新人教版八年級(jí)上整式的乘法綜合檢測試卷及答案1

1、整式的乘法同步測試一、選擇題：1下列各式中，正確的是（ ）At2·t3 = t5      Bt4+t2 = t 6      Ct3·t4 = t12      Dt5·t5 = 2t5 2下列計(jì)算錯(cuò)誤的是（ ）Aa2·(a)2 = a4               

2、     B(a)2·(a)4 = a6C(a3)·(a)2 = a5                   D(a)·(a)2 = a33下列計(jì)算中，運(yùn)算正確的個(gè)數(shù)是（ ）5x3x3 = x3             &

3、#160;       3m·2n = 6m+nam+an = am+n                   xm+1·xm+2 = xm·xm+3A1      B 2      C3    

4、60; D44計(jì)算a6(a2)3的結(jié)果等于（ ）Aa11      Ba 12      Ca14      Da365下列各式計(jì)算中，正確的是（ ）A(a3)3 = a6    B(a5)4 = a 20    C(a)53 = a15    D(a)23 = a66下列各式計(jì)算中，錯(cuò)誤的是（ ）A(m6)6 = m36   

5、 B(a4)m = (a 2m) 2    Cx2n = (xn)2    Dx2n = (x2)n7下列計(jì)算正確的是（ ）A(xy)3 = xy3            B(2xy)3 = 6x3y3C(3x2)3 = 27x5         D(a2b)n = a2nbn8下列各式錯(cuò)誤的是（ ）A(23)4 = 212  

6、                B( 2a)3 = 8a3C(2mn2)4 = 16m4n8            D(3ab)2 = 6a2b29下列計(jì)算中，錯(cuò)誤的是（ ）Amn·m2n+1 = m3n+1          &

7、#160;      B(am1)2 = a 2m2C(a2b)n = a2nbn                     D(3x2)3 = 9x6 10下列計(jì)算中，錯(cuò)誤的是（ ）A(2ab2)2·( 3a2b)3 = 108a8b7B(2xy)3·(2xy)2 = 32x5y5C(m2n)(mn2)2 =m4n4D(xy)2

8、(x2y) = x4y311下列計(jì)算結(jié)果正確的是（ ）A(6ab2 4a2b)3ab = 18ab2 12a2bB(x)(2x+x21) = x32x2+1C(3x2y)(2xy+3yz1) = 6x3y29x2y2z2+3x2yD(a3b)2ab =a4bab212若(x2)(x+3) = x2+a+b，則a、b的值為（ ）Aa = 5，b = 6             Ba = 1，b = 6Ca = 1，b = 6     &#

9、160;       Da = 5，b = 6二、解答題：1計(jì)算(1)( 5a3b2)·(3ab 2c)·( 7a2b)；(2) 2a2b3·(mn)5·ab2·(nm)2+a2(mn)·6ab2；(3) 3a2(ab2b)( 2a2b23ab)( 3a)；(4)(3x25y)(x2+2x3)2當(dāng)x = 3時(shí)，求8x2(x2)(x+1)3(x1)(x2)的值3把一個(gè)長方形的長減少3，寬增加2，面積不變，若長增加1，寬減少1，則面積減少6，求長方形的面積4(x+my1

10、)(nx2y+3)的結(jié)果中x、y項(xiàng)的系數(shù)均為0，求 3m+n之值參考答案：一、選擇題1.A說明： t4與t2不是同類項(xiàng)，不能合并，B錯(cuò)；同底數(shù)冪相乘，底不變，指數(shù)相加，所以t3·t4 = t3+4 = t7t12，C錯(cuò)；t5t5 = t5+5 = t102t5，D錯(cuò)；t2t3 = t2+3 = t5，A正確；答案為A2.C說明：a2·(a)2 = a2·a2 = a2+2 = a4，A計(jì)算正確；(a)2·(a)4 = a2·a4 = a2+4 = a6，B計(jì)算正確；(a3)·(a)2 = a3·a2 = a5a5，C計(jì)算錯(cuò)誤

11、；(a)·(a)2 = a·a2 = a3，D計(jì)算正確；所以答案為C3.A說明：5x3x3 = (51)x3 = 4x3 x3 ，錯(cuò)誤； 3m與2n 不是同底數(shù)冪，它們相乘把底數(shù)相乘而指數(shù)相加顯然是不對(duì)的，比如m = 1，n = 2，則 3m·2n = 31·22 = 3·4 = 12，而 6m+n = 61+2 = 63 = 21612，錯(cuò)誤；am與an只有在m = n時(shí)才是同類項(xiàng)，此時(shí)am+an = 2amam+n，而在mn時(shí)，am與an無法合并，錯(cuò)；xm+1·xm+2 = xm+1+m+2 = xm+m+3 = xm·

12、xm+3，正確；所以答案為A4.B說明：a6(a2)3 = a6·a2×3 = a6·a6 = a6+6 = a12，所以答案為B5.D說明：(a3)3 = a3×3 = a9，A錯(cuò)；(a5)4 = a5×4 = a20，B錯(cuò)；(a)53 = (a)5×3 = (a)15 = a15，C錯(cuò)；(a)23 = (a)2×3 = (a)6 = a6，D正確，答案為D6.D說明：(m6)6 = m6×6 = m36，A計(jì)算正確；(a4)m = a 4m，(a 2m)2 = a 4m，B計(jì)算正確；(xn)2 = x2n，C計(jì)

13、算正確；當(dāng)n為偶數(shù)時(shí)，(x2)n = (x2)n = x2n；當(dāng)n為奇數(shù)時(shí)，(x2)n = x2n，所以D不正確，答案為D7.D說明：(xy)3 = x3y3，A錯(cuò)；(2xy)3 = 23x3y3 = 8x3y3，B錯(cuò)；(3x2)3 = (3)3(x2)3 = 27x6，C錯(cuò)；(a2b)n = (a2)nbn = a2nbn，D正確，答案為D8.C說明：(23)4 = 23×4 = 212，A中式子正確；( 2a)3 = (2) 3a3 = 8a3，B中式子正確；(3ab)2 = 32a2b2 = 9a2b2，C中式子錯(cuò)誤；(2mn2)4 = 24m4(n2)4 = 16m4n8，D

14、中式子正確，所以答案為C9.D說明：mn·m2n+1 = mn+2n+1 = m3n+1，A中計(jì)算正確；(am1)2 = a2(m1) = a 2m2，B中計(jì)算正確； (a2b)n = (a2)nbn = a2nbn，C中計(jì)算正確；(3x2)3 = (3)3(x2)3 = 27x6，D中計(jì)算錯(cuò)誤；所以答案為D10.C說明：(2ab2)2·( 3a2b)3 = (2) 2a2(b2)2·(3)3(a2)3b3 = 4a2b4·(27)a6b3 = 108a2+6b4+3 = 108a8b7，A中計(jì)算正確；(2xy)3·(2xy)2 = (2xy)

15、3·(2xy)2 = (2xy)3+2 = (2xy)5 = 25x5y5 = 32x5y5，B中計(jì)算正確；(m2n)( mn2)2 =m2n() 2m2(n2)2 =m2n·m2n4 =m2+2n1+4 =m4n5，C中計(jì)算錯(cuò)誤；(xy)2(x2y) = ()2x2y2·x2y =x2y2·x2y = x4y3，D中計(jì)算正確，所以答案為C11.D說明：(6ab2 4a2b)3ab = 6ab2·3ab 4a2b·3ab = 18a2b3 12a3b，A計(jì)算錯(cuò)誤；(x)(2x+x21) = x·2x+(x)·x2(

16、x) = 2x2x3+x = x32x2+x，B計(jì)算錯(cuò)誤；(3x2y)(2xy+3yz1) = (3x2y) (2xy)+(3x2y) 3yz(3x2y) = 6x3y29x2y2z+3x2y，C計(jì)算錯(cuò)誤；(a3b)2ab = (a3) 2ab(b)2ab =a4bab2，D計(jì)算正確，所以答案為D12.B說明：因?yàn)?x2)(x+3) = xx2x+3x6 = x2+x6，所以a = 1，b = 6，答案為B二、解答題1.解：(1)( 5a3b2)·(3ab 2c)·( 7a2b) = (5)×(3)×(7)(a3·a·a2)(b2&#

17、183;b2·b)c = 105a6b 5c(2) 2a2b3·(mn)5·ab2·(nm)2+a2(mn)·6ab2 = (2·)·(a2·a)·(b3·b2)(mn)5·(mn)2+( ·6)(a2·a)(mn)b2 = a3b5(mn)7+ 2a3b2(mn)(3) 3a2(ab2b)( 2a2b23ab)( 3a) = 3a2·ab2 3a2b+ 2a2b2· 3a3ab· 3a= a3b2 3a2b+ 6a3b2 9a2b = 7a3b2 12a2b(4)(3x25y)(x2+2x3) = 3x2·x25y·x2+3x2·2x5y·2x+3x2·(3)5y·(3)= 3x45x2y+6x310xy9x2+15y= 3x4+6x35x2y9x210xy+15y2. 解：8x2(x2)(x