北京市第三中学2014—2015学年初二上数学期中试卷及答案

北京三中（初中部）2014—2015学年度第一学期初二数学期中试卷2014.11一、精心选一选（本题共30分，每小题3分）以下各题均是四个选项，其中只有一个是符合题意的．1．下列各式中，从左到右的变形是因式分解的是（）.A.a2－4ab+4b2=(a－2b)2B.1)(122yxxyxyyxC.224)2)(2(yxyxyxD.ax+ay+a=a(x+y)2．根据下列条件，能判定△ABC≌△A′B′C′的是（）．A．AB=A′B′，BC=B′C′，∠A=∠A′B．∠A=∠A′，∠B=∠B′，AC=B′C′C．∠A=∠A′，∠B=∠B′，∠C=∠C′D．AB=A′B′，BC=B′C′，△ABC的周长等于△A′B′C′的周长3．下列各式中，正确的是（）．A．2121ababB．21422aaaC．22)1(111aaaaD．abab114．以下二次根式：①12；②22；③23；④27中，与3是同类二次根式的是（）．A．①和②B．②和③C．①和④D．③和④5．下列各式中，一定成立的是（）．A．2)(ba＝a＋bB．22)1(a＝a2＋1C．12a＝1a·1aD．ba＝b1ab学校班级姓名学号DACEBOjBADC6．到三角形三条边距离相等的点是（）．A.三条高线的交点B.三个内角平分线的交点C.三条中线的交点D.三边垂直平分线的交点7．已知三角形的两边长分别为3和7，则第三边的中线长x的取值范围是()．A.2＜x＜5B.4＜x＜10C.3＜x＜7D.无法确定8．如图,OA=OB，OC=OD，AD，BC相交于E，则图中全等的三角形的对数是（）．A．2B．3C．4D．59．如图,在△ABC中,AD是它的角平分线,AB=8cm，AC=6cm，则S△ABD:S△ACD=().A．4:3B．3:4C．16:9D．9:16第8题第9题第10题10．如图，在四边形ABCD中，对角线AC平分∠BAD，AB＞AD，下列结论正确的是（）．A.AB-AD＞CBCDB.AB-AD=CBCDC.AB-AD＜CBCDD.AB-AD与CBCD的大小关系不确定二、细心填一填（本题共16分,每空2分）11．当x时，31x在实数范围内有意义．当x时，分式有意义．ABCD1xxECBAD12．一种细菌的半径为0.0004m,用科学记数法表示为m．13．不改变分式的值，使分式的分子、分母中各项系数都为整数，且结果为最简分式05.0012.02.0xx____________________．14．22169ymxyx是一个完全平方式,则m的值为.15．计算2222mnmnmn=．16．如图，把Rt△ABC（∠C=90°）折叠，使A、B两点重合，得到折痕ED,再沿BE折叠，C点恰好与D点重合,则∠A等于度．三、认真算一算（本题共27分，17-21每小题3分）17．将下列多项式分解因式．（1）2228mamb（2）3222aabab解:解:（3))()(22xybyxa（4）2233yxyx．解:解:学校班级姓名学号18．计算：（1）1284(72)2；（2）21(23)(73)(73)．19．计算：2210352abbbaa．20．先化简，再求值：2112()3369mmmmm，其中9m．解：21．解分式方程：3111xxx．解：四、列方程解应用题（本题共5分）．22．甲、乙两地相距19千米，某人从甲地到乙地，先步行7千米，然后改骑自行车，到达乙地共用了2小时，已知这人骑车速度是步行速度的4倍，求他的步行速度.五、作图题（本题共2分）23．尺规画图(不用写作法,要保留作图痕迹．．．．．．)如图1，在一次军事演习中，红方侦察员发现蓝方指挥部在A区内，到铁路与到公路的距离相等，且离铁路与公路交叉处B点400米，如果你是红方的指挥员，请你在图．2．所示的作战图上标出蓝方指挥部的位置点P.图1图2六、证明与计算题（本题共22分，24、25每小题5分,26、27每小题6分）24．已知：如图，A、B、C、D四点在同一直线上，AB=CD，AE∥BF且AE=BF．求证：EC=FD．证明：学校班级姓名学号BEACBDF25．已知：AC⊥BC，BD⊥AD，AC与BD交于O，AC=BD．求证：CBDDAC．26．已知如图，将一大、一小两个等腰直角三角尺ABC与DBE拼接（A、B、D三点共线，AB=CB，EB=DB，∠ABC=∠EBD=90°），连接AE、CD．问：AE与CD的位置关系和数量关系，并证明你的结论．ABCDOBCDEA27．在ABC△中，AB=AC，点D是直线BC上一点（不与BC、重合），以AD为一边在AD的右侧作△ADE，使BACDAEAEAD,，连接CE．（1）如图1，当点D在线段BC上时，如果90BAC°，则BCE度；（2）设BAC，BCE．①如图2，当点D在线段BC上移动，则，之间有怎样的数量关系？请说明理由；②当点D在直线BC上时，则，之间有怎样的数量关系？写出所有可能的结论并说明条件．答：（2）①数量关系_________________．理由：②数量关系________________________________________________．学校班级姓名学号图1：图2：备用图：北京市第三中学2014—2015学年度第一学期期中测试初二数学试卷答案一、选择题：将下列各题答案填入表中（每题3分，共30分）12345678910ADCCBBACAA二、填空题（每空2分，7个小题，共16分）11．31x；1x．12．4104．13．255006100xx．14．24m．15．nm1．16．030A．三、计算题（本题共27分，17-21每小题3分）17．将下列多项式分解因式．（1）2228mamb)(222bam····················································································2分=2m(a+b)(a-b)·······················································································3分（2）3222aabab)2(22babaa··············································································2分2)(baa·······················································································3分（3))()(22xybyxa=))(()()(2222bayxyxbyxa················································2分=))()((babayx···········································································3分（4）2233yxyx学校班级姓名学号学校班级姓名学号)33)(33(yxyxyxyx························································2分))((8)22)(44(yxyxyxyx．··················································3分18．（1）1284(72)2=27-22472··································································2分=237····················································································3分（2）21(23)(73)(73)=3-733441················································································2分=32···························································································3分19．2210352abbbaa=2222103104baabbaab···········································································1分=22107baab·························································································2分=ab107····························································································3分20．解：2112()3369mmmmm＝22(3)(3)(3)2mmmmm··································································1分＝33mm．·····················································································2分当9m时，原式＝931932．···························································3分21．解：方程两边同乘(1)(1)xx，得(1)3(1)(1)(1)xxxxx．·························································1分化简，得331xx．解得2x．·················································································2分检验：当2x时，(1)(1)0xx，∴2x是原分式方程的解．······························································3分四、列方程解应用题（本题共5分）22．解：设他的步行速度为x千米/时247197xx·············································································2分解得x=5．·························································································4分经检验，x=5为分式方程的解．答：他的步行速度为5千米/时．···································································5分五、作图题（本题2分）23．作AOB平分线1分，结论1分．六、证明与计算题（每小题5分，共10分）24．∵AE∥BF，∴∠A=∠FBD．·························