2022年数与形教学设计与反思 .pdf
翻转教学中的探究性学习数与形教学设计五通桥区牛华镇二码头小学杨文容教学内容:人教版六年级上册107 页例 1 教材与学情分析教材分析:数形结合思想是通过发现规律解决问题帮助学生建立数形结合的数学思想,把抽象的数学语言与直观的图表结合起来思索,使抽象的思维与形象思维结合,通过“以形助教”或“以数解形”,使得复杂问题简单化,抽象问题具体化,从而起到优化解题途径的目的。教学目标:1.经历观察、操作、归纳等活动,能借助“形”来直观感受数与数之间的关系,体会有时“形”与“数”能互相解释。2.通过数形结合分析思考问题,从而感悟数形结合的思想,经历解决问题的过程,体验迁移类推的学习方法。3.感受数学在解决实际问题中的作用,培养热爱数学、乐学数学的情感。教学重点:引导学生探索,在数与形之间建立联系,发现其中的变化规律,正确运用规律进行计算。教学难点:经历探索规律及验证规律的过程。教学准备:课件,不同颜色的小正方形。学具准备:不同颜色的小正方形,作业纸。一、谈话导入,出示课题教师:同学们,你们想和老师一起比赛吗?生:想。最近老师发现,我有一项非常神奇的本领。什么本领呢?我发现只要从1开始的连续奇数相加,比如,1+3,1+3+5,像这样的算式,我都算得特别快。你们信吗?教师:不信也没关系,我们现场来比一比。师出示题目:1=1+3=1+3+5=1+3+5+7+9 1+3+5+7+9+11=师生比赛,看谁算得快。教师:我已经算完了。(学生刚拿起笔,不服输)师:有同学说老师是先做过的,那谁来出一道这样的题目,我来算。学生出题,师马上说出得数。教师:你们想知道老师为什么这么快吗?其实这样的算式还可以用一种更快、更奇妙的方法来算出结果,那就是借助图形来解决!今天这节课我们就来研究数与形(板书)。【设计意图:从谈话导入,通过设置悬念,激发学生学习兴趣,从而顺理成章地引出课题。】二、动手实践,以形解数1.构造直观。师:(指着刚才的算式)数在这儿,形在哪儿呢?形在这!(出示一个红色正方形)老师先摆出一个正方形代表1。教师:我先根据算式中的数摆出了一个正方形。现在问题交给你们了,小组合作摆出 1+3。师:(观察学生)老师发现大家摆的有好几种形状,我选其中的两个小组上黑板来摆一摆。预设 1:我把红色小正方形摆在前面,后面依次摆了 3 个黄色小正方形,摆成一个大的长方形。(学生一边说一边操作)预设 2:我在红色小正方形的外围摆上3 个黄色的小正方形,摆成了一个更大的正方形。(学生一边说一边操作)师:我们看两组同学摆出的图形。(指着其中的大正方形)老师也是这样摆的,它到底蕴藏着什么秘密呢?大家观察一下,它有什么特点?预设:它也是一个正方形。师:真不错!再观察,这个图形中每行有几个小正方形?一共有几行?文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8生:每行有 2 个小正方形,一共有2 行。师:所以小正方形的个数我们可以怎么算?预设:22,也就是 22。(随着学生的回答师板书:1+3=22=22)师:更有挑战性的问题来了,再加一个数,1+3+5,又怎么摆?学生小组合作,动手摆图形。师巡视观察。师:我请一个小组的代表上来摆一摆。预设:摆出一个更大的正方形。师:现在我来采访一下,你摆的这个图形和我的算式1+3+5 有什么关系呢?预设:我摆出了1 个红色小正方形代表1,在它的外围摆3 个黄色小正方形代表 3,接着再在外围摆 5 个小正方形代表5,这时我摆出了一个更大的正方形。师:因此小正方形的总数可以怎么算?预设:在这个大正方形中,每行有3 个小正方形,一共有3 行,所以小正方形的个数可以 33,也就是 32。(板书:1+3+5=33=32)师:非常不错,你的思路很清晰,表达能力也很强!师:同学们猜猜,如果现在我还想拼成一个更大的正方形,还要在外围加几个小正方形?预设:7 个。师:我们来验证一下。(请学生来摆)师:哪个同学能上来把这个图形对应的算式写一写?学生板演:1+3+5+7=4 4=42。师:这个正方形的每条边上有几个小方块?有几行?(课件演示不同的颜色),这些不同的颜色分别表示几?为什么1+3+5+7可以用 4 的平方来算?预设:因为这几个不同颜色的方块拼在一起就组成了大大的正方形,这个正方形可以拼成 4 行,每行有 4 个,可以用 4 的平方来计算。【设计意图:充分让学生动手实践,感受如何将数和形结合,体会数和形之间的紧密联系,同时让学生感受到“形”可以展示“数”的特点,通过“形”使解决“数”的问题变得更加容易。】2.发现规律。文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8师:同学们,如果继续这样拼下去,再加上一个奇数9,现在有几个奇数?又怎么算呢?预设:有 5 个奇数,小正方形每条边上的个数也变成5 个,而且有这样的5行,所以它的和可以用5 的平方来算。师:那,继续这样拼下去,再增加一个奇数11 呢?它的总和可以用6 的平方来算。再来一行呢?可以用7 的平方,以此类推,如果有 n 个这样的连续奇数,那就可以用什么来计算呢?师板书算式:1+3+5+7+9+,=n 2n 个师:我们写了这么多算式,仔细观察,能发现这些算式有什么特点吗?先同桌相互说说,再全班交流。预设:我发现左边有几个连续的奇数相加,右边就是几的平方。而且左边的奇数必须都是从“1”开始的。师:你们的观察力太让我惊喜了!大家都发现了左边的加数是从“1”开始的连续奇数?那为什么加数要满足这个条件呢?预设 1:因为我们在第 1 个红色小正方形的基础上,每次加上半圈图形,就能拼成一个更大的正方形,如果不是从“1”开始,也就是说没有第一个红色小正方形的话,就会缺一个角,我们就不能拼成一个更大的正方形了。而组成每一个“半圈”的小正方形都比前一个“半圈”形的小正方形多2,这样就形成了连续的奇数。3.解释规律。师:(课件演示)同学们看,我们首先摆出一个小正方形,如果我想再拼出一个更大的正方形,再加1 个小正方形够吗?生:不够。师:还要再增加几个?预设:2 个。师:这时我们加到了几?预设:3。师:想再拼成一个更大的正方形,3 是不够的,还要再增加几个小正方形?预设:2 个。师:此时是 1+3+5,再往下要再加7,才能拼成一个更大的正方形,以此类文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8推加到了 9,就拼成了一个每行每列的小正方形个数为5 的大正方形,小正方形的个数和也就是 5 的平方。师:这时加数的个数是几?预设:5。师:也就是说,从“1”开始的几个连续奇数相加,就能摆成每行每列个数是几的大正方形,和就是几的平方。掌握了这个规律,你们能说出几个类似的算式吗?预设 1:1+3+5+7+9+11=6 2预设 2:1+3+5+7+9+11+13+15=82。师:说得非常好,我们还能一直这样加下去,都可以很快地算出来。【设计意图:学生经过充分的操作和交流活动已经积累了一些感性认识,这时组织学生对学习活动进行交流,进而用自己的语言表达所发现的规律。在学生的相互倾听和交流中提升对连续奇数的和等于奇数个数的平方这一规律的认识,清晰规律】三、巩固应用1.填空。(1)1+3+5+7+9+11+13=()。(2)()=92。(3)1+3+5+7+,+99=()。预设:答案应该是50 的平方,因为从 1 到 99一共有 50个奇数。2.教材“做一做”的第一小题。1+3+5+7+5+3+1=();1+3+5+7+9+11+13+11+9+7+5+3+1=()。学生先独立思考,再小组讨论,相互协作完成这两道题。师巡视指导。汇报交流做法。师:真棒,一题多解,锻炼了我们的思维。四、解决问题师:同学们的表现太棒了!原来,数字的问题借助图形来解决可以很简单。那形的变化背后是不是也隐藏着数的规律呢?文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8师课件出示:小明的爷爷过几天就80 大寿啦,他准备摆长桌宴邀请客人们来村里吃饭。这种桌子,四面坐人只可以坐4 个人,如果两个桌子拼到一起就可以坐 6 个人,3 张桌子拼到一起可以坐8 个人,这样的 100 张桌子拼到一起可以坐多少个人呢?师:你听懂了吗?你们有没有好的方法?预设:画图。师:如果画出来的话,(课件演示)1 张桌子可以坐 4 个人,2 张桌子可以坐6 个人,3 张桌子可以坐8 个人,100 张桌子可以坐多少人?小组讨论交流,把答案写在作业纸上。(小组讨论交流。)师:小组同学来说一说你们的做法。师:请你借助图形来说一说你为什么这样做?预设 1:我们组算的是一共有202 人。100 张桌子拼在一起,这一边也就是它的长边一共有 200 个人,再加上两头有2 个人,一共有 202 人。预设 2:它每张桌子的两边坐2 个人,他有 100张桌子,再加上边上就是它的宽分别坐 2 人,200+2=202人。师:算式就是 1002,100 2 的意思就是每张桌子两边都坐2 个人,100 张桌子就做 200 个人,旁边还有 2 人,所以需要在加上2,等于 202人。师:还有其他做法吗?预设 3:我们小组是这样想的,把第一张桌子去掉的话,每增加一张桌子就增加 2 个人,4+299=202人。师:算式是这样的,4 先不看,多了 99 张桌子,每多一张桌子就多2 个人,所以多了 299 这些人,然后再加上4 人等于 202 人师:我想问一下,这是一个图形的问题,为什么你们不去画图,却用数来算呢?预设 2:老师我感觉画图太麻烦了,因为它有100张桌子。师:对,画图太麻烦了,这时候需要借助数的力量,把形的计算问题用数来做会更加的快速、简便而且准确。那我们把这样的过程叫做化形为数,然后以数来解形。(板书)文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8【设计意图:此练习与例题相对应,让学生把几何直观与抽象的数学语言结合起来,体会数形结合不仅可以“以形助数”还可以“以数解形”。使复杂问题简单化,抽象问题具体化,从而达到优化解题途径的目的。】五、温故知新师:同学们,回顾这两个例子,在第一个例子当中,数的问题可以借助图形来思考,而第二个例子当中,形的知识可以借助数来计算,数和形各有优点,它们一一对应而且可以互相转化,互为补充,这就意味着要求我们在解决问题的时候要把数和形结合起来,这在数学上是一种重要的思想,就叫“数形结合思想”。想一想,你们曾经学过哪些知识也是数与形相联系的?预设 1:我们以前学过线段图,就是用画图的方法来解决问题的。预设 2:我们学习分数的时候,是用一个圆代表单位“1”,然后把这个圆平均分成几份,其中的一份就是它的几分之一。师:同学们说得非常好,(课件出示图片)像线段图、统计图等,都是我们曾经学过的数与形相结合的例子,而在以后的学习中,数形结合思想将和你们朝夕相伴。板书设计 数与形1+3+5=321+3+5+7+9=421+3+5+7+9+11=6 21+3+5+7+9+,=n2n 个文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1F7G6M1 HD4H2V6R10V7 ZI6C5T9A1I8文档编码:CS10V1