北京市平谷区2022-2023学年七年级上学期期末考试数学

这是一份北京市平谷区2022-2023学年七年级上学期期末考试数学，共10页。试卷主要包含了 单项式的系数和次数分别为， 下列方程变形中，正确的是，01，所得到的近似数为___．， 3等内容，欢迎下载使用。
北京市平谷区2022～2023学年度第一学期期末检测                                    七年级数学试卷              2023.1（考试时间120分钟  满分100分）学校               班级             姓名               考号              考生须知1．本试卷共6页，共三道大题，28道小题．2．在试卷和答题卡上认真填写学校、班级、姓名、考号．3．试题答案一律填涂或书写在答题卡上，在试卷上作答无效．4．在答题卡上，选择题、作图题用2B铅笔作答，其他试题用黑色字迹签字笔作答．5．考试结束，请将本试卷、答题卡和草稿纸一并交回．一、选择题（本题共16分，每小题2分）下面1-8题均有四个选项，其中符合题意的选项只有一个．1下列各组数中，互为倒数的是 A．  B C.  D.2. 2022年我国夏粮生产喜获丰收，为稳定全年粮食生产奠定了良好的基础，为稳物价保民生、稳定经济大盘、应对外部环境的不确定性提供了坚实的支撑。据统计，2022年全国夏粮播种面积397950000亩，比上年增长了0.3％，两年实现增长. 将397950000用科学记数法表示应为A．    B.     C.     D.3. 下列运算正确的是A．     B．C．    D．4. 如图,点 C,D 在线段 AB 上,若 AD=CB,则A. AC=CD      B.AC=DB C. AD=2DB     D. CD=CB5. 单项式的系数和次数分别为A．－3，2    B．－3，3  C．3，3   D．3，26. 下列方程变形中，正确的是A．方程，移项得B．方程，系数化为1得C．方程，去括号得D．方程，去分母得7． 你见过一种折叠灯笼吗？它看起来是平面的，可是提起来后却变成了美丽的灯笼，这个过程可近似地用哪个数学原理来解释（　　）A．点动成线 B．线动成面C．面动成体 D．面与面相交的地方是线8. 如图是一组有规律的图案，它们是由边长相等的小正方形组成，其中部分小正方形涂有阴影，按照这样的规律，第n个图案中涂有阴影的小正方形为（用含有n的代数式表示）   A．4+5（n-1） B．4+4nC．5+4(n-1) D．5+4n二、填空题（本题共16分，每小题2分）9．用四舍五入法把3,1415926精确到0.01，所得到的近似数为___．10.比较大小         ______        （填“>”，“<” 或“=”）11. 若∠α=15°35′,∠β=10°25′,则∠α+∠β=　_____.12.若,,则a+b=_____13. 若代数式与是同类项，那么，14. 如图所示的网格是正方形网格,点 A,B,C,D,O 是网格线交点,那么∠AOB_____∠COD15. 一种商品每件成本为a元，按成本增加25%定价，售出60件，可盈利_____________元（用含a的式子表示）．16．黑板上写着7个数，分别为：-8，a,1,13，b，0,-6，它们的和为-10，若每次从中任意擦除两个数，同时写上一个新数（新数为所擦除的两个数的和加上1），这样操作若干次，直至黑板上只剩下一个数，则所剩的这个数是____________.三、解答题(第17题20分，第18题10分，第19题6分，第20-21每小题5分，第22题6分，第23-24每小题5分，第25题6分，共68分)17．计算：(1)      (2)     (3)        (4)  .      18．解方程：.               .     19．按要求画图，并回答问题： 如图,已知平面上四个点 A,B,C,D,请按要求回答下列问题:(1)画直线 AB,射线 BD,连接 AC; (2)取线段 AD 中点E;(3)请在直线AB上确定一点F,使点 F 到点 E 与点 C 的距离之和最短,并写出画图依据 (保留作图痕迹) .    20．已知 x=-1 是方程 2a+4x=x+5a 的解.(1)求 a 的值;(2)求关于 y 的方程 ay+6=6a+2y 的解.      21.先化简,再求值:已知 a-b=5,求 3(a2b+a-2b)-2(a2b+a)- (a2b-5b-1)的值. 22.按要求补全图形并证明.如图,∠AOB=150°，OC垂直OB,OD平分∠AOC,OE平分∠BOC（1）利用三角板依题意补全图形（2）求∠DOE的度数      23. 列方程解应用题：    某车间有88名工人生产甲、乙两种零件,每名工人每天平均能生产甲种零件 24个或乙种零件10 个. 已知 2 个甲种零件和 1个乙种零件配成一套,问应分配多少名工人生产甲种零件,多少名工人生产乙种零件,才能使每天生产的这两种零件刚好配套?        24．如图：数轴上点A,B表示的数分别是a,b，其中a＞0，b＜0. （1）当 a=4，b=-2 时，线段 AB 的中点对应的数是  ___________ ．（2）若该数轴上另有一点C表示的数是 5，且 a＞5，当 BC=2AC时，求2a+b+2023  的值．     25.如果两个方程的解相差k，且k为正整数，则称解较大的方程为另一个方程的“k—后移方程”．例如：方程 x  3  0的解是x=3，方程 x 1  0的解是x=1所以：方程 x  3  0 是方程 x  1  0 的“2—后移方程”．（1）判断方程2x-3  0 是否为方程2x -1 0 的k--后移方程  ________（填“是”或“否”）;（2）若关于 x 的方程2x  m  n  0 是关于 x 的方程2x  m  0 的“2—后移方程”，求m的值（3）当时，如果方程是方程的“3—后移方程”求代数式的值．                              平谷区2022—2023学年度第一学期期末质量监控              初一数学参考答案及评分标准      2023．1一、选择题（本题共16分，每小题2分）题号12345678答案CCDBBCCC   二、填空题（9-16每题2分，本题共16分）9.  3.14     10.  <     11.     12. 0  13.  2  ，  3    14.  >    15.     16.  -4三、解答题(第17题20分，第18题10分，第19题6分，第20-21每小题5分，第22题6分，第23-24每小题5分，第25题6分，共68分)17．（1）解：=························································2分=························································3分=························································4分=························································5分（2）解： = ·······················································3分=- ·······················································4分= ·······················································5分（3）.解： = ·······················································2分=- ·······················································4分= ·······················································5分（4）解： = ·······················································2分=- ·······················································4分= ·······················································5分18．（1）解：去括号，得  ················································2分移项，得  ·················································3分合并同类项，得  ············································4分系数化为1，得  ·············································5分（2）解： 去分母（方程两边同乘以6），得  ．·······················································1分去括号，得．··············································2分移项，得  ．···············································3分合并同类项，得  ．··········································4分系数化为1，得  ．···········································5分19.   图略 ························································5分（3）依据：两点之间线段最短·································6分20.(1)解： 把x=-1代入2a+4x=x+5a中2a-4=-1+5a·················································1分2a-5a=-1+4-3 a =3a =-1·····················································3分(2)把a =-1代入ay+6=6a+2y-y+6=-6+2y·········································4分y =4···············································5分21. 3(a2b+a-2b)-2(a2b+a)- (a2b-5b-1) =3 a2b+3a-6b-2a2b-2a- a2b+5b+1···································3分=a - b+1·························································4分 ∵a-b=5原式=5+1=6······················································5分22． 图略························································1分∵垂直∴ ·····························································2分∵∴······························································3分∵平分∴······························································4分∵平分∴······························································5分∴······························································6分23．解：设应分配x名工人生产甲种零件,(88-x)名工人生产乙种零件， 根据题意列方程，得    …………………………………… 1分            .  ……………………………………….………3分   解得：    ∴ 88-x=44      ………………………………………………5分答：．应分配40名工人生产甲种零件,44名工人生产乙种零件,才能使每天生产的这两种零件刚好配套.24. 解：（1）1………………………………………………….………1分（2）∵C表示的数是5∴BC=5-b    AC=5-a……………………………………………….………3分         当BC=2AC时 5-b =2（5-a）  2a+b=15……………………………………………….……… 4分 把2a+b=15代入2a+b+2023原式=15+2023=2038……………………………………………….……… 5分25.解：（1）是…………………………………………………….……… 1分（2）解方程2x  m  n  0，……………………………………………….……… 2分解方程2x  m  0，……………………………………………….……… 3分∵关于 x 的方程2x  m  n  0 是关于 x 的方程2x  m  0 的“2—后移方程” ∴∴n=-4……………………………………………….……… 4分（3）解方程得解方程得∵方程是方程的“3—后移方程”∴∴……………………………………………….……… 5分  把c=3a+b代入6a+2b-2（c+3）原式=-6……………………………………………….……… 6分