书签分享收藏举报版权申诉 / 25

立即下载

当前位置：首页 > 教育专区 > 高考资料 > 2017年山东省泰安市中考数学试卷解析版.pdf

2017年山东省泰安市中考数学试卷解析版.pdf

上传人：H****o

文档编号：55709582

上传时间：2022-10-31

格式：PDF

页数：25

大小：762.62KB

( 4.5 )

《2017年山东省泰安市中考数学试卷解析版.pdf》由会员分享，可在线阅读，更多相关《2017年山东省泰安市中考数学试卷解析版.pdf（25页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、第1页共 25页2017 年山东省泰安市中考数学试卷参考答案与试题解析一、选择题本大题共20 小题，每题 3 分，共 60 分1 3 分 2017?泰安以下四个数：3，1，其中最小的数是AB3 C 1 D【分析】将四个数从大到小排列，即可判断【解答】解：13，最小的数为，故选 A【点评】此题考查实数的大小比较，记住任意两个实数都可以比较大小，正实数都大于 0，负实数都小于0，正实数大于一切负实数，两个负实数绝对值大的反而小2 3 分 2017?泰安以下运算正确的选项是Aa2?a2=2a2Ba2+a2=a4C 1+2a2=1+2a+4a2D a+1 a+1=1a2【分析】根据整式的乘法、加法法

2、则及完全平方公式和平方差公式逐一计算可得【解答】解：A、a2?a2=a4，此选项错误；B、a2?a2=2a2，此选项错误；C、1+2a2=1+4a+4a2，此选项错误；D、a+1 a+1=1a2，此选项正确；故选：D【点评】此题主要考查同底数幂的乘法、整式的加法及完全平方公式和平方差公式，熟练掌握整式的运算法则是解题的关键3 3 分 2017?泰安以下图案第2页共 25页其中，中心对称图形是ABCD【分析】根据中心对称图形的概念求解【解答】解：不是中心对称图形；不是中心对称图形；是中心对称图形；是中心对称图形故选：D【点评】此题主要考查了中心对称图形的概念中心对称图形是要寻找对称中心，旋转

3、180 度后两部分重合4 3 分 2017?泰安“2014 年至 2016 年，中国同一带一路沿线国家贸易总额超过 3 万亿美元”，将数据 3 万亿美元用科学记数法表示为A31014美元 B31013美元 C 31012美元 D31011美元【分析】科学记数法的表示形式为a10n的形式，其中 1|a|10，n 为整数确定 n 的值时，要看把原数变成a 时，小数点移动了多少位，n 的绝对值与小数点移动的位数相同当原数绝对值1 时，n 是正数；当原数的绝对值1 时，n是负数【解答】解：3 万亿=3 0000 0000 0000=3 1012，故选：C【点评】此题考查科学记数法的表示方法科学记数

4、法的表示形式为a10n的形式，其中 1|a|10，n 为整数，表示时关键要正确确定a 的值以及 n 的值5 3 分 2017?泰安化简 1 1的结果为ABCD文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9

5、 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10

6、 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文

7、档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6

8、J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W

9、10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A

10、2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2第3页共 25页【分析】原式括号中两项通分并利用同分母分式的减法法则计算，同时利用除法法则变形，约分即可得

11、到结果【解答】解：原式=?=，故选 A【点评】此题考查了分式的混合运算，熟练掌握运算法则是解此题的关键6 3 分 2017?泰安下面四个几何体：其中，俯视图是四边形的几何体个数是A1 B2 C 3 D4【分析】根据俯视图是分别从物体上面看，所得到的图形进行解答即可【解答】解：俯视图是四边形的几何体有正方体和三棱柱，故选：B【点评】此题考查了几何体的三视图，掌握定义是关键注意所有的看到的棱都应表现在三视图中7 3 分 2017?泰安一元二次方程x26x6=0 配方后化为A x32=15 B x32=3 C x+32=15 D x+32=3【分析】方程移项配方后，利用平方根定义开方即可求出解【解

12、答】解：方程整理得：x26x=6，配方得：x26x+9=15，即 x32=15，故选 A【点评】此题考查了解一元二次方程配方法，熟练掌握完全平方公式是解此题的关键8 3 分 2017?泰安袋内装有标号分别为1，2，3，4 的 4 个小球，从袋内文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6

13、J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W

14、10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A

15、2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2

16、B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I

17、7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B

18、7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2第4页共 25页随机

19、取出一个小球，让其标号为一个两位数的十位数字，放回搅匀后，再随机取出一个小球，让其标号为这个两位数的个位数字，则组成的两位数是3 的倍数的概率为ABC D【分析】画树状图展示所有 16 种等可能的结果数，再找出所成的两位数是3 的倍数的结果数，然后根据概率公式求解【解答】解：画树状图为：共有 16 种等可能的结果数，其中所成的两位数是3 的倍数的结果数为5，所以成的两位数是3 的倍数的概率=故选 B【点评】此题考查了列表法与树状图法：利用列表法或树状图法展示所有等可能的结果 n，再从中选出符合事件A 或 B 的结果数目 m，然后利用概率公式求事件A或 B的概率 9 3 分 2017?泰安不等式

20、组的解集为 x2，则 k 的取值范围为Ak1 Bk1 Ck1 Dk1【分析】求出每个不等式的解集，根据已知得出关于k 的不等式，求出不等式的解集即可【解答】解：解不等式组，得不等式组的解集为 x2，文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5

21、B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I

22、4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5

23、L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3

24、N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD

25、4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5

26、K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2第5页共 25页k+12，解得 k1故选：C【点评】此题考查了解一

27、元一次不等式组的应用，解此题的关键是能根据不等式的解集和已知得出关于k 的不等式，难度适中10 3 分 2017?泰安某服装店用10000 元购进一批某品牌夏季衬衫假设干件，很快售完；该店又用 14700元钱购进第二批这种衬衫，所进件数比第一批多40%，每件衬衫的进价比第一批每件衬衫的进价多10 元，求第一批购进多少件衬衫？设第一批购进x 件衬衫，则所列方程为A10=B+10=C10=D+10=【分析】根据题意表示出衬衫的价格，利用进价的变化得出等式即可【解答】解：设第一批购进 x 件衬衫，则所列方程为：+10=故选：B【点评】此题主要考查了由实际问题抽象出分式方程，正确找出等量关系是解题关键

28、11 3 分 2017?泰安为了解中考体育科目训练情况，某校从九年级学生中随机抽取部分学生进行了一次中考体育科目测试把测试结果分为A，B，C，D四个等级，并将测试结果绘制成了如下图的两幅不完整统计图，根据统计图中提供的信息，结论错误的选项是文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9

29、 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10

30、 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文

31、档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6

32、J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W

33、10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A

34、2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2第6页共 25页A本次抽

35、样测试的学生人数是40B在图 1 中，的度数是 126C该校九年级有学生500 名，估计 D 级的人数为 80D从被测学生中随机抽取一位，则这位学生的成绩是A 级的概率为【分析】利用扇形统计图以及条形统计图分别分析得出总人数以及结合的度数、利用样本估计总体即可【解答】解：A、本次抽样测试的学生人数是：1230%=40人，正确，不合题意；B、360=126，的度数是 126，故此选项正确，不合题意；C、该校九年级有学生500 名，估计 D 级的人数为：500=100人，故此选项错误，符合题意；D、从被测学生中随机抽取一位，则这位学生的成绩是A 级的概率为：，正确，不合题意；故选：C【点评】此题主

36、要考查了概率公式以及利用样本估计总体、扇形统计图与条形统计图等知识，由图形获取正确信息是解题关键12 3 分 2017?泰安如图，ABC内接于 O，假设 A=，则 OBC等于文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4

37、I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K

38、5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V

39、3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:C

40、D4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK

41、5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK

42、8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2第7页共 25页A180 2B2 C90+D90【分析】首先连接 OC，由圆周角定理，可求

43、得BOC的度数，又由等腰三角形的性质，即可求得 OBC的度数【解答】解：连接 OC，ABC内接于 O，A=，BOC=2 A=2 ，OB=OC，OBC=OCB=90 故选 D【点评】此题考查了圆周角定理与等腰三角形的性质此题比较简单，注意掌握辅助线的作法，注意数形结合思想的应用13 3 分 2017?泰安已知一次函数y=kxm2x 的图象与 y 轴的负半轴相交，且函数值 y 随自变量 x 的增大而减小，则以下结论正确的选项是Ak2，m0 B k2，m0 C k2，m0 Dk0，m0【分析】由一次函数 y=kxm2x 的图象与 y 轴的负半轴相交且函数值y 随自变量 x 的增大而减小，可得出k20

44、、m0，解之即可得出结论【解答】解：一次函数 y=kxm2x 的图象与 y 轴的负半轴相交，且函数值y随自变量 x 的增大而减小，k20，m0，k2，m0文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 H

45、K5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 Z

46、K8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编

47、码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9

48、 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10

49、 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文

50、档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2文档编码:CD4I4K2B6J9 HK5K5L5I7W10 ZK8V3N5B7A2第8页共 25页故选 A【点评】此题考查了一次函数的性质，根据一次函数的性质找出k20、m0 是解题的关键14

文档加载中……请稍候！
如果长时间未打开，您也可以点击刷新试试。

下载文档到电脑，查找使用更方便

4.3 金币

版权申诉 word格式文档无特别注明外均可编辑修改；预览文档经过压缩，下载后原文更清晰！ 立即下载

配套讲稿：: 如PPT文件的首页显示word图标，表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
特殊限制：: 部分文档作品中含有的国旗、国徽等图片，仅作为作品整体效果示例展示，禁止商用。设计者仅对作品中独创性部分享有著作权。
关键词：: 2017 山东省泰安市中考数学试卷解析

淘文阁 - 分享文档赚钱的网站所有资源均是用户自行上传分享，仅供网友学习交流，未经上传用户书面授权，请勿作他用。

限制150内

关于本文

本文标题：2017年山东省泰安市中考数学试卷解析版.pdf
链接地址：https://www.taowenge.com/p-55709582.html