山东省济南市高新区2021-2022学年七年级上学期期中考试数学【试卷+答案】
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这是一份山东省济南市高新区2021-2022学年七年级上学期期中考试数学【试卷+答案】,共11页。试卷主要包含了3×104 B.3,6)元 D.元,计算,阅读材料等内容,欢迎下载使用。
绝密★启用前2021至2022学年第一学期期中学业水平测试高新初中数学七年级试题本试题分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.第Ⅰ卷共2页,满分为48分;第Ⅱ卷共4页,满分为102分.本试题共6页,满分为150分.考试时间为120分钟.答卷前,考生务必用0.5毫米黑色墨水签字笔将自己的考点、姓名、准考证号、座号填写在答题卡上和试卷规定的位置上.考试结束后,将本试卷和答题卡一并交回.本考试不允许使用计算器.第I卷(选择题 共48分)注意事项:第Ⅰ卷为选择题,每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑;如需改动,用橡皮擦干净后,再选涂其他答案标号.答案写在试卷上无效.一、选择题(本大题共12个小题,每小题4分,共48分.在每小题给出的四个选项中,只有一项是符合题目要求的.)1.﹣2021的相反数是( )A. B. C. D.2.下列几何体中,属于棱柱的是( )A. B. C. D.3.2021年4月底,印度爆发式的疫情冲击,全球面临新冠病毒变异危机,我国将再出手拯救全球疫情.据卫生局4月26日公布,在过去的一天内,印度新增确诊病例超过353000例,至此,印度已经连续五天新增病例超过30万例,并多次突破全球每日新增病例的最高记录.数据353000用科学记数法表示为( )A.35.3×104 B.3.53×105 C.0.353×106 D.353×1034.下列图形中,不能围成正方体的是( )A. B. C. D.5.济南市2021年2月15日的最高气温是13℃,最低气温是℃,济南这一天的温差是( )A.℃ B.℃ C.℃ D.℃6.下列说法中,正确的是( )A.单项式3πxy的系数是3 B.22ab3的次数是6次 C.多项式3x﹣2x2y+8xy是三次三项式 D.多项式x2+y2﹣1的常数项是17.某正方体的每个面上都有一个汉字,如图是它的一种展开图,那么在原正方体中,与“中” 字所在面相对的面上的汉字是( )A.梦 B.聚 C.力 D.凝8.下列计算正确的是( )A. B. C. D.9.某地居民生活用水收费标准:每月用水量不超过17立方米,每立方米a元;超过部分每立方米(a+1.2)元.该地区某用户上月用水量为20立方米,则应缴水费为( )A. 20a元 B.(20a+24)元 C.(17a+3.6)元 D.(20a+3.6)元10.如图是有理数在数轴上的位置,下列结论:①;②;③;④,其中正确的是( )A.①② B.①②③ C.①②④ D.①②③④11.当x=1时,多项式ax3+bx﹣2的值为2,则当x=﹣1时,该多项式的值是( )A.﹣6 B.﹣2 C.0 D.212. 图1是长为a,宽为b(a>b)的小长方形纸片将6张如图1的纸片按图2的方式不重叠地放在长方形ABCD内,已知CD的长度固定不变,BC的长度可以变化,图中阴影部分(即两个长方形)的面积分别表示为S1,S2,若S=S1-S2,且S为定值,则a,b之间的数量关系是( )A.a=2b B.a=3b C.a=4b D.a=5b第Ⅱ卷(非选择题 共102分)注意事项:1.第II卷必须用0.5毫米黑色签字笔作答,答案必须写在答题卡各题目指定区域内相应的位置,不能写在试卷上;如需改动,先划掉原来的答案,然后再写上新的答案;不能使用涂改液、胶带纸、修正带.不按以上要求作答的答案无效.2.填空题请直接填写答案,解答题应写出文字说明、证明过程或演算步骤.二、填空题:(本大题共6个小题,每小题4分,共24分.)13.用“>”或“<”符号填空:﹣3 ﹣2.14.如果,那么代数式的值是 .15.若a、b互为相反数,c、d互为倒数,且m的绝对值是1,则的值是 .16. 七年级小莉同学在学习完第二章《有理数及其运算》后,对运算产生了浓厚的兴趣.为庆祝“国庆节”,她借助有理数的运算,定义了一种新运算“”,规则如下:.则= .17. 某小区一块宽为a、长为b的长方形绿化带如图所示,其中4个同样大小的半圆为花园,其余部分种植草坪,则草坪面积为 .(用含a、b的式子表示) 18.如图所示,在这个数据运算程序中,若开始输入的x的值为2,结果输出的是1,返回进行第二次运算则输出的是﹣4,…,则第2021次输出的结果是 .三、解答题:(本大题共9个小题,共78分.解答应写出文字说明、证明过程或演算步骤.)19.(本题4分)计算:20.(本题4分)计算:21.(本题4分)合并同类项:22.(本题5分)计算:23.(本题5分)化简:24.(本题6分)如图,是由6个大小相同的小立方体块搭建的几何体,其中每个小正方体的棱长为1厘米.请按要求在方格内分别画出从这个几何体的三个不同方向看到的形状图. 25.(本题6分)将﹣2.5,,,在数轴上表示出来,并用“>”把他们连接起来. 26.(本题6分)先化简,再求值:﹣3a2b+3(4ab2﹣a2b)﹣2(2ab2﹣a2b),其中a=1,b=﹣1 27.(本题8分)为了有效控制酒后驾车,济南市交警的汽车在一条东西方向的公路上巡逻,若规定向东为正,向西为负,从出发点开始所走的路程为:+15,﹣4,+13,﹣10,﹣12,+3,﹣13,﹣17(单位:千米)(1)此时,该交警如何向队长描述他所处的位置?(2)如果队长命令他马上返回出发点,这次巡逻(含返回)共耗油多少升?(已知每千米耗油0.4升) 28.(本题8分)阅读材料:求1+2+22+23+24+…+22017.首先设S=1+2+22+23+24+…+22017①,则2S=2+22+23+24+25+…+22018②,②﹣①得S=22018﹣1, 即1+2+22+23+24+…+22017=22018﹣1.以上解法,在数列求和中,我们称之为:“错位相减法”.请你根据上面的材料,解决下列问题:(1)求1+3+32+33+34+…+32020的值;(2)若a为正整数且a≠1,求1+a+a2+a3+a4+…+a2020. 29.(本题10分)已知点A在数轴上对应的数为a,点B在数轴上对应的数为b,A、B之间的距离记为|AB|=|a﹣b|或|b﹣a|,请回答问题:(1)当a=,b=2时,|AB|= .(2)设点P在数轴上对应的数为x,若|x﹣3|=5,则x= .(3)如图,点M,N,P是数轴上的三点,点M表示的数为4,点N表示的数为﹣1,动点P表示的数为x. ①若点P在点M、N之间,则|x+1|+|x-4|= .②若|x+1|+|x﹣4|=10,则x= .③若点P表示的数是﹣5,现在有一蚂蚁从点P出发,以每秒1个单位长度的速度向右运动,当经过多少秒时,蚂蚁所在的点到点M、点N的距离之和是8? 30.(本题12分)把正整数1,2,3,4,…,排列成如图1所示的一个表,从上到下分别称为第1行、第2行、…,从左到右分别称为第1列、第2列、…,用图2所示的方框在图1中框住16个数,把其中没有被阴影覆盖的四个数分别记为A、B、C、D.设A=x. (1)在图1中,2017排在第 行第 列;(2)A﹣B+C﹣D的值是否为定值?如果是,请求出它的值;如果不是,请说明理由;(3)将图1中的奇数都改为原数的相反数,偶数不变.①设此时图1中排在第m行第n列的数(m、n都是正整数)为w,请用含m、n的式子表示w;②此时A+B﹣C﹣D的值能否为3918?如果能,请求出A所表示的数;如果不能,请说明理由.
2021至2022学年第一学期期中学业水平测试高新初中数学七年级参考答案及评分标准一、选择题:(本大题共12个小题,每小题4分,共48分.)题号123456789101112答案ADBCCCDBDBAA二、填空题:(本大题共6个小题,每小题4分,共24分.)13.< 14.﹣1 15.2020 16.24 17. 18.-6三、解答题:(本大题共12个小题,共78分.解答应写出文字说明、证明过程或演算步骤.)19.(本题4分)解:原式 ············································2分 ··································································4分 20.(本题4分)解:原式 ·······················································2分 ····································································4分 21.(本题4分)解:原式 ···································3分························································4分 22.(本题5分)解:原式 ·············································3分 ························································4分····································································5分23.(本题5分)解:原式 ········································3分 ························································5分24.(本题6分)解:所画图形如下所示: ···································6分 25.(本题6分)解:如图所示:·················4分用“>”将它们连接起来:|-5|>3> -(-2) >﹣2.5.·····································6分(本题6分)解:原式 ····················3分 ························································4分当a=1,b=﹣1时,原式 ······································5分 ····································································6分27.(本题8分)解:(1) ······ ··········2分 · ································································3分答:此时,该交警在出发点西边25千米处.················································4分(2)交警共行驶的路程为|+15|+|-4|+|+13|+|-10|+|-12|+|+3|+|-13|+|-17|+|-25|=112(千米).这次巡逻(含返回)共耗油112×0.4=44.8(升).······································7分答:这次巡逻(含返回)共耗油44.8升.····················································8分28.(本题8分)解:(1)设S=1+3+32+33+34+…+32020 ①,则3S=3+32+33+34+35+…+32021 ②,····························································2分②﹣①得2S=32021﹣1,所以S= ,即1+3+32+33+34+…+32020=;································4分(2)设S=1+a+a2+a3+a4+…+a2020 ①,则aS=a+a2+a3+a4+..+a2019+a2021 ②,··························································6分②﹣①得:(a﹣1)S=a2021﹣1,所以S=,即1+a+a2+a3+a4+…+a2020=.································8分 29.(本题10分)解:(1)5.·····································································1分(2)8或﹣2.······················································································3分(3)① 5; ·························································································4分②﹣3.5或6.5;·····················································································6分③t秒后,点P表示的数是t﹣5,NP=|t﹣5+1|=|t﹣4|,MP=|t﹣5﹣4|=|t﹣9|,当t﹣5<﹣1时,|t﹣4|+|t﹣9|=4﹣t+9﹣t=13﹣2t=8,·····························7分解得t=2.5,····················································································8分当t﹣5>4时,|t﹣4|+|t﹣9|=t﹣4+t﹣9=2t﹣13=8,································9分解得t=10.5,····················································································10分答:经过2.5秒或10.5秒时,蚂蚁所在的点到点M、点N的距离之和是8.30.(本题12分)解:(1)∵2017÷8=252······1,∴在图1中,2017排在第253行第1列;故答案为253,1;····················································································2分(2)是定值;·························································································3分由题知,∴的值为定值,这个定值为0.················································5分(3)① 法一:当n是奇数时,;当n是偶数时,.·············· ······························9分法二:.·································································9分② 不能;······························································································10分理由如下:如果结果等于3918,说明此时A、B都是正数,C、D都是负数.∵A=x,所以B=x+24,D=,C=.∴A+B﹣C﹣D=x+x+24+x+3+x+27=4x+54=3918,解得x=966,∴A所表示的数应为966.∵966=8×120+6,∴此时A在第121行,第6列.此时图2的方框只能框到3列数,C、D都框不到数了,所以A+B﹣C﹣D的值不能为3918.···························································12分
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