七年級(jí)數(shù)學(xué)下冊(cè)第1章整式的乘除1.4整數(shù)的乘法1.4.2單項(xiàng)式乘多項(xiàng)式精練（新版）北師大版.docx

第2課時(shí)單項(xiàng)式乘多項(xiàng)式測(cè)試時(shí)間:25分鐘一、選擇題1.化簡(jiǎn)x(x2+3)+x2(x-3)-3x(x2-x-1)的結(jié)果為()A.-x3+6xB.-x3-6xC.-x3-6x2+6xD.-x3-6x21.答案A原式=x3+3x+x3-3x2-3x3+3x2+3x=-x3+6x.2.計(jì)算-2a(a2-1)的結(jié)果是()A.-2a3-2aB.-2a3+aC.-2a3+2aD.-a3+2a2.答案C3.下列計(jì)算結(jié)果正確的是()A.(6ab2-4a2b)3ab=18ab2-12a2bB.(-x)(2x+x2-1)=-x3-2x2+1C.(-3x2y)(-2xy+3yz-1)=6x3y2-9x2y2z2+3x2yD.34a3-12b2ab=32a4b-ab23.答案DA.(6ab2-4a2b)3ab=18a2b3-12a3b2,此選項(xiàng)計(jì)算錯(cuò)誤;B.(-x)(2x+x2-1)=-x3-2x2+x,此選項(xiàng)計(jì)算錯(cuò)誤;C.(-3x2y)(-2xy+3yz-1)=6x3y2-9x2y2z+3x2y,此選項(xiàng)計(jì)算錯(cuò)誤;D.34a3-12b2ab=32a4b-ab2,此選項(xiàng)計(jì)算正確.故選D.4.要使(-6x3)(x2+ax+5)+3x4的結(jié)果中不含x4項(xiàng),則a的值是()A.0B.12C.-12D.24.答案B原式=-6x5-6ax4-30x3+3x4=-6x5+(3-6a)x4-30x3,由(-6x3)(x2+ax+5)+3x4的結(jié)果中不含x4項(xiàng),得3-6a=0,解得a=12,故選B.5.計(jì)算(-3x)(2x2-5x-1)的結(jié)果是()A.-6x3-15x2-3xB.-6x3+15x2+3xC.-6x3+15x2D.-6x3+15x2-15.答案B(-3x)(2x2-5x-1)=-3x2x2+3x5x+3x=-6x3+15x2+3x.故選B.6.化簡(jiǎn)x(2x-1)-x2(2-x)的結(jié)果是()A.-x3-xB.x3-xC.-x2-1D.x3-16.答案B原式=2x2-x-2x2+x3=x3-x,故選B.7.下列計(jì)算正確的是()A.(-4x)(2x2+3x-1)=-8x3-12x2-4xB.(6xy2-4x2y)3xy=6xy2-12x3y2C.(-x)(2x+x2-1)=-x3-2x2+1D.(-3x2y)(-2xy+3yz+1)=6x3y2-9x2y2z-3x2y7.答案D(-4x)(2x2+3x-1)=-8x3-12x2+4x,A錯(cuò)誤;(6xy2-4x2y)3xy=18x2y3-12x3y2,B錯(cuò)誤;(-x)(2x+x2-1)=-x3-2x2+x,C錯(cuò)誤;(-3x2y)(-2xy+3yz+1)=6x3y2-9x2y2z-3x2y,D正確.故選D.8.要使(y2-ky+2y)(-y)的展開式中不含y2項(xiàng),則k的值為()A.-2B.0C.2D.38.答案C(y2-ky+2y)(-y)=-y3+ky2-2y2,展開式中不含y2項(xiàng),k-2=0,解得k=2.故選C.9.已知xy2=-2,則-xy(x2y5-xy3-y)的值為()A.2B.6C.10D.149.答案Cxy2=-2,-xy(x2y5-xy3-y)=-x3y6+x2y4+xy2=-(xy2)3+(xy2)2+xy2=-(-2)3+(-2)2+(-2)=8+4-2=10,故選C.二、填空題10.計(jì)算:a(a+1)=.10.答案a2+a解析a(a+1)=aa+a1=a2+a.11.計(jì)算:(-2a)14a3-1=.11.答案-12a4+2a解析(-2a)14a3-1=(-2a)14a3+(-2a)(-1)=-12a4+2a.12.計(jì)算:12b2-4a2(-4ab)=. 12.答案-2ab3+16a3b解析原式=-2ab3+16a3b.13.計(jì)算:12m2n3-2mn2+(2m2n)2=. 13.答案-m3n5+2m6n5解析12m2n3-2mn2+(2m2n)2=12m2n3(-2mn2+4m4n2)=-m3n5+2m6n5.14.已知一圓柱體的底面半徑為x,高為2x+4,則它的體積為(結(jié)果保留).14.答案2x3+4x2解析圓柱體的體積為x2(2x+4)=2x3+4x2.15.一個(gè)長(zhǎng)方體的長(zhǎng)為2m,寬為3n,高為4mn-1,則這個(gè)長(zhǎng)方體的體積是.15.答案24m2n2-6mn解析一個(gè)長(zhǎng)方體的長(zhǎng)為2m,寬為3n,高為4mn-1,這個(gè)長(zhǎng)方體的體積是2m3n(4mn-1)=6mn(4mn-1)=24m2n2-6mn.16.下面規(guī)定一種運(yùn)算:ab=a(a-b),則x2yxy2的計(jì)算結(jié)果是.16.答案x4y2-x3y3解析ab=a(a-b),x2yxy2=x2y(x2y-xy2)=x4y2-x3y3.17.若-2x2y(-xmy+3xy3)=2x5y2-6x3yn,則m=,n=.17.答案3;4解析-2x2y(-xmy+3xy3)=2xm+2y2-6x3y4=2x5y2-6x3yn,m+2=5,n=4,m=3,n=4.三、解答題18.先化簡(jiǎn),再求值:3a(2a2-4a+3)-2a2(3a+4),其中a=-2.18.解析3a(2a2-4a+3)-2a2(3a+4)=6a3-12a2+9a-6a3-8a2=-20a2+9a,當(dāng)a=-2時(shí),原式=-20(-2)2+9(-2)=-98.19.計(jì)算:(1)-43ab292a2b-12ab+34b2;(2)a2