2018-2019学年度福州市八年级第一学期质量调研数学试卷

八年级数学—1—（共4页）1准考证号：姓名：1（在此卷上答题无效）2018-2019学年度福州市八年级第一学期质量调研数学试卷本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，完卷时间120分钟，满分150分．注意事项：1．答题前，考生务必在试题卷、答题卡规定位置填写本人准考证号、姓名等信息．考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致．2．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．非选择题答案用0.5毫米黑色墨水签字笔在答题卡上相应位置书写作答，在试题卷上答题无效．3．作图可先使用2B铅笔画出，确定后必须用0.5毫米黑色墨水签字笔描黑．4．考试结束，考生必须将试题卷和答题卡一并交回．第Ⅰ卷一、选择题（本题共10小题，每小题4分，共40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．下列所给的四个小篆字中为轴对称图形的是()ABCD2．下列式子是最简二次根式的是()A．2B．4C．8D．123．若一个多边形的内角和与它的外角和相等，则这个多边形是()A．三角形B．四边形C．五边形D．六边形4．下列计算正确的是()A．5510aaaB．33(2)6aaC．824aaaD．3412()aa八年级数学—2—（共4页）5．如图，已知ABC△，点D，E在边BC上，ABD△≌ACE△．下列结论不．一定．．成立的是()A．ABACB．BDCEC．CACDD．BAECAD6．下列多项式中，能分解出因式m1的是()A．221mmB．21mC．2mmD．2(1)2(1)1mm7．下列代数式变形正确的是()A．221xyxyxyB．22xyxyC．11111()xyxyyxD．222()xyxyxyxy8．计算22(3)(4)xx的结果是()A．72xB．1C．27xD．19．如图，ABC△，AB＞AC＞BC，边AB上存在一点P，使得PAPCAB．下列描述正确的是()A．P是AC的垂直平分线与AB的交点B．P是BC的垂直平分线与AB的交点C．P是∠ACB的平分线与AB的交点D．P是以点B为圆心，AC长为半径的弧与边AB的交点10．若2()21acb，2()2019acb，则2222abcab的值是()A．1020B．1998C．2019D．2040第Ⅱ卷注意事项：1．用0.5毫米黑色墨水签字笔在答题卡上相应位置书写作答，在试题卷上作答，答案无效．2．作图可先用2B铅笔画出，确定后必须用0.5毫米黑色墨水签字笔描黑．二、填空题（本题共6小题，每小题4分，共24分）11．若根式2x在实数范围内有意义，则x的取值范围是．12．若分式211xx的值是0，则实数x的值是．13．点（3，2）关于x轴的对称点的坐标是．BACDEABC八年级数学—3—（共4页）14．如图，E是矩形ABCD中AD边上一点，将AEB△沿BE折叠得到FEB△．若119BED，则CBF是度．15．若330xy，则82xy的结果是．16．如图，D是等边三角形ABC中BA延长线上一点，连接CD，E是BC上一点，且DEDC，若63BDBE，23CE，则这个等边三角形的边长是．三、解答题（本题共9小题，共86分．解答应写出文字说明、证明过程或演算步骤）17．（本小题满分8分）分解因式：（1）29xyy；（2）244mm．18．（本小题满分8分）计算：23532339mmmm．19．（本小题满分8分）先化简，再求值：22()()()2mnmnmnm，其中21m，21n．20．（本小题满分8分）如图，ABC△≌BAD△，BC与AD交于点E．求证：ABE△是等腰三角形．21．（本小题满分8分）如图，在RtABC△中，∠ACB90°，30ABC，CD平分ACB．（1）尺规作图：作线段AB的垂直平分线l；（要求：保留作图痕迹，不写作法）（2）记直线l与AB，CD的交点分别是点E，F．当4AC时，求EF的长．22．（本小题满分10分）某商场11月初花费15000元购进一批某品牌英语点读笔，因深受顾客喜爱，销售一空．该商场于12月初又花费24000元购进一批同品牌英语点读笔，且所购数量是11月初的1.5倍，但每支进价涨了10元．（1）求商场11月初购进英语点读笔多少支？（2）11月份商场该品牌点读笔每支的售价是270元，若12月份购买的点读笔全部售完，且所获利润是11月份利润的1.2倍，求12月份该品牌点读笔每支的售价？ABCDEACDBABCDEABCDEF八年级数学—4—（共4页）23．（本小题满分10分）求证：两边分别相等且其中一组等边的对角相等的两个锐角三角形全等．（要求：根据题意写出已知，求证，并证明）（友情提醒：可将锐角三角形的问题转化为直角三角形的问题处理）24．（本小题满分13分）已知：12xM，21xNx．（1）当x＞0时，判断MN与0的关系，并说明理由；（2）设2yNM．①当3y时，求x的值；②若x是整数，求y的正整数值．25．（本小题满分13分）已知60ABC，ABBC，D是BC边上一点，延长AD到点E，使得ADDE，连接CE，过点D作BC的垂线，交CE的垂直平分线于点F，连接BF．（1）如图1，当点D与点C重合时，证明：2BFDF；（2）如图2，当点D不与B，C两点重合时，（1）中的结论是否还成立？并说明理由．ABCABCABCDEFABC()DEF图1图2八年级数学—5—（共4页）2018-2019学年度福州市八年级第一学期质量调研数学试题答案及评分标准评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分参考制定相应的评分细则．2．对于计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．4．只给整数分数．选择题和填空题不给中间分．一、选择题：每小题4分，满分40分．1．A2．A3．B4．D5．C6．C7．D8．A9．B10．A二、填空题：每小题4分，满分24分．11．x≥212．113．（3，2）14．3215．1816．1033三、解答题：本题共9小题，共86分．解答应写出文字说明、证明过程和演算步骤．17．解：（1）原式22(3)yx·········································································1分(3)(3)yxx．···································································4分（2）原式2(44)mm····································································1分2(2)m．········································································4分【注：直接写出最后结果只得3分．】18．解：（解法一）原式22(3)3(3)9[](3)(3)(3)(3)53mmmmmmmm···························3分八年级数学—6—（共4页）2(3)3(3)(3)(3)(3)(3)53mmmmmmm····································4分(3)(3)53(3)(3)53mmmmmm··········································6分1．············································································8分（解法二）原式2392()3353mmmm··················································1分(3)(3)(3)(3)32353353mmmmmmmm·······················2分2(3)3(3)5353mmmm·······················································4分263953mmm····························································6分5353mm·······································································7分1．············································································8分19．解：原式2222222mnmmnnm·······················································4分2mn．······················································································5分当21m，21n时，································································6分原式2(21)(21)····································································7分222[(2)1]2(21)2．··························································································8分【注：直接代入求值，正确得3分．】20．证明：∵ABC△≌BAD△，∴CBADAB，·········································································4分八年级数学—7—（共4页）∴EAEB，··················································································8分∴ABE△是等腰三角形．【注：只给出“EAEB”，没有给出最后结论，不扣分；但由“CBADAB”直接得到“ABE△是等腰三角形”，扣2分】21．（1）···················································································································2分如图所示，直线l是所求作的线段AB的垂直平分线．·································3分（2）解：连接EC．∵90ACB，30B，4AC，∴12ACAB，60A，∴8AB．·····················································4分∵EF是AB的垂直平分线，∴1