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八年数学试卷共四页第1页2021年苏科版数学八年级下册期末复习试卷一、选择题1.下列图形中，既是轴对称图形，又是中心对称图形的是（）A.B.C.D.2.下列各式：a－b2，x－3x，5＋yπ，a＋ba－b，1n(x－y)中，是分式的共有（）A.1个B.2个C.3个D.4个3.下列式

子从左到右变形一定正确的是（）A.ab=a2b2B.ab=a＋1b＋1C.ab=a－1b－1D.a2ab=ab4.若2x－1在实数范围内有意义，则x的取值范围是„（）A.x≥12B.x≥－12C.x＞12D.x≠125.下列

计算：（1）(2)2=2，（2）(－2)2=2，（3）(－23)2=12，（4）(2＋3)(2－3)=－1，其中结果正确的个数为（）A.1B.2C.3D.46.一只不透明的袋子中装有4个黑球、2个白球，每个球除颜色外都

相同，从中任意摸出3个球，下列事件为必然事件的是（）A.至少有1个球是黑球B.至少有1个球是白球C.至少有2个球是黑球D.至少有2个球是白球7.已知P1(x1，y1)，P2(x2，y2)，P3(x3，y3)是反

比例函数y=6x的图像上三点，且y1＜y2＜0＜y3，则x1，x2，x3的大小关系是（）A.x1＜x2＜x3B.x3＜x2＜x1C.x2＜x1＜x3D.x2＜x3＜x18.关于x的分式方程7xx－1＋5=2m－1x－1有增根，则m的值为（）A.

5B.4C.3D.1八年数学试卷共四页第2页9.如图，在菱形ABCD中，∠BCD=110°，AB的垂直平分线交对角线AC于点F，E为垂足，连接DF，则∠CDF等于„（）A.15°B.25°C.45°D.

55°10.如图，在平面直角坐标系中，直线y=33x＋2与x轴交于点A，与y轴交于点B，将△ABO沿直线AB翻折,点O的对应点C恰好落在双曲线y=kx(k≠0)上，则k的值为（）A.－4B.－2C.－23D.－33二、填空题：11.若分式x－3x值为0，则x的值为.12.若

最简二次根式2a－3与5是同类二次根式，则a的值为.13.若反比例函数y=k－2x的图像经过第二、四象限，则k的取值范围是.14.关于x的分式方程x＋mx－2＋2m2－x=3的解为正实数，则实数m的取值范围是.15.如图

，点O是矩形ABCD的对角线AC的中点，OM∥AB交AD于点M，若OM=2，BC=6，则OB的长为.16.如图，正方形ABCD的边长为6，点G在对角线BD上（不与点B、D重合），GF⊥BC于点F，连接AG，若∠AGF=105°，则线段BG=.CFEDBA（第9题）yxOABCMDABOC八年数学

试卷共四页第3页17.如图，在平面直角坐标系中，点A的坐标为（1，0），等腰直角三角形ABC的边AB在x轴的正半轴上，∠ABC=90°，点B在点A的右侧，点C在第一象限.将△ABC绕点A逆时针旋转75°，若点C的对应点E恰好落在y轴上，则边AB的长为.18.如图

，已知点A是一次函数y=23x(x≥0)图像上一点，过点A作x轴的垂线，B是上一点（B在A上方），在AB的右侧以AB为斜边作等腰三角形ABC，反比例函数y=kx(x＞0)的图像过点B、C，若△OAB的面积为5，则△ABC的面积是.三、解答题1

9.计算：（1）6×33－(12)－2＋|1－2|；（2）(312－213＋48)÷3；（3）1m－2－4m2－4；（4）解方程：1x－2－1－x2－x=－3.ADGBFCCEOAByxBOACxy八年数学试卷共四页第4页20.先

化简，再求值：x－1x÷(x－1x)，其中x=3－1.21.今年4月23日是第23个“世界读书日”.某校围绕学生日人均阅读时间这一问题，对初二学生进行随机抽样调查.如图是根据调查结果绘制成的统计图（不完整），

请你根据图中提供的信息解答下列问题：（1）本次抽样调查的样本容量是多少？（2）请将条形统计图补充完整.（3）在扇形统计图中，计算出日人均阅读时间在1~1.5小时对应的圆心角度数.（4）根据本次抽样调查，试估计我市12000名初二学生

中日均阅读时间在0.5~1.5小时的有多少人.90756045301500.511.5人数t180°1≤t＜1.520%0≤t＜0.50.5≤t＜1日人均阅读时间各时间段人数所占的百分比八年数学试卷共四页第5页

22.如图，在□ABCD中，E、F为对角线BD上的两点，且∠BAE=∠DCF.求证：BF=DE.23.如图，方格纸中每个小正方形的边长都是1个单位长度.Rt△ABC的三个顶点A（－2，2），B（0，5），C（0，2）.（1）将△ABC以点C为旋转中心旋

转180°，得到△A1B1C，请画出的图形△A1B1C.（2）平移△ABC，使点A的对应点A2坐标为（－2，－6），请画出平移后对应的△A2B2C2.（3）请用无刻度的直尺在第一、四象限内画出一个以A1B2为边，面积是7的矩

形A1B1EF.（保留作图痕迹，不写作法）（4）若将△A1B1C绕某一点旋转可得到△A2B2C2，请直接写出旋转中心的坐标.FEABCDBACOxy八年数学试卷共四页第6页24.某公司在工程招标时，接到甲、乙两个工程队的投标书.工程领导小组根据甲、乙两队的投标书测算：每

施工一天，需付甲工程队工程款1.5万元，付乙工程队工程款1.1万元.甲队单独完成此工程刚好如期完工，乙队单独完成此工程要比规定工期多用5天，若甲、乙两队合作4天，剩下的工程由乙独做也正好如期完工.（1）求甲、乙两队单独完成此项工程各需要多少天？（2）由于任务紧迫，公

司要求工程至少提前7天完成，问怎样安排甲、乙两个工程队施工所付施工费最少？最少施工费是多少万元？（施工天数不满一天以一天计）25.如图，在平面直角坐标系中，菱形ABCD的顶点C与原点O重合，点B在y轴的正半轴上，点A在反比例函数y=

kx（k＞0，x＞0）的图像上，点D的坐标为（2，32），设AB所在直线解析式为y=kx＋b（a≠0），（1）求k的值，并根据图像直接写出不等式ax＋b＞kx的解集；（2）若将菱形ABCD沿x轴正方向平移m个单位，

①当菱形的顶点B落在反比例函数的图像上时，求m的值；②在平移中，若反比例函数图像与菱形的边AD始终有交点，求m的取值范围.DOBAxy（C）八年数学试卷共四页第7页26.在矩形ABCD中，AB=4，AD=3，现将纸片折叠，点D的对应点记为点P，折痕为EF（点E、F是折痕与矩形的边的交

点），再将纸片还原.（1）若点P落在矩形ABCD的边AB上（如图1）.①当点P与点A重合时，∠DEF=°，当点E与点A重合时，∠DEF=°.②当点E在AB上时，点F在DC上时（如图2），若AP=72，求四边形EPFD的周长.（2）若点F与点C重合，点E在AD上，线段BA

与线段FP交于点M（如图3），当AM=DE时，请求出线段AE的长度.（3）若点P落在矩形的内部（如图4），且点E、F分别在AD、DC边上，请直接写出AP的最小值.APBCFDEAEPDFCBDCEMAPBDFCEPAB（图1）（图2）（图3）（图4）八年数学试卷共四页第8页参考答案1．

C2．C3．D4．A5．D6．A7．C8．B9．A10．D11．312．413．2k14．62mm且15．1316．3117．218．5319.（本题满分16分）解：(1)原式2421·································

··································································3分225．······························

··············································································4分(2)原式2(63343)33·······································

··········································3分283··········································································

·········································4分（3）原式142(2)(2)mmm·······························································

···················1分24(2)(2)mmm····································································································2分12m····

············································································································

····4分（4）1）1(1)3(2)xx···························································································2分∴2x经检验是原方程

的增根，原方程无解·································································4分20.（本题满分4分）解:原式=xxxx112=)1)(1(1xxxxx················

·······················································1分=11x···················································

········································································2分当13x时，原式=1131=31=33·················

···················································4分21.（本题满分8分）解：（1）样本容量是：30÷20%=150；····················································

·································2分（2）日人均阅读时间在0.5～1小时的人数是：150-30-45=75（人）．画图略····················4分（3）人均阅读时间在1～1.5小时对应的圆心角度数是：360°×45150=108°

；···············6分（4）12000×75＋45150=9600（人）．·························································································8分22.（本题满分8分）

证明：∵□ABCD∴AB∥CD，AB=CD···················································································2分∴∠ABE=

∠CDF·························································································································

····4分在△ABE和△DCF中，BAEDCFABCDABECDF∠∠∠∠∴△ABE≌△DCF（ASA），······························································

·································6分∴BE=DF···················································································

············································7分∴BE+EF=DF+EF即BF=DE····························································

····································8分八年数学试卷共四页第9页23.（本题满分8分）（1）如图；（2）如图；（3）如图；(4)（0，-2）；·································

················各2分xyFEC2B2A2B1A1CBAO或xyFEC2B2A2B1A1CBAO24.（本题满分8分）解：⑴设甲队单独完成此项工程需x天，则乙队单独完成此项工程需（x+5）天．由题意，得：114

4155xxxx··············································································2分解得：x=20．···············

·················································································································3分经检验：x=20是原分式

方程的解．∴（x+5）=25．答：甲队单独完成此项工程需20天，则乙队单独完成此项工程需25天；························4分（2）设甲队施工a天，乙队施工b天，需支付工程费w万元由题意，得：12025ab······························

············································································5分当a=13，b=9时，w=29.4；当a=12，b=1

0时，w=29；当a=11，b=12时，w=29.7；当a=10，b=13时，w=29.3························································7分∴当甲

施工12天，乙施工10天，即在要求的13天内甲队施工12天，乙队施工10天，支付工程费最少为29万元.···················································

························································8分25.（本题满分10分）解：（1）延长AD交x轴于F，由题意得AF⊥x轴∵点D的坐标为（2，32），∴OF=2，DF=3

2，∴OD=52，∴AD=52····················································································································1分∴点

A坐标为（2，4），∴k=xy=2×4=8，··················································································

3分由图像得解集：2x；·············································································································5分（2）①将菱

形ABCD沿x轴正方向平移m个单位,则平移后B′坐标为(m,52),因B′落在函数8yx（x＞0）的图象上,则165m.······································

······················7分②将菱形ABCD沿x轴正方向平移m个单位，使得点D落在函数8yx（x＞0）的图象D′八年数学试卷共四页第10页点处，∴点D′的坐标为3(2,)2m····································

··················································8分∵点D′在8yx的图象上∴3822m，解得：103m，·······································

·········9分∴1003m.·························································································

··································10分26.（本题满分12分）(1)①90，45··············································

·················································································2分②设EF与PD交于点O，由折叠知EF垂直平分PD

∴DO=PO,EF⊥PD··························································································

·······························3分∵矩形ABCD∴DC∥AB∴∠FDO=∠EPO∵∠DOF=∠EOP∴△DOF≌△POE∴DF=PE∵DF∥PE∴四边形DEPF是平行四边形·············································

····································4分∵EF⊥PD∴四边形DEPF是菱形········································································

·····················5分当AP=72时，设菱形边长为x，则72AEx，DE=x在Rt△ADE中，222ADAEDE∴22273()2xx····································

·················6分∴8528x∴菱形的周长=857······································································

·························7分(2)连接EM，设AE=x由折叠知PE=DE,∠CDB=∠EPM=90°,CD=CP=4∵AM=DE∠A=90°EM=EM∴Rt△AEM≌Rt△PME(HL)·····················

··················································································8分∴AE=PM=x，∴CM=4-x，BM=AB-AM=AB-DE=4-(3-x)=1+x在Rt△BCM中，222BMBCCM

∴2223(1)(4)xx得x=0.6····························································································10分(3)AP的最小值=5

-4=1············································································································12分

.