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2012年中考数学卷精析版聊城卷.pdf

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1、2012 年中考数学卷精析版聊城卷（本试卷满分120分，考试时间120 分钟）一、选择题（本题共12 小题，每题3分，共 36 分）1（2012 山东聊城3 分）计算|地结果是【】ABC 1D1【答案】A.【考点】绝对值，有理数地减法.【分析】根据绝对值地性质去掉绝对值符号，然后根据有理数地减法运算，减去一个数等于加上这个数地相反数进行计算即可得解：.故选 A.2（2012 山东聊城3 分）下列计算正确地是【】Ax2+x3=x5Bx2?x3=x6C（x2）3=x5Dx5 x3=x2【答案】D.【考点】合并同类项，同底数幂地乘法和除法，幂地乘方.故选 D.3（2012 山东聊城3 分）“抛一枚均

2、匀硬币，落地后正面朝上”这一事件是【】A必然事件B随机事件C确定事件D不可能事件【答案】B.【考点】随机事件.【分析】根据随机事件地定义，随机事件就是可能发生，也可能不发生地事件，即可判断：抛 1枚均匀硬币，落地后可能正面朝上，也可能反面朝上，故抛1枚均匀硬币，落地后正面朝上是随机事件.故选 B.4（2012 山东聊城3 分）用两块完全相同地长方体搭成如图所示地几何体，这个几何体地主视图是【】【答案】C.【考点】简单组合体地三视图.【分析】根据主视图地定义，找到从正面看所得到地图形即可：从物体正面看，左边1 列、右边1列上下各一个正方形，且左右正方形中间是虚线.故选 C.5（2012 山东聊城

3、3 分）函数y=中自变量x 地取值范围是【】Ax 2Bx2Cx 2Dx 2.【答案】A.【考点】函数自变量地取值范围，二次根式和分式有意义地条件.【分析】根据二次根式被开方数必须是非负数和分式分母不为0 地条件，要使在实数范围内有意义，必须.故选 A.6（2012 山东聊城3 分）将一副三角板按如图所示摆放，图中地度数是【】A75B 90C105D120【答案】C.【考点】三角形地外角性质，三角形内角和定理.【分析】如图，先根据直角三角形地性质得出BAE 及 E 地度数，再由三角形内角和定理及对顶角地性质即可得出结论：图中是一副直角三角板，BAE=45 ，E=30.AFE=180 BAE E=

4、105.=105.故选 C.7（2012 山东聊城3 分）某排球队12 名队员地年龄如下表所示：年龄/岁18 19 20 21 22 文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6

5、HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6

6、 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C

7、6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6

8、C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X

9、6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10

10、X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R10X6C6 HM8N1A1D8X6 ZB7W10K4F6T10文档编码:CT4Q5R1

11、0X6C6 HM8N1A1D8X6 ZB7W10K4F6T10人数/人1 4 3 2 2 该队队员年龄地众数与中位数分别是【】A19 岁，19岁B19 岁，20岁C20 岁，20 岁D20 岁，22 岁【答案】B.【考点】众数，中位数.8（2012 山东聊城3 分）如图，四边形ABCD 是平行四边形，点E在边 BC 上，如果点F是边 AD 上地点，那么 CDF 与ABE 不一定全等地条件是【】ADF=BEBAF=CECCF=AEDCFAE【答案】C.【考点】平行四边形地性质，全等三角形地判定.【分析】根据平行四边形地性质和全等三角形地判定方法逐项分析即可：A、当 DF=BE 时，由平行四边形地

12、性质可得：AB=CD，B=D，利用 SAS 可判定 CDF ABE；B、当 AF=CE 时，由平行四边形地性质可得：BE=DF，AB=CD，B=D，利用 SAS 可判定 CDF ABE；C、当 CF=AE 时，由平行四边形地性质可得：AB=CD，B=D，利用 SSA 不能可判定CDF ABE；D、当 CF AE 时，由平行四边形地性质可得：AB=CD，B=D，AEB=CFD，利用 AAS可判定 CDF ABE.故选 C.9（2012 山东聊城3 分）如图，在方格纸中，ABC 经过变换得到DEF，正确地变换是【】文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档

13、编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8

14、H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档

15、编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8

16、H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档

17、编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8

18、H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档

19、编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3A把 ABC 绕点 C 逆时针方向旋转90，再向下平移2 格B把 ABC 绕点 C 顺时针方向旋转90，再向下平移5 格C把 ABC 向下平移4格，再绕点C 逆时针方向旋转180D把 ABC 向下平移5格，再绕点C 顺时针方向旋转180【答案】B.【考点】几何变换地类型.【分析】根据图象，ABC 绕点 C 顺时针方向旋转90，再向下平移5 格即可与 DEF 重合.故选 B.10（2012 山东聊城 3 分）在如图所示地数轴上，点B 与点 C

20、关于点 A 对称，A、B 两点对应地实数分别是和 1，则点 C 所对应地实数是【】A1+B2+C 21D2+1【答案】D.【考点】实数与数轴，一元一次方程地应用.【分析】设点C 所对应地实数是x根据中心对称地性质，对称点到对称中心地距离相等，则有，解得.故选 D.11（2012 山东聊城 3 分）如图，在ABC 中，点 D、E 分别是 AB、AC 地中点，则下列结论不正确地是【】ABC=2DEBADE ABCCDSABC=3S ADE【答案】D.文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6

21、D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8

22、F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6

23、D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8

24、F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6

25、D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8

26、F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6

27、D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3BC=2DE.故 A 正确.DEBC，ADE ABC，故 B 正确.ADE ABC，故 C 正确.DE 是ABC 地中位线，AD：BC=1：2，SABC=4S ADE，故 D 错误.故选 D.12（2012 山东聊城 3 分）如图，在直角坐标系中，以原点O 为圆心地同心圆地半径由内向外依次为1，2，3，4，同心圆与直线y=x 和 y=x 分别交于A1，A2，A3，A4，则点 A30 地坐标是【】A（30，30）B（8，8）C（4，4）D（4，4）【答案】C.【考点】分类归纳（图形地变化类），一次函数综合

28、题，解直角三角形.【分析】A1，A2，A3，A4 四点一个周期，而30 4=7 余 2，A30 在直线 y=x 上，且在第二象限.即射线 OA30 与 x 轴地夹角是45，如图 OA=8，AOB=45 ，在直角坐标系中，以原点O 为圆心地同心圆地半径由内向外依次为1，2，3，4，OA30=8.A30 地横坐标是 8sin45=4，纵坐标是4，即 A30 地坐标是（4，4）.故选 C.二、填空题（本题共5 个小题，每小题3 分，共 15 分）文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6

29、S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 H

30、J2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6

31、S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 H

32、J2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6

33、S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 H

34、J2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6

35、S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S313（2012 山东聊城 3 分）一元二次方程x22x=0 地解是14（2012 山东聊城 3 分）在半径为6cm 地圆中，60 地圆心角所对地弧长等于cm（结果保留）【答案】.【考点】弧长地计算.【分析】根据弧长公式，把半径和圆心角代入公式计算就可以求出弧长：.15（2012 山东聊城 3 分）计算：=【答案】.【考点】分式地混合运算.【分析】将式子括号内部分通分，然后根据分式除法地运算法则，将其转化为乘法，再将分母中地式子因式分解，即可得到结果：.16（2012 山东聊城 3 分）我市初中毕业男生体育测试工

36、程有四项，其中“立定跳远”“1000M 跑”“肺活量测试”为必测工程，另一项“引体向上”或“推铅球”中选一项测试小亮、小明和大刚从“引体向上”或“推铅球”中选择同一个测试工程地概率是【答案】.【考点】列表法或树状图法，概率.【分析】分别用A，B 代表“引体向上”与“推铅球”，画树状图得：文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5

37、D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9

38、 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5

39、D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9

40、 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5

41、D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9

42、 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3共有 8 种等可能地结果，小亮、小明和大刚从“引体向上”或“推铅球”中选

43、择同一个测试工程地有 2 种情况，小亮、小明和大刚从“引体向上”或“推铅球”中选择同一个测试工程地概率是.17（2012 山东聊城 3 分）如图，在直角坐标系中，正方形地中心在原点O，且正方形地一组对边与x轴平行，点P（3a，a）是反比例函数（k0）地图象上与正方形地一个交点若图中阴影部分地面积等于9，则这个反比例函数地解读式为【答案】.【考点】待定系数法，曲线上点地坐标与方程地关系，反比例函数图象地对称性，正方形地性质.【分析】由反比例函数地对称性可知阴影部分地面积和正好为小正方形面积地，设小正方形地边长为b，图中阴影部分地面积等于9 可求出 b 地值，从而可得出直线AB 地表达式，再根据点

44、P（3a，a）在直线AB 上可求出a地值，从而得出反比例函数地解读式：反比例函数地图象关于原点对称，阴影部分地面积和正好为小正方形地面积.设正方形地边长为b，则 b2=9，解得 b=6.正方形地中心在原点O，直线AB 地解读式为：x=3.点 P（3a，a）在直线 AB 上，3a=3，解得 a=1.P（3，1）.点 P在反比例函数（k0）地图象上，k=3 1=3.此反比例函数地解读式为：.三、解答题（本题共8 小题，除第24 题 10 分，25 题 12 分，其余每小题7 分）18（2012 山东聊城 7 分）解不等式组文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6

45、S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 H

46、J2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6

47、S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 H

48、J2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6

49、S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 H

50、J2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6S3文档编码:CU2D3X7M8F9 HJ2K8H9F1W4 ZD4I6D5D6

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关键词：: 2012 年中数学卷精析版聊城

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本文标题：2012年中考数学卷精析版聊城卷.pdf
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