【文档说明】苏教版数学七年级下册期中复习试卷一（含答案）.doc，共(12)页，136.500 KB，由MTyang资料小铺上传

转载请保留链接：https://www.ichengzhen.cn/view-30419.html

以下为本文档部分文字说明：

-1-2021年苏教版数学七年级下册期中复习试卷一、选择题1．计算(－a3)2的结果是（）A．a6B．－a6C．－a5D．a52．下列运算正确的是（）A．a＋2a=3a2B．a3·a2=a5C．(a4)2=a6D．a3＋a4=a73．每到四月，许多地方杨絮、柳絮如雪花般漫天飞舞

，人们不堪其扰，据测定，杨絮纤维的直径约为0.0000105m，该数值用科学记数法表示为（）A．1.05×105B．1.05×10－5C．－1.05×105D．105×10－74．下列图形中，由AB∥CD，能得到∠1=∠2的是（）21ABCDA．21

DCBAB．21CADBC．21ABCDD．5．下列从左到右的变形，属于因式分解的是（）A．(x＋3)(x－3)=x2－9B．x2－2x－1=x(x－2)－1C．8a2b3=2a2·4b3D．x2－2x＋1=(x－1)26.下列整式乘法中，能运用平方差公式进行运算的是（）A．(2a＋b)(

2b－a)B．(m＋b)(m－b)C．(a－b)(b－a)D．(－x－b)(x＋b)7．下列命题中的真命题．．．是（）A．相等的角是对顶角B．内错角相等C．如果a3=b3，那么a2=b2D．两个角的两边分别平行，则这两个角相等8.比

较255、344、433的大小（）-2-A.255＜344＜433B．433＜344＜255C．255＜433＜344D．344＜433＜255二、填空题9．计算：（13）-2=．10．计算：(x＋1)(x－5)的结果是．

11．因式分解：2a2－8=．12.若am=3，an=2，则am－2n的值为．13.命题“两直线平行，同旁内角互补”的逆命题是．14．若2a＋b=－3，2a－b=2，则4a2－b2=.15．将两张长方形纸片按

如图所示摆放，使其中一张长方形纸片的一个顶点恰好落在另一张长方形纸片的一条边上，则∠1＋∠2=°．16．如图，将边长为6cm的正方形ABCD先向上平移3cm，再向右平移1cm，得到正方形A′B′C′D′，此时阴影部分的面积为cm2.17常见的“幂的运算”有：①同底数幂的乘

法，②同底数幂的除法，③幂的乘方，④积的乘方．在“(a3·a2)2=(a3)2(a2)2=a6·a4=a10”的运算过程中，运用了上述幂的运算中的．18．如图a是长方形纸带，∠DEF=28°，将纸带沿EF折叠成

图b，再沿BF折叠成图c，则图c中的∠CFE=°．ADACBAEACABAFADACDBAEAFCAGBAABAEAFCAGBAA图a图b图c(第18题)（第16题）ABCDA′D′C′B′（第14题）12-3--4-三

、解答题19．计算：（1）(－2a2)3＋2a2·a4－a8÷a2；（2）2a(a－b)(a＋b).20．因式分解：（1）xy2－x；（2）3x2－6x＋3.21．先化简，再求值：4(x－1)2－(2x＋3)(2x－3)，其中x=－1．22．画图并填空：如图，方格纸中每个小正方形的边长

都为1．在方格纸内将△ABC经过一次平移后得到△A′B′C′，图中标出了点D的对应点D′．（1）根据特征画出平移后的△A′B′C′；（2）利用网格的特征，画出AC边上的高BE并标出画法过程中的特征点；（3）△A′B′C′的面积为．BCAD'D(第22题)-5-23．在下

列解题过程的空白处填上适当的内容（推理的理由或数学表达式）如图，在△ABC中，已知∠ADE=∠B，∠1=∠2，FG⊥AB于点G.求证CD⊥AB.证明:∵∠ADE=∠B（已知），∴（），∵DE∥BC（已证），∴（），又∵∠1=∠2（已知），∴（），∴CD∥FG（），∴（两直线平行同位角相等），∵F

G⊥AB（已知），∴∠FGB=90°（垂直的定义）.即∠CDB=∠FGB=90°，∴CD⊥AB.（垂直的定义）.24．证明：平行于同一条直线的两条直线平行．已知：如图，．求证：．证明：B(第23题)ACDEFG12cba（第24题）-6-25．发现与探索。（1

）根据小明的解答将下列各式因式分解①a2－12a＋20②（a－1）2－8（a－1）＋7③a2－6ab＋5b2（2）根据小丽的思考解决下列问题：①说明：代数式a2－12a＋20的最小值为－16．②请仿照小丽的思考解释代数式－（a＋1）2＋8的最大值为8，并求代数式－a2＋12a－8的最大值．小明的解

答：a2－6a＋5=a2－6a＋9－9＋5=（a－3）2－4=（a－5）（a－1）小丽的思考：代数式（a－3）2＋4无论a取何值（a－3）2都大于等于0，再加上4，则代数式（a－3）2＋4大于等于4，则（a－3）2＋4有最小值为4.-7-26．模型与应用。【模型】（1）如图①，已知AB∥CD，求证

∠1＋∠MEN＋∠2=360°.【应用】（2）如图②，已知AB∥CD，则∠1+∠2+∠3+∠4+∠5+∠6的度数为．如图③，已知AB∥CD，则∠1+∠2+∠3+∠4+∠5+∠6＋„＋∠n的度数为．（3）如图④，已知AB∥CD，∠AM1M2的角平分线M1O与∠CMnMn－1的角平分线MnO交于点

O，若∠M1OMn=m°．在（2）的基础上，求∠2+∠3+∠4+∠5+∠6＋„„＋∠n－1的度数．（用含m、n的代数式表示）③12n－1nABCD124356ABCDEFGHMN②①DCEBA21MN-8-参考答案一、选择题题号12345678答案ABB

BDBCC二、填空题9．910．x2－4x－511．2（a－2）（a＋2）12.3413.同旁内角互补，两直线平行14．－615．90°16．1517④③①18．96°三、解答题19．（8分）计算：（1）(－2a2)3＋2

a2·a4－a8÷a2；解原式=－8a6＋2a6－a6„„„„„„„„„„„„„„„„„„„„„„„„„3分=－7a6„„„„„„„„„„„„„„„„„„„„„„„„„„4分（2）2a(a－b)(a＋b).解原式=2a（a2－b2）„„„„„

„„„„„„„„„„„„„„„„„„„„„2分=2a3－2ab2„„„„„„„„„„„„„„„„„„„„„„„„„„4分20．（8分）因式分解：（1）xy2－x；解原式=x（y2－1）„„„„„„„„„„„„„„„„„„„„„„„„„

„2分=x（y－1）（y＋1）„„„„„„„„„„„„„„„„„„„„„„„„„„4分（2）3x2－6x＋3.解原式=3（x2－2x＋1）„„„„„„„„„„„„„„„„„„„„„„„„„„2分=3（x－1）2„„„„„„„„„„„„„„„„„„„„„„„„„„4分21．（6

分）先化简，再求值：4(x－1)2－(2x＋3)(2x－3)，其中x=－1．4(x－1)2－(2x＋3)(2x－3)解原式=4(x2－2x＋1)－(4x2－9)„„„„„„„„„„„„„„„„„„„„2分=4x2－8x＋4－4x2＋9„„„„„„„

„„„„„„„„„„„„„4分=－8x＋13„„„„„„„„„„„„„„„„„„„„„„„„„„5分当x=－1时，原式=21„„„„„„„„„„„„„„„„„„„„6分-9-22．（6分）（1）作图正确2分，„„„

„„„„„„„„„„„„2分（2）作图正确2分，„„„„„„„„„„„„„„„4分（3）3．„„„„„„„„„„„„„„„6分23．（8分）证明:∵∠ADE=∠B（已知），∴DE∥BC（同位角相等两直

线平行），„„„„„„„2分∵DE∥BC（已证），∴∠1=∠DCF（两直线平行内错角相等），„„„„„„„4分又∵∠1=∠2（已知），∴∠DCF=∠2（等量代换），„„„„„„„6分∴CD∥FG（同位角相等两直线平行），„„„„„„„7分∴∠BDC=∠BGF（两直线平行同位角相等），„„„„„

„„8分∵FG⊥AB（已知），∴∠FGB=90°（垂直的定义）.即∠CDB=∠FGB=90°，∴CD⊥AB.（垂直的定义）.24．（8分）证明：平行于同一条直线的两条直线平行．已知：如图，已知b∥a，c∥a．„„„„„„„„„„„„„„„„„„„„1分求证：b∥c．„„„„

„„„„„„„„„„„„„„„„2分证明：作直线DF交直线a、b、c，交点分别为D、E、F，„„„„„„„„„„„„„3分∵a∥b，∴∠1=∠2，„„„„„„„„„„„„„5分又∵a∥c，∴∠1=∠3，„„„„„„„

„„„„„„7分∴∠2=∠3，∴b∥c．„„„„„„„„„„„„„„„„„„8分cba（第24题）DEF123-10-25．（10分）发现与探索。（1）①a2－12a＋20解原式=a2－12a＋36－36＋20=（a－

6）2－42=（a－10）（a－2）„„„„„„„„„„„„„„„„„„„„1分②（a－1）2－8（a－1）＋12解原式=（a－1）2－8（a－1）＋16－16＋12=（a－5）2－22=（a－7）（a－3）„„„„„„„„„„„„„„„„„„„„„„3分③a2－6ab＋5b2解原式

=a2－6ab＋9b2－9b2＋5b2=（a－3b）2－4b2=（a－5b）（a－b）„„„„„„„„„„„„„„„„„„„„„„5分（2）根据小明的发现结合小丽的思考解决下列问题.①说明：代数式a2－12a＋20的最小值为﹣16.a2－12a＋20解原式=a2－12a＋36－36＋20

=（a－6）2－16„„„„„„„„„„„„„„„„„„„„6分无论a取何值（a－6）2都大于等于0，再加上﹣16，则代数式（a－6）2－16大于等于－16，则a2－12a＋20的最小值为－16.„„„„„„„„„„„„„„„7分②无论a取何值－（a＋1）2都小于等于0

，再加上8，则代数式－（a＋1）2＋8小于等于8，则－（a＋1）2＋8的最大值为8.„„„„„„„„„„„„„„„„„„8分﹣a2＋12a－8.解原式=﹣（a2－12a＋8）=﹣（a2－12a＋36－36＋8

）=﹣（a－6）2＋36－8-11-=﹣（a－6）2＋28无论a取何值﹣（a－6）2都小于等于0，再加上28，则代数式﹣（a－6）2＋28小于等于28，则﹣a2＋12a－8.的最大值为28.„„„„„„„„„„„„„„„„10分26．（10分）模型与应用。【模型】（1）如图①，已知AB∥CD，求

证∠1＋∠2＋∠MEN=360°.证明：过点E作EF∥CD，„„„„„„„„1分∵AB∥CD，∴EF∥AB，„„„„„„„„2分∴∠1＋∠MEF=180°，同理∠2＋∠NEF=180°„„„„„„„„3分∴∠1＋∠2＋∠

MEN=360°„„„„„„„„„„„„„„„„„4【应用】（2）900°„„„„„„„„„„„„„„„„„„5分180°(n－1)„„„„„„„„„„„„„„„„„„6分（3）解：过点O作SR∥AB，∵AB∥CD，∴SR∥CD，∴∠AM1O=∠AM1OR同理∠C

MnO=∠MnOR∴∠AM1O＋∠CMnO=∠M1OR＋∠MnOR，∴∠AM1O＋∠CMnO=∠M1OMn=m°，„„„„„„„„„„„„„8分∵M1O平分∠AM1M2，∴∠AM1M2=2∠AM1O，同

理∠CMnMn-1=2∠CMnO，∴∠AM1M2＋∠CMnMn-1=2∠AM1O＋2∠CMnO=2∠M1OMn=2m°，„„„9分又∵∠AM1E＋∠2+∠3+∠4+∠5+∠6＋„„＋∠n－1＋∠CMnMn-1=180°(n－1)，∠2+∠

3+∠4+∠5+∠6＋„＋∠n－1=(180n－180－2m)°„„„„„„„„10分①DCEBA21MFN④M2Mn－1M1Mn2n－1ABCDOSR-12-