【人教版】2016年八年级上:第14章《整式的乘除与因式分解》全章检测题(含答案)01

【人教版】2016年八年级上:第14章《整式的乘除与因式分解》全章检测题(含答案)01
【人教版】2016年八年级上:第14章《整式的乘除与因式分解》全章检测题(含答案)01

金题教育

第十四章检测题

一、选择题(每小题3分,共30分)

1.下列运算正确的是( )

A .3a 2-2a 2=1

B .(a 2)3=a 5

C .a 2·a 4=a 6

D .(3a)2=6a 2

2.下列计算错误的是( )

A .(5-2)0=1

B .28x 4y 2÷7x 3=4xy 2

C .(4xy 2-6x 2y +2xy)÷2xy =2y -3x

D .(a -5)(a +3)=a 2-2a -15

3.下列因式分解正确的是( )

A .a 4b -6a 3b +9a 2b =a 2b(a 2-6a +9)

B .x 2-x +14=(x -12

)2 C .x 2-2x +4=(x -2)2 D .4x 2-y 2=(4x +y)(4x -y)

4.将(2x)n -81分解因式后得(4x 2+9)(2x +3)(2x -3),则n 等于( )

A .2

B .4

C .6

D .8

5.若m =2100,n =375,则m ,n 的大小关系是( )

A .m>n

B .m

C .m =n

D .无法确定

6.已知a +b =3,ab =2,则a 2+b 2的值为( )

A .3

B .4

C .5

D .6

7.计算:(a -b +3)(a +b -3)=( )

A .a 2+b 2-9

B .a 2-b 2-6b -9

C .a 2-b 2+6b -9

D .a 2+b 2-2ab +6a +6b +9

8.在边长为a 的正方形中挖去一个边长为b 的小正方形(a>b)(如图甲),把余下的部分拼成一个长方形(如图乙),根据两个图形中阴影部分的面积相等,可以验证( )

A .(a +b)2=a 2+2ab +b 2

B .(a -b)2=a 2-2ab +b 2

C .a 2-b 2=(a +b)(a -b)

D .(a +2b)(a -b)=a 2+ab -2b 2

9.若x 2+mx -15=(x -3)(x +n),则m ,n 的值分别是( )

A .4,3

B .3,4

C .5,2

D .2,5

10.观察下列各式及其展开式:

(a +b)2=a 2+2ab +b 2

(a +b)3=a 3+3a 2b +3ab 2+b 3

(a +b)4=a 4+4a 3b +6a 2b 2+4ab 3+b 4

(a +b)5=a 5+5a 4b +10a 3b 2+10a 2b 3+5ab 4+b 5

请你猜想(a +b)10的展开式第三项的系数是( )

A .36

B .45

C .55

D .66

二、填空题(每小题3分,共24分)

11.计算:(x -y)(x 2+xy +y 2)=

12.分解因式:(a -b)2-4b 2=

13.若(2x +1)0=(3x -6)0,则x 的取值范围是.

14.已知a m =3,a n =2,则a 2m -3n = .

15.若一个正方形的面积为a 2+a +14

,则此正方形的周长为 . 16.已知实数a ,b 满足a 2-b 2=10,则(a +b)3·(a -b)3的值是 .

17.已知△ABC 的三边长为整数a ,b ,c ,且满足a 2+b 2-6a -4b +13=0,则c 为

18.观察下列各式,探索发现规律:22-1=1×3;32-1=2×4;42-1=3×5;52-1=4×6;….按此规律,第n 个等式为

三、解答题(共66分)

19.(8分)计算:

(1)y(2x -y)+(x +y)2; (2)(-2a 2b 3)÷(-6ab 2)·(-4a 2b).

20.(8分)用乘方公式计算:

(1)982; (2)899×901+1.

21.(12分)分解因式:

(1)18a 3-2a ; (2)ab(ab -6)+9; (3)m 2-n 2+2m -2n.

22.(10分)先化简,再求值:

(1)(2+a)(2-a)+a(a -5b)+3a 5b 3÷(-a 2b)2,其中ab =-12

(2)[(x +2y)(x -2y)-(x +4y)2]÷4y ,其中x =-5,y =2.

23.(8分)如图,某市有一块长为(3a +b)米,宽为(2a +b)米的长方形地块,规划部门计划将阴影部分进行绿化,中间修建一座雕像,求绿化的面积是多少平方米?并求出当a =3,b =2时的绿化面积.

24.(8分)学习了分解因式的知识后,老师提出了这样一个问题:设n为整数,则(n+7)2-(n-3)2的值一定能被20整除吗?若能,请说明理由;若不能,请举出一个反例.

25.(12分)阅读材料并回答问题:

课本中多项式与多项式相乘是利用平面几何图形中的面积来表示的,例如:(2a+b)(a +b)=2a2+3ab+b2就可以用如图①②所示的图形的面积来表示.

(1)请写出如图③所示的图形的面积表示的代数恒等式;

(2)试画出一个几何图形,使它的面积能表示为(a+b)(a+3b)=a2+4ab+3b2;

(3)请仿照上述方法另写一个含有a,b的代数恒等式,并画出与之对应的几何图形.

C C B B B C C C

D B

x 3-y 3 (a +b )(a -3b ) __x ≠-12且x ≠2__ __98

__ __4a +2__

__1000__ __(n +1)2-1=n (n +2)__

解:原式=x 2+4xy 解:原式=-43

a 3

b 2 解:原式=9604 原式=810000

解:原式=2a (3a +1)(3a -1) 解:原式=(ab -3)2 解:原式=(m -n )(m +n +2)

解:原式=4-2ab ,当ab =-12

时,原式= 解:原式=-2x -5y ,当x =-5,y =2时,原式=0

解:绿化面积为(3a +b )(2a +b )-(a +b )2=5a 2+3ab (平方米).当a =3,b =2时,5a 2+3ab =63,即绿化面积为63平方米

解:(n +7)2-(n -3)2=(n +7+n -3)(n +7-n +3)=20(n +2),∴一定能被20整除 解:(1)(a +2b )(2a +b )=2a 2+5ab +2b 2

(2)如图④

(3)(答案不唯一)(a +2b )(a +3b )=a 2+5ab +6b 2,如图⑤

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