济南高新区2021-2022学年第二学期八年级学业质量抽测

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