精校版人教版数学八年级上整式的乘法综合检测试卷及答案【1】

上传人:仙*** 文档编号:43277148 上传时间:2021-11-30 格式:DOC 页数:6 大小:215.50KB
返回 下载 相关 举报
精校版人教版数学八年级上整式的乘法综合检测试卷及答案【1】_第1页
第1页 / 共6页
精校版人教版数学八年级上整式的乘法综合检测试卷及答案【1】_第2页
第2页 / 共6页
精校版人教版数学八年级上整式的乘法综合检测试卷及答案【1】_第3页
第3页 / 共6页
点击查看更多>>
资源描述
最新资料最新资料最新资料最新资料最新资料整式的乘法同步测试一、选择题:1下列各式中,正确的是( )At2·t3 = t5      Bt4+t2 = t 6      Ct3·t4 = t12      Dt5·t5 = 2t5 2下列计算错误的是( )Aa2·(a)2 = a4                    B(a)2·(a)4 = a6C(a3)·(a)2 = a5                   D(a)·(a)2 = a33下列计算中,运算正确的个数是( )5x3x3 = x3                     3m·2n = 6m+nam+an = am+n                   xm+1·xm+2 = xm·xm+3A1      B 2      C3      D44计算a6(a2)3的结果等于( )Aa11      Ba 12      Ca14      Da365下列各式计算中,正确的是( )A(a3)3 = a6    B(a5)4 = a 20    C(a)53 = a15    D(a)23 = a66下列各式计算中,错误的是( )A(m6)6 = m36    B(a4)m = (a 2m) 2    Cx2n = (xn)2    Dx2n = (x2)n7下列计算正确的是( )A(xy)3 = xy3            B(2xy)3 = 6x3y3C(3x2)3 = 27x5         D(a2b)n = a2nbn8下列各式错误的是( )A(23)4 = 212                  B( 2a)3 = 8a3C(2mn2)4 = 16m4n8            D(3ab)2 = 6a2b29下列计算中,错误的是( )Amn·m2n+1 = m3n+1                 B(am1)2 = a 2m2C(a2b)n = a2nbn                     D(3x2)3 = 9x6 10下列计算中,错误的是( )A(2ab2)2·( 3a2b)3 = 108a8b7B(2xy)3·(2xy)2 = 32x5y5C(m2n)(mn2)2 =m4n4D(xy)2(x2y) = x4y311下列计算结果正确的是( )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,则a、b的值为( )Aa = 5,b = 6             Ba = 1,b = 6Ca = 1,b = 6             Da = 5,b = 6二、解答题:1计算(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当x = 3时,求8x2(x2)(x+1)3(x1)(x2)的值3把一个长方形的长减少3,宽增加2,面积不变,若长增加1,宽减少1,则面积减少6,求长方形的面积4(x+my1)(nx2y+3)的结果中x、y项的系数均为0,求 3m+n之值参考答案:一、选择题1.A说明: t4与t2不是同类项,不能合并,B错;同底数幂相乘,底不变,指数相加,所以t3·t4 = t3+4 = t7t12,C错;t5t5 = t5+5 = t102t5,D错;t2t3 = t2+3 = t5,A正确;答案为A2.C说明:a2·(a)2 = a2·a2 = a2+2 = a4,A计算正确;(a)2·(a)4 = a2·a4 = a2+4 = a6,B计算正确;(a3)·(a)2 = a3·a2 = a5a5,C计算错误;(a)·(a)2 = a·a2 = a3,D计算正确;所以答案为C3.A说明:5x3x3 = (51)x3 = 4x3 x3 ,错误; 3m与2n 不是同底数幂,它们相乘把底数相乘而指数相加显然是不对的,比如m = 1,n = 2,则 3m·2n = 31·22 = 3·4 = 12,而 6m+n = 61+2 = 63 = 21612,错误;am与an只有在m = n时才是同类项,此时am+an = 2amam+n,而在mn时,am与an无法合并,错;xm+1·xm+2 = xm+1+m+2 = xm+m+3 = xm·xm+3,正确;所以答案为A4.B说明:a6(a2)3 = a6·a2×3 = a6·a6 = a6+6 = a12,所以答案为B5.D说明:(a3)3 = a3×3 = a9,A错;(a5)4 = a5×4 = a20,B错;(a)53 = (a)5×3 = (a)15 = a15,C错;(a)23 = (a)2×3 = (a)6 = a6,D正确,答案为D6.D说明:(m6)6 = m6×6 = m36,A计算正确;(a4)m = a 4m,(a 2m)2 = a 4m,B计算正确;(xn)2 = x2n,C计算正确;当n为偶数时,(x2)n = (x2)n = x2n;当n为奇数时,(x2)n = x2n,所以D不正确,答案为D7.D说明:(xy)3 = x3y3,A错;(2xy)3 = 23x3y3 = 8x3y3,B错;(3x2)3 = (3)3(x2)3 = 27x6,C错;(a2b)n = (a2)nbn = a2nbn,D正确,答案为D8.C说明:(23)4 = 23×4 = 212,A中式子正确;( 2a)3 = (2) 3a3 = 8a3,B中式子正确;(3ab)2 = 32a2b2 = 9a2b2,C中式子错误;(2mn2)4 = 24m4(n2)4 = 16m4n8,D中式子正确,所以答案为C9.D说明:mn·m2n+1 = mn+2n+1 = m3n+1,A中计算正确;(am1)2 = a2(m1) = a 2m2,B中计算正确; (a2b)n = (a2)nbn = a2nbn,C中计算正确;(3x2)3 = (3)3(x2)3 = 27x6,D中计算错误;所以答案为D10.C说明:(2ab2)2·( 3a2b)3 = (2) 2a2(b2)2·(3)3(a2)3b3 = 4a2b4·(27)a6b3 = 108a2+6b4+3 = 108a8b7,A中计算正确;(2xy)3·(2xy)2 = (2xy)3·(2xy)2 = (2xy)3+2 = (2xy)5 = 25x5y5 = 32x5y5,B中计算正确;(m2n)( mn2)2 =m2n() 2m2(n2)2 =m2n·m2n4 =m2+2n1+4 =m4n5,C中计算错误;(xy)2(x2y) = ()2x2y2·x2y =x2y2·x2y = x4y3,D中计算正确,所以答案为C11.D说明:(6ab2 4a2b)3ab = 6ab2·3ab 4a2b·3ab = 18a2b3 12a3b,A计算错误;(x)(2x+x21) = x·2x+(x)·x2(x) = 2x2x3+x = x32x2+x,B计算错误;(3x2y)(2xy+3yz1) = (3x2y) (2xy)+(3x2y) 3yz(3x2y) = 6x3y29x2y2z+3x2y,C计算错误;(a3b)2ab = (a3) 2ab(b)2ab =a4bab2,D计算正确,所以答案为D12.B说明:因为(x2)(x+3) = xx2x+3x6 = x2+x6,所以a = 1,b = 6,答案为B二、解答题1.解:(1)( 5a3b2)·(3ab 2c)·( 7a2b) = (5)×(3)×(7)(a3·a·a2)(b2·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+1)3(x1)(x2) = 8x2(x22x+x2)3(x2x2x+2)= 8x2x2+x+23x2+9x6 = 4x2+10x4当x = 3时,原式 = 4·(3)2+10·(3)4 = 36304 = 23. 解:设长方形的长为x,宽为y,则由题意有即解得xy = 36答:长方形的面积是364. 解:(x+my1)(nx2y+3) = nx22xy+3x+mnxy2my2+3mynx+2y3= nx2(2mn)xy2my2+(3n)x+( 3m+2)y3x、y项系数为0,得故 3m+n = 3·()+3 = 1最新精品资料
展开阅读全文
相关资源
正为您匹配相似的精品文档
相关搜索

最新文档


当前位置:首页 > 压缩资料 > 基础医学


copyright@ 2023-2025  zhuangpeitu.com 装配图网版权所有   联系电话:18123376007

备案号:ICP2024067431-1 川公网安备51140202000466号


本站为文档C2C交易模式,即用户上传的文档直接被用户下载,本站只是中间服务平台,本站所有文档下载所得的收益归上传人(含作者)所有。装配图网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对上载内容本身不做任何修改或编辑。若文档所含内容侵犯了您的版权或隐私,请立即通知装配图网,我们立即给予删除!