2019-2020学年广东省惠州一中八年级(下)开学数学试卷 ( 解析版)

2019-2020学年广东省惠州一中八年级(下)开学数学试卷 ( 解析版)
2019-2020学年广东省惠州一中八年级(下)开学数学试卷 ( 解析版)

2019-2020学年广东省惠州一中八年级(下)开学数学试卷

一.选择题(共10小题)

1.下列各组数中,能作为一个三角形三边边长的是()

A.1,1,2B.1,2,4C.2,3,4D.2,3,5

2.下列平面图形中,不是轴对称图形的是()

A.B.C.D.

3.一个多边形的内角和比外角和多540°,这个多边形为()

A.五边形B.六边形C.七边形D.八边形

4.在平面直角坐标系中,点P(2,﹣3)关于y轴对称的点的坐标是()A.(﹣2,﹣3)B.(﹣2,3)C.(2,3)D.(2,﹣3)

5.在式子,,,中,分式的个数是()

A.1B.2C.3D.4

6.下列运算一定正确的是()

A.(m+n)2=m2+n2B.(mn)3=m3n3

C.(m3)2=m5D.m?m2=m2

7.如图,在△ABC中,DE是AC的垂直平分线,且分别交BC,AC于点D和E,∠B=60°,∠C=25°,则∠BAD为()

A.50°B.70°C.75°D.80°

8.若(2x+5)﹣3有意义,则x满足的条件是()

A.B.C.x≠0D.

9.已知﹣=1,则代数式的值为()

A.3B.1C.﹣1D.﹣3

10.如图,已知点B、C、D在同一条直线上,△ABC和△CDE都是等边三角形.BE交AC

于F,AD交CE于G.则下列结论中错误的是()

A.AD=BE B.BE⊥AC

C.△CFG为等边三角形D.FG∥BC

二.填空题(共7小题)

11.若a、b、c为三角形的三边,且a、b满足+(b﹣2)2=0,第三边c为偶数,则c=.

12.如图:∠DAE=∠ADE=15°,DE∥AB,DF⊥AB,若AE=8,则DF等于.

13.0.000000301用科学记数法表示是.

14.若分式的值为零,则x的值为.

15.若多项式x2﹣mx+9是一个完全平方式,那么m=.

16.按一定规律排列的一列数:21,22,23,25,28,213,…,若x、y、z表示这列数中的连续三个数,猜想x、y、z满足的关系式是.

17.若关于x的分式方程﹣2m=无解,则m的值为.

三.解答题(共8小题)

18.解方程:.

19.分解因式:

(1)

(2)a3﹣2a2b+ab2

20.如图,两条公路相交于点O,在交角侧有A、B两个,现在要建一加油站P,使得加油站P到两条公路的距离和到A、B两个村庄的距离相等,请画出加油站P的位置.(用尺规作图,保留作图痕迹,不写作法和证明过程).

21.先化简,再求值:(﹣y)÷﹣(x﹣2y)(x+y),其中x=﹣1,y=2.

22.阅读探索题:

(1)如图1,OP是∠MON的平分线,以O为圆心任意长为半径作弧,分别交射线ON、OM于C、B两点,在射线OP上任取一点A(点O除外),连接AB、AC.求证:△AOB ≌△AOC.

(2)请你参考以上方法,解答下列问题:

如图2,在Rt△ABC中,∠ACB=90°,∠A=60°,CD平分∠ACB,试判断BC和AC、AD之间的数量关系并证明.

23.某公司计划购买A,B两种型号的机器人搬运材料.已知A型机器人比B型机器人每小时多搬运30kg材料,且A型机器人搬运1000kg材料所用的时间与B型机器人搬运800kg 材料所用的时间相同.

(1)求A,B两种型号的机器人每小时分别搬运多少材料;

(2)该公司计划采购A,B两种型号的机器人共20台,要求每小时搬运材料不得少于2800kg,则至少购进A型机器人多少台?

24.在△ABC中,∠ABC和∠ACB的平分线相交于点O,

(1)若∠ABC=60°,∠ACB=40°,求∠BOC的度数;

(2)若∠ABC=60°,OB=4,且△ABC的周长为16,求△ABC的面积.

25.【知识生成】我们已经知道,通过计算几何图形的面积可以表示一些代数恒等式.例如图1可以得到(a+b)2=a2+2ab+b2,基于此,请解答下列问题:

(1)根据图2,写出一个代数恒等式:.

(2)利用(1)中得到的结论,解决下面的问题:若a+b+c=10,ab+ac+bc=35,则a2+b2+c2=.

(3)小明同学用图3中x张边长为a的正方形,y张边长为b的正方形,z张宽、长分别为a、b的长方形纸片拼出一个面积为(2a+b)(a+2b)长方形,则x+y+z=.【知识迁移】(4)事实上,通过计算几何图形的体积也可以表示一些代数恒等式,图4表示的是一个边长为x的正方体挖去一个小长方体后重新拼成一个新长方体,请你根据图4中图形的变化关系,写出一个代数恒等式:.

参考答案与试题解析

一.选择题(共10小题)

1.下列各组数中,能作为一个三角形三边边长的是()

A.1,1,2B.1,2,4C.2,3,4D.2,3,5

【分析】根据三角形中任意两边之和大于第三边,任意两边之差小于第三边.即可求解.【解答】解:A、1+1=2,不满足三边关系,故错误;

B、1+2<4,不满足三边关系,故错误;

C、2+3>4,满足三边关系,故正确;

D、2+3=5,不满足三边关系,故错误.

故选:C.

2.下列平面图形中,不是轴对称图形的是()

A.B.C.D.

【分析】结合选项根据轴对称图形的概念求解即可.

【解答】解:A、不是轴对称图形,本选项正确;

B、是轴对称图形,本选项错误;

C、是轴对称图形,本选项错误;

D、是轴对称图形,本选项错误.

故选:A.

3.一个多边形的内角和比外角和多540°,这个多边形为()

A.五边形B.六边形C.七边形D.八边形

【分析】根据多边形的内角和公式(n﹣2)?180°,外角和等于360°列出方程求解即可.【解答】解:设多边形的边数是n,

根据题意得,(n﹣2)?180°﹣540°=360°,

解得n=7.

故选:C.

4.在平面直角坐标系中,点P(2,﹣3)关于y轴对称的点的坐标是()A.(﹣2,﹣3)B.(﹣2,3)C.(2,3)D.(2,﹣3)

【分析】根据关于y轴对称点的坐标特点:横坐标互为相反数,纵坐标不变可得答案.【解答】解:点P(2,﹣3)关于y轴对称的点的坐标是(﹣2,﹣3),

故选:A.

5.在式子,,,中,分式的个数是()

A.1B.2C.3D.4

【分析】根据分式的概念:一般地,如果A,B表示两个整式,并且B中含有字母,那么式子叫做分式可得答案.

【解答】解:式子,是分式,共2个,

故选:B.

6.下列运算一定正确的是()

A.(m+n)2=m2+n2B.(mn)3=m3n3

C.(m3)2=m5D.m?m2=m2

【分析】分别根据完全平方公式,积的乘方运算法则,幂的乘方运算法则以及同底数幂的乘法法则逐一判断即可.

【解答】解:(m+n)2=m2+2mn+n2,故选项A不合题意;

(mn)2=m2n2,正确,故选项B符合题意;

(m3)2=m6≠m5,故选项C不符合题意;

m?m2=m3≠m2,故选项D不符合题意.

综上,故选:B.

7.如图,在△ABC中,DE是AC的垂直平分线,且分别交BC,AC于点D和E,∠B=60°,∠C=25°,则∠BAD为()

A.50°B.70°C.75°D.80°

【分析】根据线段垂直平分线的性质得到DA=DC,根据等腰三角形的性质得到∠DAC =∠C,根据三角形内角和定理求出∠BAC,计算即可.

【解答】解:∵DE是AC的垂直平分线,

∴DA=DC,

∴∠DAC=∠C=25°,

∵∠B=60°,∠C=25°,

∴∠BAC=95°,

∴∠BAD=∠BAC﹣∠DAC=70°,

故选:B.

8.若(2x+5)﹣3有意义,则x满足的条件是()

A.B.C.x≠0D.

【分析】根据负整数指数幂的意义即可求出答案.

【解答】解:由题意可知:2x+5≠0,

x=,

故选:B.

9.已知﹣=1,则代数式的值为()

A.3B.1C.﹣1D.﹣3

【分析】由﹣=1利用分式的加减运算法则得出m﹣n=﹣mn,代入原式=计算可得.

【解答】解:∵﹣=1,

∴﹣=1,

则=1,

∴mn=n﹣m,即m﹣n=﹣mn,

则原式=

=﹣3,

故选:D.

10.如图,已知点B、C、D在同一条直线上,△ABC和△CDE都是等边三角形.BE交AC

于F,AD交CE于G.则下列结论中错误的是()

A.AD=BE B.BE⊥AC

C.△CFG为等边三角形D.FG∥BC

【分析】A、证明△ACD≌△BCE即可得出答案;

B、根据等边三角形性质得出AB=BC,只有F为AC中点时,才能推出AC⊥BE.

C、由△ACG≌△BCF,推出CG=CF,根据∠ACG=60°即可证明;

D、根据等边三角形性质得出∠CFG=∠ACB=60°,根据平行线的判定推出即可.【解答】解:A、∵△ABC和△CDE均为等边三角形,

∴AC=BC,EC=DC,

∠ACB=∠ECD=60°,

∴∠ACD=∠ECB,

在△ACD与△BCE中,

∵,

∴△ACD≌△BCE(SAS),

∴AD=BE,正确,故本选项错误;

B、根据已知不能推出F是AC中点,即AC和BF不垂直,所以AC⊥BE错误,故本选项正确;

C、△CFG是等边三角形,理由如下:

∵∠ACG=180°﹣60°﹣60°=60°=∠BCA,

∵△ACD≌△BCE,

∴∠CBE=∠CAD,

在△ACG和△BCF中

∵,

∴△ACG≌△BCF(ASA),

∴CG=CF,

又∵∠ACG=60°

∴△CGH是等边三角形,正确,故本选项错误;

D、∵△CFG是等边三角形,

∴∠CFG=60°=∠ACB,

∴FG∥BC,正确,故本选项错误;

故选:B.

二.填空题(共7小题)

11.若a、b、c为三角形的三边,且a、b满足+(b﹣2)2=0,第三边c为偶数,则c=10.

【分析】先根据非负数的性质求出a和b的值,再根据三角形三边关系求出c的取值范围,进而求出c的值.

【解答】解:∵a、b满足+(b﹣2)2=0,

∴a=10,b=2,

∵a、b、c为三角形的三边,

∴8<c<12,

∵第三边c为偶数,

∴c=10.

故答案为:10.

12.如图:∠DAE=∠ADE=15°,DE∥AB,DF⊥AB,若AE=8,则DF等于4.

【分析】作DG⊥AC,根据DE∥AB得到∠BAD=∠ADE,再根据∠DAE=∠ADE=15°得到∠DAE=∠ADE=∠BAD,求出∠DEG=15°×2=30°,再根据30°的角所对的直角边是斜边的一半求出GD的长,然后根据角平分线的性质求出DF.

【解答】解:作DG⊥AC,垂足为G.

∵DE∥AB,

∴∠BAD=∠ADE,

∵∠DAE=∠ADE=15°,

∴∠DAE=∠ADE=∠BAD=15°,

∴∠DEG=15°×2=30°,

∴ED=AE=8,

∴在Rt△DEG中,DG=DE=4,

∴DF=DG=4.

故答案为:4.

13.0.000000301用科学记数法表示是 3.01×10﹣7.

【分析】绝对值小于1的负数也可以利用科学记数法表示,一般形式为a×10﹣n,与较大数的科学记数法不同的是其所使用的是负指数幂,指数由原数左边起第一个不为零的数字前面的0的个数所决定.

【解答】解:0.000000301=3.01×10﹣7.

故答案为:3.01×10﹣7.

14.若分式的值为零,则x的值为3.

【分析】分式的值为零:分子等于零,且分母不等于零,由此得到3﹣|x|=0且x+3≠0,从而得到x的值.

【解答】解:依题意得:3﹣|x|=0且x+3≠0,

解得x=3.

故答案是:3.

15.若多项式x2﹣mx+9是一个完全平方式,那么m=±6.

【分析】根据首末两项是x和3的平方可得,中间一项为加上或减去它们乘积的2倍.【解答】解:∵多项式x2﹣mx+9是一个完全平方式,

∴mx=±2?x?3,

∴m=±6.

16.按一定规律排列的一列数:21,22,23,25,28,213,…,若x、y、z表示这列数中的连续三个数,猜想x、y、z满足的关系式是xy=z.

【分析】首项判断出这列数中,2的指数各项依次为1,2,3,5,8,13,…,从第三个数起,每个数都是前两数之和;然后根据同底数的幂相乘,底数不变,指数相加,可得这列数中的连续三个数,满足xy=z,据此解答即可.

【解答】解:∵21×22=23,22×23=25,23×25=28,25×28=213,…,

∴x、y、z满足的关系式是:xy=z.

故答案为:xy=z.

17.若关于x的分式方程﹣2m=无解,则m的值为或.【分析】根据方程无解的两种可能:①分母为0,由此可得x=3,②分母不等于0,化简后所得的整式方程无解.

【解答】解:①分母为0,即是x=3,

将方程可转化为x﹣2m(x﹣3)=3m﹣1,

当x=3时,m=.

②分母不为0,整理得:x﹣2mx+6m=3m﹣1,

x=,

因为方程无解,所以2m﹣1=0,

解得:m=.

故答案为:或.

三.解答题(共8小题)

18.解方程:.

【分析】观察可得2﹣x=﹣(x﹣2),所以可确定方程最简公分母为:(x﹣2),然后去分母将分式方程化成整式方程求解.注意检验.

【解答】解:方程两边同乘以(x﹣2),

得:x﹣3+(x﹣2)=﹣3,

解得x=1,

检验:x=1时,x﹣2≠0,

∴x=1是原分式方程的解.

19.分解因式:

(1)

(2)a3﹣2a2b+ab2

【分析】(1)原式利用平方差公式计算即可求出值;

(2)原式提取公因式,再利用完全平方公式分解即可.

【解答】解:(1)原式=(x+y)(x﹣y);

(2)原式=a(a2﹣2ab+b2)=a(a﹣b)2.

20.如图,两条公路相交于点O,在交角侧有A、B两个,现在要建一加油站P,使得加油站P到两条公路的距离和到A、B两个村庄的距离相等,请画出加油站P的位置.(用尺规作图,保留作图痕迹,不写作法和证明过程).

【分析】作线段AB的垂直平分线MN交两条公路所成的角的角平分线OT于点P,点P 即为所求.

【解答】解:如图,点P即为所求.

21.先化简,再求值:(﹣y)÷﹣(x﹣2y)(x+y),其中x=﹣1,y=2.【分析】原式利用分式的混合运算顺序和运算法则化简,再将x、y的值代入计算可得.【解答】解:原式=(﹣)÷﹣(x2+xy﹣2xy﹣2y2)

=?(x+y)﹣x2+xy+2y2

=﹣xy﹣x2+xy+2y2

=﹣x2+2y2,

当x=﹣1、y=2时,

原式=﹣(﹣1)2+2×22

=﹣1+8

=7.

22.阅读探索题:

(1)如图1,OP是∠MON的平分线,以O为圆心任意长为半径作弧,分别交射线ON、OM于C、B两点,在射线OP上任取一点A(点O除外),连接AB、AC.求证:△AOB ≌△AOC.

(2)请你参考以上方法,解答下列问题:

如图2,在Rt△ABC中,∠ACB=90°,∠A=60°,CD平分∠ACB,试判断BC和AC、AD之间的数量关系并证明.

【分析】(1)根据以O为圆心任意长为半径作弧,交射线ON,OM为C,B两点,OP 是∠MON的平分线,运用SAS判定△AOB≌△AOC即可;

(2)先截取CE=CA,连接DE,根据SAS判定△CAD≌△CED,得出AD=DE,∠A=∠CED=60°,AC=CE,进而得出结论BC=AC+AD;

【解答】(1)证明:在△AOB和△AOC中,,

∴△AOB≌△AOC(SSS).

(2)在CB上截取CE=CA,

∵CD平分∠ACB,

∴∠ACD=∠BCD,

在△ACD和△ECD中,,

∴△ACD≌△ECD(SAS),

∴∠CAD=∠CED=60°,

∵∠ACB=90°,

∴∠B=30°,

∴∠EDB=30°,

即∠EDB=∠B,

∴DE=EB,

∵BC=CE+BE,

∴BC=AC+DE,

∴BC=AC+AD.

23.某公司计划购买A,B两种型号的机器人搬运材料.已知A型机器人比B型机器人每小时多搬运30kg材料,且A型机器人搬运1000kg材料所用的时间与B型机器人搬运800kg 材料所用的时间相同.

(1)求A,B两种型号的机器人每小时分别搬运多少材料;

(2)该公司计划采购A,B两种型号的机器人共20台,要求每小时搬运材料不得少于2800kg,则至少购进A型机器人多少台?

【分析】(1)设B型机器人每小时搬运x千克材料,则A型机器人每小时搬运(x+30)千克材料,根据A型机器人搬运1000kg材料所用的时间与B型机器人搬运800kg材料所用的时间相同建立方程求出其解就可以得出结论.

(2)设购进A型机器人a台,根据每小时搬运材料不得少于2800kg列出不等式并解答.【解答】解:(1)设B型机器人每小时搬运x千克材料,则A型机器人每小时搬运(x+30)千克材料,

根据题意,得=,

解得x=120.

经检验,x=120是所列方程的解.

当x=120时,x+30=150.

答:A型机器人每小时搬运150千克材料,B型机器人每小时搬运120千克材料;

(2)设购进A型机器人a台,则购进B型机器人(20﹣a)台,

根据题意,得150a+120(20﹣a)≥2800,

解得a≥.

∵a是整数,

∴a≥14.

答:至少购进A型机器人14台.

24.在△ABC中,∠ABC和∠ACB的平分线相交于点O,

(1)若∠ABC=60°,∠ACB=40°,求∠BOC的度数;

(2)若∠ABC=60°,OB=4,且△ABC的周长为16,求△ABC的面积.

【分析】(1)利用角平分线的定义得到∠OBC=30°,∠OCB=20°,然后根据三角形内角和计算∠COB的度数;

(2)过O作OD⊥AB于D点,OE⊥AC于E,OF⊥BC于F,连接AO,如图,先利用含30度的直角三角形三边的关系得到OD=2,再根据角平分线的性质得到OE=OF=2,然后根据三角形面积公式计算.

【解答】解:(1)∵BO、CO分别平分∠ABC和∠ACB,

∵∠ABC=60°,∠ACB=40°

∴∠OBC=30°,∠OCB=20°,

∴∠COB=180°﹣(30°+20°)=130°;

(2)过O作OD⊥AB于D点,OE⊥AC于E,OF⊥BC于F,连接AO,如图,

∵∠ABC=60°,OB=4

∴∠OBD=30°,

∴OD=OB=2,

∵∠ABC和∠ACB的平分线相交于点O,

∴OE=OF=2,

∵S△ABC=S△AOB+S△AOC+S△BOC

=×2×AB+×2×AC+×2×BC

=AB+BC+AC,

又∵△ABC的周长为16,

∴S△ABC=16.

25.【知识生成】我们已经知道,通过计算几何图形的面积可以表示一些代数恒等式.例如图1可以得到(a+b)2=a2+2ab+b2,基于此,请解答下列问题:

(1)根据图2,写出一个代数恒等式:(a+b+c)2=a2+b2+c2+2ab+2ac+2bc.

(2)利用(1)中得到的结论,解决下面的问题:若a+b+c=10,ab+ac+bc=35,则a2+b2+c2=30.

(3)小明同学用图3中x张边长为a的正方形,y张边长为b的正方形,z张宽、长分别为a、b的长方形纸片拼出一个面积为(2a+b)(a+2b)长方形,则x+y+z=9.【知识迁移】(4)事实上,通过计算几何图形的体积也可以表示一些代数恒等式,图4表示的是一个边长为x的正方体挖去一个小长方体后重新拼成一个新长方体,请你根据图4中图形的变化关系,写出一个代数恒等式:x3﹣x=(x+1)(x﹣1)x..

【分析】(1)依据正方形的面积=(a+b+c)2;正方形的面积=a2+b2+c2+2ab+2ac+2bc,可得等式;

(2)依据a2+b2+c2=(a+b+c)2﹣2ab﹣2ac﹣2bc,进行计算即可;

(3)依据所拼图形的面积为:xa2+yb2+zab,而(2a+b)(a+2b)=2a2+4ab+ab+2b2=2a2+5b2+2ab,即可得到x,y,z的值.

(4)根据原几何体的体积=新几何体的体积,列式可得结论.

【解答】解:(1)由图2得:正方形的面积=(a+b+c)2;正方形的面积=a2+b2+c2+2ab+2ac+2bc,

∴(a+b+c)2=a2+b2+c2+2ab+2ac+2bc,…(2分)

故答案为:(a+b+c)2=a2+b2+c2+2ab+2ac+2bc;

(2)∵(a+b+c)2=a2+b2+c2+2ab+2ac+2bc,

∵a+b+c=10,ab+ac+bc=35,

∴102=a2+b2+c2+2×35,

∴a2+b2+c2=100﹣70=30,

故答案为:30;…(4分)

(3)由题意得:(2a+b)(a+2b)=xa2+yb2+zab,

∴2a2+5ab+2b2=xa2+yb2+zab,

∴,

∴x+y+z=9,

故答案为:9;…(6分)

(4)∵原几何体的体积=x3﹣1×1?x=x3﹣x,新几何体的体积=(x+1)(x﹣1)x,

∴x3﹣x=(x+1)(x﹣1)x.

故答案为:x3﹣x=(x+1)(x﹣1)x.…(8分)

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