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1、2011 年教师招聘小学数学专业知识模拟试题及答案一、填空题（本大题共10 个小题，每小题 2 分，共 20 分）1.用 0-9 这十个数字组成最小的十位数是_，四舍五入到万位 _，记作万_。2.在一个边长为 6 厘米的正方形中剪一个最大的圆，它的周长是厘米_。面积是_。3.+=44+=64 那么=_，=_。4.汽车站的 1 路车 20 分钟发一次车 _，5 路车 15 分钟发一次车，车站在8：00 同时发车后 _，再遇到同时发车至少再过_。5.2/7 的分子增加 6，要使分数的大小不变 _，分母应该增加 _。6.有一类数 _，每一个数都能被 11 整除_，并且各位数字之和是20_，问这类数中
2、 _，最小的数是 _。7.在 y 轴上的截距是 1，且与 x 轴平行的直线方程是 _。8.函数 y=1x+1 的间断点为 x=_。9.设函数 f(x)=x，则 f（1）=_。10.函数 f(x)=x3在闭区间-1，1上的最大值为 _。二、选择题（在每小题的四个备选答案中，选出一个符合题意的正确答案，并将其字母写在题干后的括号内。本大题共10 小题，每小题 3 分，共 30 分）1.自然数中，能被 2 整除的数都是（）。A.合数B.质数C.偶数D.奇数2.下列图形中，对称轴只有一条的是（）。A.长方形B.等边三角形C.等腰三角形D.圆3.把 5 克食盐溶于 75 克水中，盐占盐水的（）。A.1/
3、20 B.1/16 C.1/15 D.1/14 4.设三位数 2a3 加上 326，得另一个三位数5b9，若 5b9 能被 9 整除，则 a+b等于（）。A.2 B.4 C.6 D.8 5.一堆钢管，最上层有5 根，最下层有 21 根，如果自然堆码，这堆钢管最多能堆（）根。A.208 B.221 C.416 D.442 6.“棱柱的一个侧面是矩形”是“棱柱为直棱柱”的（）。A.充要条件B.充分但不必要条件C.必要但不充分条件D.既不充分又不必要条件7.有限小数的另一种表现形式是（）。A.十进分数B.分数C.真分数D.假分数8.设 f(x)=xln(2-x)+3x2-2limx1f(x)，则 l
4、imx 1f(x)等于（）。A.-2 B.0 C.1 D.2 9.如果曲线 y=f(x)在点（x,y）处的切线斜率与 x2 成正比，并且此曲线过点（1，-3）和（2，11），则此曲线方程为（）。A.y=x3-2 B.y=2x3-5 C.y=x2-2 D.y=2x2-5 10.设 A与 B为互不相容事件，则下列等式正确的是（）。A.P（AB）=1 B.P（AB）=0 C.P（AB）=P（A）P（B）D.P（AB）=P（A）+P（B）三、解答题（本大题共18 分）1.脱式计算（能简算的要简算）：（4 分）112+（3.6-115）1170.8 2.解答下列应用题（4 分）前进小学六年级参加课外活动
5、小组的人数占全年级总人数的48%，后来又有 4 人参加课外活动小组，这时参加课外活动的人数占全年级的52%，还有多少人没有参加课外活动？3.计算不定积分：x1+xdx。（4 分）4.设二元函数 z=x2ex+y，求（1）唞唜；（2）唞唝；（3）dz。（6 分）四、分析题（本大题共1 个小题，6 分）分析下题错误的原因，并提出相应预防措施。文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3
6、V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI1
7、0C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C
8、10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10
9、HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7
10、P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档
11、编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10
12、S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8“12能被 0.4 整除”成因：预防措施：五、论述题（本题满分5 分）举一例子说明小学数学概念形成过程。六、案例题（本大题共2 题，满分共 21 分）1.下面是两位老师分别执教接近整百、整千数加减法的简便计算的片断，请你从数学思想方法的
13、角度进行分析。（11 分）张老师在甲班执教：1.做凑整（十、百）游戏；2.抛出算式 323+198和 323-198，先让学生计算，再小组内部交流，班内汇报讨论，讨论的问题是：把198 看作什么数能使计算简便？加上（或减去）200后，接下去要怎么做？为什么？然后师生共同概括速算方法。练习反馈表明，学生错误率相当高。主要问题是：在“323+198=323+200-2”中，原来是加法计算，为什么要减2？在“323-198=323-200+2”中，原来是减法计算，为什么要加2？李老师执教乙班：给这类题目的速算方法找了一个合适的生活原型生活实际中收付钱款时常常发生的“付整找零”活动，以此展开教学活动。
14、1.创设情境：王阿姨到财务室领奖金，她口袋里原有 124元人民币，这个月获奖金 199 元，现在她口袋里一共有多少元？让学生来表演发奖金：先给王阿姨 2 张 100 元钞（200元），王阿姨找还 1 元。还表演：小刚到商场购物，他钱包中有217 元，买一双运动鞋要付 198 元，他给“营业员”2 张 100 元钞，“营业员”找还他2 元。2.将上面发奖金的过程提炼为一道数学应用题：王阿姨原有124 元，收入 199 元，现在共有多少元？3.把上面发奖金的过程用算式表示：124+199=124+200-1，算出结果并检验结果是否正确。4.将上面买鞋的过程加工提炼成一道数学应用题：小刚原有 217
15、 元，用了 198 元，现在还剩多少元？结合表演，列式计算并检验。5.引导对比，小结整理，概括出速算的法则。练习反馈表明，学生“知其然，也应知其所以然”。2.根据下面给出的例题，试分析其教学难点，并编写出突破难点的教学片段。（10分）例：小明有 5 本故事书，小红的故事书是小明的2 倍，小明和小红一共有多少本故事书？参考答案及解析（下一页）一、填空题1.1023456789102346 解析 越小的数字放在越靠左的数位上得到的数字越小，但零不能放在最左边的首数位上。故可得最小的十位数为1023456789，四舍五入到万位为 102346万。2.6 9平方厘米 解析 正方形中剪一个最大的圆，即为
16、该正方形的内切圆。故半径 r=126=3(厘米)，所以它的周长为2r=23=6（厘米），面积为r2=32=9（厘米 2）。3.1710解析由题干知+2=44(1)3+2=64(2)，（2）-（1）得 2=20，则=10，从而 2=44-10，解得=17。4.60 分钟解析由题干可知，本题的实质是求20 与 15 的最小公倍数。因为文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档
17、编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10
18、S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7
19、 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P
20、7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8
21、Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:C
22、G7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3
23、V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V820=225，15=35，所以它们的最小公倍数为223560。即再遇到同时发车至少再过 60 分钟。5.21 解析设分母应增加 x，则 2+67+x27，即：2x+1456，解得 x21。6.1199解析略7.y=1 解析与 x 轴平行的直线的
24、斜率为0，又在 y 轴上的截距为 1，由直线方程的斜截式可得，该直线的方程为y=1。8.-1 解析间断点即为不连续点，显然为x+1=0 时，即 x=-1。9.12解析 由 f(x)=x可知，f(x)=(x)=(x12)=12x-12=12x，故 f（1）12112。10.1 解析因为 f(x)=3x2 0，所以 f（x）在定义域 R上单调递增，所以在-1，1上也递增，故最大值在x=1 处取得，即为 f(1)1。二、选择题1.C解析 2 能被 2 整除，但它为质数，故A错误。4 能被 2 整除，但 4 是合数而不是质数，故B错误。奇数都不能被2 整除，能被 2 整除的数都为偶数。2C解析长方形有
25、两条对称轴，A排除。等边三角形有三条对称轴，B排除。圆有无数条对称轴，D排除。等腰三角形只有一条对称轴，即为底边上的中线（底边上的高或顶角平分线）。3.B解析盐水有 5+7580（克），故盐占盐水的580=116。4.C解析由 2a3+326=5b9可得，a+2=b，又 5b9 能被 9 整除，可知 b=4，则a=2，所以 a+b=2+4=6。5.B解析 如果是自然堆码，最多的情况是：每相邻的下一层比它的上一层多1 根，即构成了以5 为首项，1 为公差的等差数列，故可知21 为第 17 项，从而这堆钢管最多能堆（5+21）172221(根)。6.C解析 棱柱的一个侧面是矩形/棱柱的侧棱垂直于底
26、面，而棱柱为直棱柱堇庵的侧棱垂直于底面堇庵的侧面为矩形。故为必要但不充分条件。7.A解析 13 为分数但不是有限小数，B排除。同样 13 也是真分数，但也不是有限小数，排除C。43 是假分数，也不是有限小数，D排除。故选 A。8.C解析对 f(x)=xln（2-x）+3x2-2limx 1f(x)两边同时取极限为：limx1f（x）=0+3-2limx 1f(x)，即 3limx 1f(x)3，故 limx 1f(x)1。故选C。9.B解析由曲线过点（1，-3）排除 A、C项。由此曲线过点（2，11）排除D，故选 B。y=2x3-5 显然过点（1,-3）和（2，11），且它在（x,y）处的切线
27、斜率为 6x2，显然满足与 x2 成正比。10.B 解析由 A与 B为互不相容事件可知，AB=h，即 P(AB)=0 且 P(A+B)P（AB）=P(A)+P(B)。故选 B。三、解答题1.解：112+(3.6-115）1170.8=32+(335-115)8745=(32+12578)45=(32+2110)45=18554=92。2.解：设全年级总人数为x 人，则x48%+4x=52%文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y1
28、0 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7
29、M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8
30、文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C
31、10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10
32、V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV
33、8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8
34、I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8解得：x=100 所以没有参加课外活动的人数为100（1-52）48（人）。3.解：x1+xdx=x+1-1x+1dx=dx-1x+1dx=x-ln|x+1|+C（C为常数）。4.解：
35、（1）唞唜=2xex+y+x2ex+y=(x2+2x)ex+y；(2)唞唝=x2ex+y；(3)dz=唞唜 dx+唞唝 dy=(x2+2x)ex+ydx+x2ex+ydy。四、分析题参考答案：成因：没有理解整除的概念，对于数的整除是指如果一个整数a，除以一个自然数 b，得到一个整数商c，而且没有余数，那么叫做a 能被 b 整除或b 能整除 a。概念要求除数应为自然数，0.4 是小数。而且混淆了整除与除尽两个概念。故错误。预防措施：在讲整除概念时，应让学生清楚被除数、除数和商所要求数字满足的条件。即被除数应为整数，除数应为自然数，商应为整数。并且讲清整除与除尽的不同。五、简答题参考答案：小学数学
36、概念的形成过程主要包括（1）概念的引入；（2）概念的形成；（3）概念的运用。例如：对于“乘法分配律”的讲解：（1）概念的引入：根据已经学过的乘法交换律，只是对于乘法的定律，在计算时，很多时候会遇到乘法和加法相结合的式子，如（21+14）3。（2）概念的形成：通过让学生计算，归纳发现乘法分配律。比较大小：（32+11）5325+115(26+17)2262+172 学生通过计算后很容易发现每组中左右两个算式的结果相等，再引导学生观察分析，可以看出左边算式是两个数的和与一个数相乘，右边算式是两个加数分别与这个数相乘，再把两个积相加。虽然两个算式不同，但结果相同。然后就可以引导学生归纳总结出“乘法分
37、配律”，即（a+b）c=ac+bc。（3）概念的运用：通过运用概念达到掌握此概念的目的。计算下题：（35+12）10(25+12.5)8 学生通过运用所学的乘法分配律会很快得到结果，比先算括号里两个数的和再乘外面的数要快的多，从而学生在以后的计算中会想到运用乘法分配律，也就掌握了概念。六、案例题1.参考答案：分析建议：张教师主要用了抽象与概括的思想方法；李老师用了教学模型的方法，先从实际问题中抽象出数学模型，然后通过逻辑推理得出模型的解，最后用这一模型解决实际问题。教师可从这方面加以论述。2.参考答案：略。文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8
38、文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C
39、10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10
40、V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV
41、8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8
42、I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码
43、:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8文档编码:CG7M7P8I8Y10 HV8P7Y1C10V7 ZI10C10S4X3V8