(完整word版)上海市八上期中数学试卷(带解析).pdf
《(完整word版)上海市八上期中数学试卷(带解析).pdf》由会员分享,可在线阅读,更多相关《(完整word版)上海市八上期中数学试卷(带解析).pdf(7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HEDCBA八年级第一学期数学学科期中试卷(考试时间:90 分钟,满分 100 分)2017.11.题号一二三四总分得分一、选择题:(本大题共 6 题,每题 2 分,满分 12 分)1下列根式中,与12为同类二次根式的是.()(A)2;(B)3;(C)5;(D)62下列二次根式中,属于最简二次根式的是()(A)21;(B)8;(C)yx2;(D)yx23已知一元二次方程:2330 xx,2330 xx.下列说法正确的是()(A)方程都有实数根;(B)方程有实数根,方程没有实数根;(C)方程没有实数根,方程有实数根;(D)方程都没有实数根.4.某种产品原来每件价格为800 元,经过两次
2、降价,且每次降价的百分率相同,现在每件售价为 578 元,设每次降价的百分率为x,依题意可列出关于x 的方程 .()(A)2800(1%)578x;(B)2800(1)578x;(C)2578(1%)800 x;(D)2578(1)800 x5.下列命题中,真命题是.()(A)两条直线被第三条直线所截,同位角相等;(B)两边及其中一边的对角对应相等的两个三角形全等;(C)直角三角形的两个锐角互余;(D)三角形的一个外角等于两个内角的和6.如图,在 ABC 中,ADBC 于 D,BEAC 于 E,AD、BE 交于点 H,且 HD=DC,那么下列结论中,正确的是.()(A)ADC BDH;(B)H
3、E=EC;(C)AH=BD;(D)AHEBHD .(第 6 题图)精品资料-欢迎下载-欢迎下载 名师归纳-第 1 页,共 7 页 -第 2 页二、填空题:(本大题共 12题,每题 3 分,满分 36分)7.化简:27_.8.如果代数式31x有意义,那么实数x的取值范围是 _.9.计算:28xyy_.10.写出1a的一个有理化因式是_.11.不等式:(32)1x的解集是 _.12.方程2xx的解为 _13.在实数范围内因式分解:241xx_ 14.如果关于x的一元二次方程02mxx有两个不相等的实数根,那么m 的取值范围是_15.如果关于x的一元二次方程22(1)10axxa的一个根是0,那么a
4、的值为 _.16.如图,已知点B、F、C、E 在一条直线上,FB=CE,AC=DF,要使 ABC DEF 成立,请添加一个条件,这个条件可以是_.17.将命题“两个全等三角形的面积相等”改写成“如果,那么 ”的形式:_.18.如图,在 ABC 中,CAB=70.在同一平面内,现将 ABC 绕点 A 旋转,使得点B 落在点 B,点 C落在点 C,如果 CC/AB,那么 BAB=_.三、解答题:(本大题共 4 题,第 1922 题,每题 6 分;第 23 题 8 分;第 2425 题每题 10分,满分 52 分)19.计算:1(32)(32)21.FEDCBA(第 16 题图)(第 18 题图)C
5、BCBA精品资料-欢迎下载-欢迎下载 名师归纳-第 2 页,共 7 页 -文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N
6、4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3
7、HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G
8、10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6
9、 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2
10、T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V
11、2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2第 3 页FEDCBA20.用公式法解方程:2530 xx.21.用配方法解方程:212302
12、xx.22.已知:如图,ACCD 于 C,BDCD 于 D,点 E 是 AB 的中点,联结CE 并延长交 BD 于点 F.求证:CE=FE.23.如图,某工程队在工地互相垂直的两面墙AE、AF 处,用 180 米长的铁栅栏围成一个长方形场地ABCD,中间用同样材料分割成两个长方形.已知墙 AE 长 120 米,墙 AF 长 40 米,要使长方形ABCD 的面积为 4000平方米,问BC 和 CD 各取多少米?A F E D B C 精品资料-欢迎下载-欢迎下载 名师归纳-第 3 页,共 7 页 -文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF
13、6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U
14、9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN
15、8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10
16、H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 Z
17、X2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3
18、R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文
19、档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2第 4 页EDCBA24我们知道:任意一个有理数与无理数的和为无理数,任意一个不为零的有理数与一个无理数的积为无理数,而零与无理数的积为零由此可得:如果 ax+b=0,其中 a、b 为有理数,x 为无理数,那么a=0 且b=0运用上述知识,解决下列问题:(1)如果0
20、32)2(ba,其中 a、b 为有理数,那么a=,b=;(2)如果5)21()22(ba,其中 a、b 为有理数,求a+2b 的值25如图,在四边形ABCD 中,AB/CD,B=ADC,点 E 是 BC 边上的一点,且AE=DC(1)求证:ABC EAD;(2)如果 ABAC,求证:BAE=2ACB精品资料-欢迎下载-欢迎下载 名师归纳-第 4 页,共 7 页 -文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2
21、K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R1
22、0V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编
23、码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G
24、5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P
25、3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F2G10H5L6 ZX2K2T3R10V2文档编码:CF6G5N4U9P3 HN8F
- 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:
- 完整 word 上海市 上期 数学试卷 解析
限制150内