《整式的乘除》基础测试

2014-5-11 0:12:28 下载本试卷

《整式的乘除》基础测试

(一)填空题(每小题2分,共计20分)

1.x10=(-x32·_________=x12÷x(  )【答案】x4;2.

2.4(mn3÷(nm2=___________.【答案】4(mn).

3.-x2·(-x3·(-x2=__________.【答案】x7

4.(2ab)()=b2-4a2.【答案】-2ab

5.(ab2=(ab2+_____________.【答案】-4ab

6.(2+p0=_________;4101×0.2599=__________.【答案】10;16.

7.20×19=(  )·(  )=___________.【答案】20+,20-,399

8.用科学记数法表示-0.=___________.

  【答案】-3.08×10-5

9.(x-2y+1)(x-2y-1)2=(  )2-(  )2=_______________.

  【答案】x-2y,1x2-4xy+4y

10.若(x+5)(x-7)=x2mxn,则m=__________,n=________.【答案】-2,35.

(二)选择题(每小题2分,共计16分)

11.下列计算中正确的是…………………………………………………………………(  )

(A)an·a2a2n  (B)(a32a5  (C)x4·x3·xx7  (D)a2n-3÷a3na3n-6

  【答案】D.

12.x2m+1可写作…………………………………………………………………………(  )

(A)(x2m+1   (B)(xm2+1   (C)x·x2m   (D)(xmm+1【答案】C.

13.下列运算正确的是………………………………………………………………(  )

(A)(-2ab)·(-3ab3=-54a4b4

(B)5x2·(3x32=15x12

(C)(-0.16)·(-10b23=-b7

(D)(2×10n)(×10n)=102n【答案】D.

14.化简(anbmn,结果正确的是………………………………………………………(  )

(A)a2nbmn  (B)  (C)  (D)

  【答案】C.

15.若ab,下列各式中不能成立的是………………………………………………(  )

(A)(ab2=(-ab2    (B)(ab)(ab)=(ba)(ba

(C)(ab2n=(ba2n    (D)(ab3=(ba3

  【答案】B.

16.下列各组数中,互为相反数的是……………………………………………………(  )

(A)(-2)3与23     (B)(-2)2与2-2 

(C)-33与(-3   (D)(-3)3与(3

  【答案】D.

17.下列各式中正确的是………………………………………………………………(  )

(A)(a+4)(a-4)=a2-4    (B)(5x-1)(1-5x)=25x2-1

(C)(-3x+2)2=4-12x+9x2   (D)(x-3)(x-9)=x2-27

  【答案】C.

18.如果x2kxab=(xa)(xb),则k应为…………………………………(  )

(A)ab  (B)ab  (C)ba  (D)-ab

  【答案】B.

(三)计算(每题4分,共24分)

19.(1)(-3xy23·(x3y2; 【答案】-x9y8

(2)4a2x2·(-a4x3y3)÷(-a5xy2);【答案】ax4y

(3)(2a-3b2(2a+3b2;【答案】16a4-72a2b2+81b4

(4)(2x+5y)(2x-5y)(-4x2-25y2); 【答案】625y4-16x4

(5)(20an-2bn-14an-1bn+1+8a2nb)÷(-2an-3b);【答案】-10abn-1+7a2bn-4an+3

(6)(x-3)(2x+1)-3(2x-1)2

    【答案】-10x2+7x-6.

20.用简便方法计算:(每小题3分,共9分)

(1)982

    【答案】(100-2)2=9604.

(2)899×901+1;

    【答案】(900-1)(900+1)+1=9002=810000.

(3)(2002·(0.49)1000

    【答案】(2·(2000·(0.7)2000

(四)解答题(每题6分,共24分)

21.已知a2+6ab2-10b+34=0,求代数式(2ab)(3a-2b)+4ab的值.

【提示】配方:(a+3)2+(b-5)2=0,a=-3,b=5,

【答案】-41.

22.已知ab=5,ab=7,求a2abb2的值.

【答案】[(ab2-2ab]=ab2ab

a2abb2=(ab2-3ab=4.

23.已知(ab2=10,(ab2=2,求a2b2ab的值.

  【答案】a2b2[(ab2+(ab2]=6,

ab[(ab2+(ab2]=2.

24.已知a2b2c2abbcac,求证abc

  【答案】用配方法,a2b2c2abbcac=0,∴ 2(a2b2c2abacbc)=0,

即(ab2+(bc2+(ca2=0.∴ abc

(五)解方程组与不等式(25题3分,26题4分,共7分)

25.

  【答案】

26.(x+1)(x2x+1)-xx-1)2<(2x-1)(x-3).

  【答案】x>-