北京市西城区2017-2018学年第一学期期末七年级数学试题(含答案)

1西城区2017-2018学年第一学期七年级期末考试数学试题2018.1一、选择题（本题共30分，每小题3分）1．据中新社2017年10月8日报道，2017年我国粮食总产量达到736000000吨，将736000000用科学记数法表示为（）.（A）673610（B）773.610（C）87.3610（D）90.736102.如图所示，将两个圆柱体紧靠在一起，从上面看这两个立体图形，得到的平面图形是（）.3.下列运算中，正确的是（）.（A）2(2)4=（B）224=（C）236=（D）3(3)274.下列各式进行的变形中，不正确．．．的是（）.（A）若3a=2b，则3a+2=2b+2（B）若3a=2b，则3a-5=2b-5（C）若3a=2b，则9a=4b（D）若3a=2b，则23ab=5．若2(1)210xy，则x+y的值为（）.（A）12（B）12（C）32（D）326.在一些商场、饭店或写字楼中，常常能看到一种三翼式旋转门在圆柱体的空间內旋转.旋转门的三片旋转翼把空间等分．．成三个部分，下图是从上面俯视旋转门的平面图，两片旋转翼之间的角度是（）.（A）100°（B）120°（C）135°（D）150°7.实数a，b，c，d在数轴上对应点的位置如图所示，正确的结论是（）（A）ac（B）b+c0（C）|a|＜|d|（D）-b＜d28.如图，在下列各关系式中，不正确．．．的是（）.（A）AD-CD=AB+BC（B）AC-BC=AD-DB（C）AC-BC=AC+BD（D）AD-AC=BD-BC9.某礼品包装商店提供了多种款式的包装纸片，将它们沿实线折叠（图案在包装纸片的外部，内部无图案），再用透明胶条粘合，就折成了正方体包装盒，小明用购买的纸片制作的包装盒如右图所示，在下列四种款式的纸片中，小明所选的款式的是（）.10.《九章算术》是中国古代数学专著，《九章算术》方程篇中有这样一道题：“今有善行者行一百步，不善行者行六十步，今不善行者先行一百步，善行者追之，问几何步及之？”这是一道行程问题，意思是说：走路快的人走100步的时候，走路慢的才走了60步；走路慢的人先走100步，然后走路快的人去追赶，问走路快的人要走多少步才能追上走路慢的人？如果走路慢的人先走100步，设走路快的人要走x步才能追上走路慢的人，那么，下面所列方程正确的是（）.（A）10060100xx（B）10010060xx（C）10060100xx（D）10010060xx3二、填空题（本题共20分，第11～14题每小题3分，第15～18题每小题2分）11．已知x=2是关于x的方程3x+a=8的解，则a=．12．一个有理数x满足:x＜0且2x，写出一个满足条件的有理数x的值:x=．13．在一面墙上用一根钉子钉木条时，木条总是来回晃动；用两根钉子钉木条时，木条就会固定不动，用数学知识解释这两种生活现象为．14．已知222xx，则多项式2243xx的值为．15．已知一个角的补角比这个角的一半多30°，设这个角的度数为x°，则列出的方程是：．16．如图是一所住宅的建筑平面图（图中长度单位：m），这所住宅的建筑面积为m.．17．如图，点A，O，B在同一条直线上，射线OD平分∠BOC，射线OE在∠AOC的内部，且∠DOE=90°，写出图中所有互为余角的角：．18．如图，一艘货轮位于O地，发现灯塔A在它的正北方向上，这艘货轮沿正东方向航行，到达B地，此时发现灯塔A在它的北偏西60°的方向上．(1)在图中用直尺、量角器画出B地的位置；(2)连接AB，若货轮位于O地时，货轮与灯塔A相距1.5千米，通过测量图中AB的长度，计算出货轮到达B地时与灯塔A的实际距离约为千米（精确到0.1千米）．45三、计算题（本题共16分，每小题4分）19．(21)(9)(8)(12)20．311()()(2)42421．31125(25)25()42422．3213(2)0.254[()]4028四、解答题（本题共20分，每小题5分）23．先化简，再求值：2223()2()3xxyxyxy，其中1x，3y．24．解方程12423xx．25．解方程组2531.xyxy，626．已知AB=10，点C在射线AB上，且12BCAB，D为AC的中点．（1）依题意，画出图形；（2）直接写出线段BD的长．解：（1）依题意，画图如下：（2）线段BD的长为．五、解答题（本题共13分，第27题6分，第28题7分）27．列方程或方程组解应用题为了备战学校体育节的乒乓球比赛活动，某班计划买5副乒乓球拍和若干盒乒乓球（多于5盒）．该班体育委员发现在学校附近有甲、乙两家商店都在出售相同品牌的乒乓球拍和乒乓球，乒乓球拍每副售价100元，乒乓球每盒售价25元．经过体育委员的洽谈，甲商店给出每买一副乒乓球拍送一盒乒乓球的优惠；乙商店给出乒乓球拍和乒乓球全部九折的优惠．（1）若这个班计划购买6盒乒乓球，则在甲商店付款元，在乙商店付款元；（2）当这个班购买多少盒乒乓球时，在甲、乙两家商店付款相同？728.如图，A，O，B三点在同一直线上，∠BOD与∠BOC互补.（1）试判断∠AOC与∠BOD之间有怎样的数量关系，写出你的结论，并加以证明；（2）OM平分∠AOC，ON平分∠AOD，①依题意，将备用图补全；②若∠MON=40°，求∠BOD的度数.解：（1）答：∠AOC与∠BOD之间的数量关系为：；理由如下：（2）①补全图形；②8北京市西城区2017—2018学年度第一学期期末试卷七年级数学参考答案及评分标准2018.1一、选择题（本题共30分，每小题3分）题号12345678910答案CADCABDCDB二、填空题（本题共20分，第11～14题每小题3分，第15～18题每小题2分）题号1112131415答案2答案不唯一，如：-1经过一点有无数条直线，两点确定一条直线11180302xx题号161718答案2218xx∠1和∠3,∠2和∠3,∠1和∠4,∠2和∠4互为余角作图位置正确1分3.0千米2分三、计算题（本题共16分，每小题4分）19．(21)(9)(8)(12)解：(21)(9)(8)(12)=-21+9-8+12·········································································1分=-29+21···················································································3分=-8···························································································4分20．311()()(2)424解：311()()(2)424319424···············································································2分314429···············································································3分16························································································4分21．31125(25)25()424解：31125(25)25()424=311252525424······························································1分=31125()424·········································································2分=25····································································································4分922．3213(2)0.254[()]4028解：3213(2)0.254[()]4028=1380.254()4048····························································1分=180.254()408·······························································2分=24840············································································3分=10·······················································································4分四、解答题（本题共21分，23～25题每小题5分，第26题6分）23．2223()2()3xxyxyxy，其中1x，3y．解：2223()2()xxyxyxy=22233223xxyxyxy···························································2分=222xy··················································································3分当1x，3y时，原式=22(1)23·······································································4分=19．··················································································5分24．解方程12423xx．解：去分母，得3(1)2(2)24xx．············································1分去括号，得332424xx．···············································2分移项，得322443xx．··················································3分合并同类项，得525x．························································4分系数化1，得5x．·································································5分25．2531.xyxy，解：由①得52xy．③·······························································1分把③代入②，得3(52)1yy．·················································2分解这个方程，得2y．···········································