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济南高新区2021-2022学年第二学期八年级学业质量抽测

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济南高新区2021-2022学年第二学期八年级学业质量抽测数学试题本试题分第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5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.要使分式2x−1有意义,x的取值应满足(  )A.x≠0B.x≠1C.x≠2D.x为任意实数2.下列等式从左到右的变形,是因式分解的是(  )A.(x+y)(x﹣y)=x2﹣y2B.x2﹣2x+1=(x﹣1)2C.x2+2x+2=(x+1)2+1D.12xy2=2x⋅6y23.在平行四边形ABCD中,若∠A+∠C=80°,则∠B的度数是(  )A.140°B.100°C.40°D.120°4.已知x=1是方程x2﹣3x+c=0的一个根,则实数c的值是(  )A.﹣1B.0C.1D.25.如图,在菱形ABCD中,两条对角线长AC=6,BD=8,则此菱形的面积为(  )A.48B.24C.20D.126.从口袋中随机摸出一球,再放回口袋中,不断重复上述过程,共摸了150次,其中有50次摸到黑球,已知口袋中有黑球10个和若干个白球,由此估计口袋中大约有多少个白球(  )A.10个B.20个C.30个D.无法确定7.下列式子中,能运用平方差公式分解因式的是(  )A.﹣4a2+b2B.x2+4C.a2+c2﹣2acD.﹣a2﹣b28.化简m−1m2÷1−mm3的结果是(  )A.mB.1mC.﹣mD.−1m9.已知关于x的一元二次方程x2﹣2x﹣k﹣1=0有两个实数根,则k的取值范围是(  )A.k>﹣2B.k≥﹣2C.k≤﹣2D.k<﹣210.如图,将△ABC沿着它的中位线DE对折,点A落在F处.若∠C=120°,∠A=20°,则∠FEB的度数是(  )A.140°B.120°C.100°D.80°11.已知a,b,c,d都是正数,如果M=(a+b+c)(b+c+d),N=(a+b+c+d)(b+c),那么M,N的大小关系是(  )A.M>NB.M=NC.M<ND.不确定12.如图,在平面直角坐标系xOy中,P(4,4),A、B分别是x轴正半轴、y轴正半轴上的动点,且△ABO的周长是8,则P到直线AB的距离是(  )第8页(共9页)学科网(北京)股份有限公司A.4B.3C.2.5D.2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.因式分解:2ab﹣4a=  .14.已知一个正n边形的每个内角都为120°,则n=  .15.随机闭合开关S1,S2,S3中的两个,能够让灯泡发亮的概率为  .16.若关于x的方程mx+4−x−1x+4=0产生增根,则m=  .17.如图,在一块长11m,宽为7m的矩形空地内修建三条宽度相等的小路,其余部分种植花草.若花草的种植面积为60m2,则小路宽为  m.18.如图,在矩形ABCD中,E,F分别是边AB,AD上的动点,P是线段EF的中点,PG⊥BC,PH⊥CD,G,H为垂足,连接GH.若AB=8,AD=6,EF=6,则GH的最小值是  .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6分)化简:x2+2x+1x2−1−xx−1.20.(本题6分)解方程:2x2﹣3x=1﹣2x.21.(本题6分)已知:如图,▱ABCD的对角线AC,BD相交于点O,点E,F分别在AO,CO上,且AE=CF,求证:∠EBO=∠FDO.22.(本题8分)第8页(共9页)学科网(北京)股份有限公司第24届冬季奥林匹克运动会(简称“冬奥会”)于2022年2月4日在北京开幕,本届冬奥会设7个大项、15个分项、109个小项.某校组织了关于冬奥知识竞答活动,随机抽取了七年级若干名同学的成绩,并整理成如下不完整的频数分布表、频数分布直方图和扇形统计图:分组频数60<x≤70470<x≤801280<x≤901690<x≤100请根据图表信息,解答下列问题:(1)本次知识竞答共抽取七年级同学  名;在扇形统计图中,成绩在“90<x≤100”这一组所对应的扇形圆心角的度数为  °;(2)该校计划对此次竞答活动成绩最高的小颖同学:奖励两枚“2022•北京冬梦之约”的邮票.现有如图所示“2022•北京冬梦之约”的四枚邮票供小颖选择,依次记为A,B,C,D,背面完全相同.将这四枚邮票背面朝上,洗匀放好,小颖从中随机抽取一枚不放回,再从中随机抽取一枚.请用列表或画树状图的方法,求小颖同学抽到的两枚邮票恰好是B(冰墩墩)和C(雪容融)的概率.23.(本题8分)如图,在正方形ABCD中,E为AB的中点,连接CE,将△CBE沿CE对折,得到△CGE,延长EG交CD的延长线于点H.(1)求证:△HCE是等腰三角形.(2)若AB=4,求HD的长度.24.(本题10分)某汽车贸易公司销售A,B两种型号的新能源汽车,A型车每台进货价格比B型车每台进货价格少3万元,该公可用24万元购买A型车的数量和用30万元购买B型车的数量相同.(1)求购买一台A型、一台B型新能源汽车的进货价格各是多少万元?(2)该公可准备用不超过300万,采购A,B两种新能源汽车共22台,问最少需要采购A型新能源汽车多少台?25.(本题10分)先阅读下面的材料,再解决问题:因式分解多项式:am+an+bm+bn,先把它的前两项分成一组,并提出a;把它的后两项分成一组,并提出b:第8页(共9页)学科网(北京)股份有限公司得:a(m+n)+b(m+n)再提公因式(m+n),得:(m+n)(a+b).于是得到:am+an+bm+bn=a(m+n)+b(m+n)=(a+b)(m+n).这种因式分解的方法叫做分组分解法.请用上面材料中提供的方法解决问题:(1)将多项式ab﹣ac+b2﹣bc分解因式;(2)若△ABC的三边a、b、c满足条件:a4﹣b4+a2c2+b2c2=0,试判断△ABC的形状.26.(本题12分)利用完全平方公式(a+b)2=a2+2ab+b2和(a﹣b)2=a2﹣2ab+b2的特点可以解决很多数学问题.下面给出两个例子:例1.分解因式:x2+2x﹣3=x2+2x+1﹣4=(x+1)2﹣4=(x+1+2)(x+1﹣2)=(x+3)(x﹣1)例2.求代数式2x2﹣4x﹣6的最小值:2x2﹣4x﹣6=2(x2﹣2x)﹣6=2(x2﹣2x+1﹣1)﹣6=2[(x﹣1)2﹣1]﹣6=2(x﹣1)2﹣8又∵2(x﹣1)2≥0∴当x=1时,代数式2x2﹣4x﹣6有最小值,最小值是﹣8.仔细阅读上面例题,模仿解决下列问题:(1)分解因式:m2﹣6m﹣7;(2)当x、y为何值时,多项式2x2+y2﹣8x+6y+20有最小值?并求出这个最小值;(3)已知△ABC的三边长a、b、c都是正整数,且满足a2+b2=8a+6b﹣25,求△ABC周长的最大值.第8页(共9页)学科网(北京)股份有限公司27.(本题12分)【问题原型】如图1,在四边形ABCD中,∠ADC=90°,AB=AC.点E、F分别为AC、BC的中点,连接EF,DE.试说明:DE=EF.【探究】如图2,在问题原型的条件下,当AC平分∠BAD,∠DEF=90°时,求∠BAD的大小.【应用】如图3,在问题原型的条件下,当AB=2,且四边形CDEF是菱形时,直接写出四边形ABCD的面积.济南高新区2021-2022学年第二学期八年级学业质量抽测数学参考答案及评分标准一、选择题题号123456789101112答案BBADBBACBCAA二、填空题:(本大题共6个小题,每小题4分,共24分.)13.2a(b﹣2).14.6.15.23.16.﹣5.17.1.18.7.三、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6分)解:原式=(x+1)2(x+1)(x−1)−xx−1=x+1x−1−xx−1··················································································4分=1x−1···························································································4分20.(本题6分)解:原方程化为2x2﹣x﹣1=0·······································································2分∵a=2,b=﹣1,c=﹣1,∴Δ=b2﹣4ac=(﹣1)2﹣4×2×(﹣1)=9>0····································································4分∴x=1±92×2=1±34,∴x1=1,x2=−12···········································································································6分21.(本题6分)证明:连接DE、BF,如图所示:∵四边形ABCD是平行四边形,第8页(共9页)学科网(北京)股份有限公司∴OB=OD,OA=OC···································································································2分∵AE=CF,∴OE=OF···················································································································3分∴四边形BEDF是平行四边形··························································································4分∴BE∥DF···················································································································5分∴∠EBO=∠FDO·········································································································6分22.(本题8分)解:(1)40,72·························································································2分(2)画树状图如下:································································································5分共有12种等可能的结果··································································································6分其中小颖同学抽到的两枚邮票恰好是B(冰墩墩)和C(雪容融)的结果有2种·························7分∴P小颖同学抽到的两枚邮票恰好是B(冰墩墩)和C(雪容融)的概率=212=16····························································8分23.(本题8分)(1)证明:在正方形ABCD中,AB∥CD,∴∠BEC=∠ECD·········································································································1分根据翻折,可得∠BEC=∠GEC························································································2分∴∠ECD=∠GEC·········································································································3分∴HE=HC,即△HCE是等腰三角形·················································································4分(2)设HD=x,∵AB=4,∴BC=CD=4,∵E为AB的中点,∴EB=2,根据翻折,GC=BC=4,EG=EB=2,∵HC=4+x··················································································································5分∴HE=4+x,∴HG=4+x﹣2=2+x·····································································································6分在Rt△HGC中根据勾股定理,得(x+4)2=42+(x+2)2································································································7分解得x=1,即HD=1·····································································································8分24.(本题10分)解:设一台B型新能源汽车的进货价格是x万元,则一台A型新能源汽车的进货价格是(x-3)万元·····1分由题意可得:30x=24x−3···································································································3分解得:x=15·················································································································4分经检验,x=15是原方程的解,且符合题意·········································································5分∴x﹣3=12(万元),∴购买一台A型新能源汽车的进货价格是12万元,购买一台B型的是15万元··························6分(2)设需要采购A型新能源汽车a台···············································································7分由题意可得:12a+15(22﹣a)≤300·················································································8分∴a≥11·······················································································································9分∴最少需要采购A型新能源汽车11台··············································································10分25.(本题10分)解:(1)ab﹣ac+b2﹣bc第8页(共9页)学科网(北京)股份有限公司=(ab﹣ac)+(b2﹣bc)·······························································································2分=a(b﹣c)+b(b﹣c)=(a+b)(b﹣c)·········································································································4分(2)由已知,得(a2﹣b2)(a2+b2)+c2(a2+b2)=0····························································6分即(a2+b2)(a2﹣b2+c2)=0∵a2+b2>0∴a2﹣b2+c2=0·············································································································8分即a2+c2=b2··············································································································9分∴△ABC是直角三角形·································································································10分26.(本题12分)解:(1)m2﹣6m﹣7=m2﹣6m+9﹣9﹣7·······································································································1分=(m﹣3)2﹣16···········································································································2分=(m﹣3+4)(m﹣3﹣4)=(m+1)(m﹣7)········································································································4分(2)2x2+y2﹣8x+6y+20=(2x2﹣8x)+y2+6y+9+11=2(x2﹣4x+4﹣4)+y2+6y+9+11=2(x﹣2)2﹣8+(y+3)2+11=2(x﹣2)2+(y+3)2+3································································································6分∵2(x﹣2)2≥0,(y+3)2≥0,··························································································7分∴当x=2,y=﹣3时,2x2+y2﹣8x+6y+20有最小值,最小值是3············································8分(3)∵a2+b2=8a+6b﹣25,∴a2﹣8a+16+b2﹣6b+9=0,∴(a﹣4)2+(b﹣3)2=0······························································································9分∴a﹣4=0,b﹣3=0,∴a=4,b=3··············································································································10分∵4﹣3<c<4+3,∴1<c<7,∵c为正整数,∴c最大取6················································································································11分∴△ABC周长的最大值=3+4+6=13,∴△ABC周长的最大值为13···························································································12分27.(本题12分)解:【问题原型】证明:在△ABC中,点E,F分别为AC,BC的中点∴EF∥AB,且EF=12AB··································································································2分在Rt△ACD中,点E为AC的中点∴DE=12AC·················································································································4分∵AB=AC,∴DE=EF···················································································································5分【探究】解:∵AC平分∠BAD,EF∥AB,DE=12AC=AE=EC第8页(共9页)学科网(北京)股份有限公司∴∠BAC=∠DAC,∠CEF=∠BAC∠DEC=2∠DAC=∠BAD······························································································7分∵∠DEF=90°∴∠CEF+∠DEC=∠BAC+2∠DAC=90°∴∠BAC=∠DAC=30°·································································································9分∴∠BAD=60°············································································································10分【应用】四边形ABCD的面积为:332··············································································12第8页(共9页)学科网(北京)股份有限公司第9页(共9页)学科网(北京)股份有限公司

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发布时间:2022-07-08 14:50:02 页数:9
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