江苏省南京市秦淮区2021-2022学年八年级下学期期末考试数学试卷

这是一份江苏省南京市秦淮区2021-2022学年八年级下学期期末考试数学试卷，文件包含八下期末pdf、八下期末答案docx、答题卷pdf等3份试卷配套教学资源，其中试卷共11页， 欢迎下载使用。
2021－2022学年度第二学期第二阶段学业质量监测试卷八年级数学参考答案及评分标准说明：本评分标准每题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，参照本评分标准的精神给分．一、选择题（每小题2分，共计12分）题号123456答案CCBAAD二、填空题（每小题2分，共计20分）7．抽样调查      8．x≥1         9．       10．k＞3            11．0.6012．②①③      13．128         14．2        15．＜x＜2         16．＋三、解答题（本大题共10小题，共计68分）17．（1）解：方程两边同乘x(x－1) ，得9(x－1)＝8xx＝9···························································3分检验：当x＝9时，x(x－1)≠0∴x＝9是原方程的解．·············································4分（2）解：方程两边同乘，得 ······························································7分检验：当时，＝0∴是增根，原方程无解．···········································8分18．（1）·（a≥0）＝····························································2分＝4a3·························································4分（2）×＋÷－＝2＋－·······················································7分＝···························································8分19．÷＝÷····························································2分＝· ····························································4分＝······························································5分当x＝6时，原式＝＝－················································6分20．解：（1）200  ··················································2分       （2）补全条形统计图··················································3分                                      D选项所对应扇形的圆心角度数=72°·······················4分    （3）3000×＝900（人）答：该学校学生中喜爱合唱团和动漫创作社的总人数为900人．··············6分21．解：设甲同学每小时做x面彩旗，则乙同学每小时做(x－5)面彩旗．根据题意，列方程得 ＝．..........................................3分解得 x＝30.....................................................4分经检验，x＝30是原方程的解，且符合题意x－5＝30－5＝25．...............................................5分答：甲同学每小时做30面彩旗，乙同学每小时做25面彩旗．..................6分22．四边形EFGH是矩形．·············································1分理由如下：连接BD，AC相交于点M，AC交EH于点N．∵点E，H分别是AB，AD的中点，∴EH＝BD，EH∥BD，··········································2分同理，FG＝BD，FG∥BD，HG∥AC，······························3分∴ EH＝FG，EH∥FG，∴ 四边形EFGH是平行四边形．···································4分又∵四边形ABCD是菱形，∴ AC⊥BD，·················································5分∴∠AMD＝90°，∵ EH∥BD，∴∠ANH＝∠AMD＝90°，∵ HG∥AC，∴∠EHG＝∠ANH＝90°，∴ 平行四边形是矩形．············································6分23．解：（1）∵反比例函数的图像经过点A（，n），B（2，3）．∴m＝2×3＝－3n，∴m＝6，n＝－2．∴反比例函数表达式为．··············· 2分∵一次函数的图像经过点A（，），B（2，3）．∴解得  ∴一次函数表达式为；··············································· 3分（2）－3≤x＜0或x≥2·············································· 5分（3）点P的坐标为（3，0）或（－5，0）．·································· 7分24．解：（1）∵，∴∵平分，∴∴，∴························································· 1分又∵，∴又∵，∴四边形是平行四边形 … 2分又∵∴平行四边形ABCD是菱形．……………………………………… 3分（2）∵四边形是菱形，对角线、交于点．∴．，，∴．在中，．∴．∵，∴．在中，．为斜边中点．∴．···························································6分25．解：（1）原式＝·················································2分                ＝······················································3分（2）4－＝＝····················································4分－4＝＝························································5分∵4＋＜＋4    ∴＞   即4－＞－4·······································6分26．解：（1） (－3，1)··················································2分（2）设反比例函数表达式为y＝．由题意得，点B′坐标为(－3＋t，1)， 点D′坐标为(－7＋t，3) ．∵点B′和点D′都在此反比例函数图像上，∴，解得．所以t的值为9，这个反比例函数的表达式为y＝．………………………………6分（3）P(，0)，Q(，4)或P(7，0)，Q(3，2)或P(－7，0)，Q(－3，－2) ………9分