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

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）

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            11．0.60

12．②①③      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)＝8x

x＝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