have rules to follow. To ease the burden on students, let students to escape from the crowd, must teach students timely sum up the basic law of problem solving from the problem solving, to achieve the purpose of the lines, one instance. Teaching, the teacher should often inspire, guide the student to reflection after problem solving this kind of law of basic problem solving mathematical problems, the method of classification. This is especially important for improve problem solving ability.   例2 已知f(x)=kx3-x2+■kx-16在R上单调递增,则k的取值范围是( )。 Case 2 known f (x) = kx3 - x2 + s kx - 16 monotone increasing on R, k value range is ().   A. k>1 B. k≥1 C. |k|>1 D. |k|≥1 B. a. k > 1 k or 1 C. | | k > 1 D. | | k 1 or higher   错解: f ′(x)=3kx2-2x+■k,依题意,对一切x Wrong solutions: 'f (x) = 3 kx2-2 x + s k, according to the question, for all x   ∈R,f ′(x)>0,选A。 '∈ R, f (x) > 0, choose A.   正解:依题意,对一切x∈R,f ′(x)≥0,应选B。   我们知道,对一切x∈R,f ′(x)>0是f(x)在R上单调递增的充分不必要条件。该题中,f(x)在R上单调递增的充要条件是:对一切x∈R,f ′(x)≥0。值得提醒的是,并不是对一切函数f(x),f(x)在R上单调递增的充要条件都是:对一切x∈R,f ′(x)≥0。f ′(x)=0所对应的情形应特别加以考虑。这些解题的小规律,通过解题后的反思,学生可以透过问题表层,充分挖掘其内在因素,掌握问题元素间的深层关系,优化自身知识结构。   三、帮助学生积累解题技巧   学生的解题能力自然包含会解难题的能力。一些难度中上的题目,一般需要一些处理过程才可应用书本的有关知识解决。也就是说,一些问题有其特殊的解题技巧,学生若不理解并熟记一些解题技巧,即使概念定理、公式学得再熟,也难以用得上。这些技巧需要教师帮助学生归纳提炼。 The problem solving ability of students will naturally contains ability of solving problems. Some difficulty of subject, generally need some process can only be used books related knowledge to solve. That is to say, some problems have their special problem-solving skills, if students don't understand and memorize some problem solving skills, even if the concept learned theorem, formula and cooked, can also be difficult to use. Requires teachers to help students inductive refining these skills.   例3 数列{an}满足a1=1,a2=2,an+2=(1+cos2■)an Example 3 sequence {an} satisfies a1 = 1, a2 = 2, the an a + 2 = 1 + cos2 (s) of the an   +sin2■,n=1,2,3,…。 Sin2 + s, n = 1, 2, 3,... .   (1)求a3,a4,并求数列{an}的通项公式。 (1) for a3, a4, and the sequence {an} general term formula.   (2)设bn=■,Sn=b1+b2+…+bn,证明:当n≥6时,Sn-2<■。 (2) establish bn = s, Sn = b1 + b2 +... + bn, prove that when n 6 or higher, Sn - 2 < s.   解:因为a1=1,a2=2,所以a3=(1+cos2■)a1+sin2■ Solution: because a1 = 1, a2 = 2, so the a3 = (1 + cos2 s) a1 + sin2 s   =a1+1=2,a4=(1+cos2?仔)a2+sin2?仔=2a2=4。 = a1 + 1 = 2, a4 = (1 + cos2? Wang) a2 + sin2? Wang = 2 a2 = 4.   一般地,当n = 2 k - 1(k∈N*)时, a 2k+1 =1+cos2■a2 k - 1+sin2■?仔=a2 k - 1+1,即a2 k + 1- Generally, when n = 2 (k ∈ n *) k - 1, a 2 k + 1 = 1 + cos2 s a2 + sin2 k - 1 s? Wang = a2 + 1 k - 1, namely, a2 - k + 1   a2k-1=1。 A2k - 1 = 1.   所以数列{a2k-1}是首项为1,公差为1的等差数列,于是有a2k-1=k。 So the sequence {} a2k - 1 is the first item is 1, the tolerance is 1 of the arithmetic progression, so have the a2k - 1 = k.   当n = 2 k(k∈N *)时,a2k+2 =( 1 + cos2■) a2k + When n = 2 k ∈ n (k *), a2k + 2 = (1 + cos2 s) a2k +   sin2■=2a2k。 Sin2 a2k s = 2.   所以数列{a2k}是首项为2,公比为2的等比数列,于是有a2k=2k。 So the sequence {a2k} is the first of 2, geometric sequence, with a common ratio 2 and there a2k = 2 k.   故数列{an}的通项公式为 So the sequence {an} for general term formula   an=■,n=2k-1,(k∈N*)■,n=2k。(k∈N*) An = s, n = k - 1, 2 (k ∈ n *) s, n = 2 k. (k ∈ N *)   对于这道题,由解题过程可知,如果不知道n=2k For this problem, by the problem solving process, if you don't know n = 2 k   -1(k∈N*)时,a2k+1=1+cos2■a2k-1+sin2■?仔=a2k-1+1,即a2k+1-a2k-1=1这个处理技巧,那么将难以 - 1 (k ∈ N *), a2k + 1 = 1 + cos2 s + sin2 a2k - 1 s? Wang = a2k - 1 + 1, namely the a2k + 1 - a2k - 1 = 1 this processing technique, then will be hard to   解答。 Answer it.   实际上,许多数学难题都是由于有好的技巧才能顺利解决。比如,当年欧几里得证明素数有无限多个,采用的就是反面假设的处理技巧。■是无理数的证明,更是堪称经典。这些技巧,学生有必要掌握,才能在解题时做到得心应手,进而真正提高解题能力。 In fact, many mathematical problems are due to have a good skill to solve properly. For example, when the Euclidean proved to have an unlimited number of primes, use is opposite assumption processing skills. S is proof of irrational Numbers, is classic. These skills, it is necessary for students to take, to do well in the problem solving, and truly improve problem solving ability.
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