2021年福建省泉州市下学期八年级期中测试数学试题卷+答案

这是一份2021年福建省泉州市下学期八年级期中测试数学试题卷+答案，文件包含2021年春泉州八下期中考数学试卷docx、2021年春泉州八下期中考数学答案及评分建议docx等2份试卷配套教学资源，其中试卷共11页， 欢迎下载使用。
2020-2021学年第二学期八年级期中测试-数学试题卷参考答案及评分建议一、选择题：本题共10小题，每小题4分，共40分．12345678910BADADBDCCB二、填空题：本题共6小题，每小题4分，共24分．11．y=3x+4 12．213．20 14．315．－2≤m≤2 16．－2三、解答题：本题共9小题，共86分．17．（本小题满分8分）解：原式=3－2+1－3···················································6分=－1．······················································8分 18．（本小题满分8分）解：方程两边同乘(x2－4)，得4+(x+3)(x+2)=(x－1)(x－2)，整理得10+5x=－3x+2，解得x=－1．检验：当x=－1时，x2－4≠0，∴x=－1是原方程的解．·············································8分 19．（本小题满分8分）解：原式====． 6分当x=时，原式．···············································8分    20．（本小题满分8分）（1）证明：∵四边形ABCD是平行四边形，∴AD∥BC，AB∥CD，AD=BC，∴∠DAE=∠F，∠D=∠ECF．∵E是CD的中点，∴DE=CE．在△ADE和△FCE中，，∴△ADE≌△FCE(AAS)，∴AD=CF，∴BC=CF，∴C是线段BF的中点．········································5分（2）解：由（1）可知：△ADE≌△FCE，∴AE=EF=3，∴AF=6．∵四边形ABCD是平行四边形，CD=8，∴CD=AB=8．由（1）可得：BF=2BC=10．在△ABF中，AB2+AF2=BF2，∴△ABF是直角三角形，且BF为斜边，∴∠BAF=90°．··············································8分 21．（本小题满分8分）解：（1）设该商家第一次购进的羽绒服有x件，则第二次购进的羽绒服有2x件．由题意得：，解得x=240．经检验，x=240是原方程的解．答：该商家第一次购进的羽绒服有240件．···························4分（2）设每件羽绒服的标价为a元．由题意得：0.6a×50+(240+240×2－50)a－(26 400+57 600)≥(26 400+57 600)×25%，解得a≥150．答：每件羽绒服的标价至少为150元．·······························8分 22．（本小题满分10分）解：（1）y=2x  0≤x≤4  ·················································3分（2）20··························································6分（3）此次消毒有效．理由如下：当y=2时，，解得x=16，当y=2时，y=2x=2，解得x=1，∵16－1=15>14，∴此次消毒有效．··············································10分23．（本小题满分10分）解：（1）n=2m·······················································4分（2）如图，过点E作EF⊥BC，垂足为F，则EF=6－3=3．∵△BDE是等边三角形，∴∠FED=30°，∴ED=2DF，∴DF=BF=，即n－m=，由（1）可知：n=2m，∴m=，n=，∴此时点D的坐标是(6，)，∴k=．······················································8分∵，∴B(6，)．∴当k=时，△BDE为等边三角形，此时点B的坐标是(6，)．··············10分 24．（本小题满分12分）解：（1）设直线AB的解析式为y=kx+b(k≠0)．将(0，－6)，(3，0)代入y=kx+b中，得，解得，∴直线AB的解析式为y=2x－6．····································3分（2）①当点C在x轴上方时，=3m－9；·················································5分当点C在x轴下方时，=9－3m．·················································7分②∵直线OC将△AOB的面积分为1∶2两部分，∴点C在线段AB上，且BC=或BC=，∴C是线段AB的三等分点．∵C(m，n)，A(0，－6)，B(3，0)，∴m=1，n=－4或m=2，n=－2，∴点C的坐标是(1，－4)或(2，－2)．······························12分 25．（本小题满分14分）解：（1）  －6·······················································4分（2）设直线与双曲线的另一个交点为C，则OA=OC，∴C(3，－2)，∵S△ABP=2S△ABO，则点P不在线段AO上，∴当点P在AO的延长线上时，S△ABP=2S△ABO，即点P与点C重合，此时P(3，－2)；当点P在OA的延长线上时，S△ABP=2S△ABO，即PA=2AO，此时P(－9，6)，综上所述，满足条件的点P的坐标是(3，－2)或(－9，6)．················8分（3）当点M在OA下方时，若∠AOM=45°，则点M在第三象限，此时不存在满足条件的点M，当点M在OA上方，如图，将OA绕点O顺时针旋转90°得到OA′，则A′(2，3)，连结AA′，取AA′的中点D，直线OD在第二象限交双曲线于点M，此时∠AOM=45°．由A(－3，2)，A′(2，3)可知D(，)，∴直线OD的解析式为y=－5x．由，解得或．∵点M在第二象限，∴点M的坐标为(，)．···········································14分