PAGEPAGE11.4整式的乘法第1課時單項式與單項式相乘學(xué)問要點(diǎn)基礎(chǔ)練學(xué)問點(diǎn)單項式與單項式相乘1.計算3a2b·2a的結(jié)果是(A)A.6a3b B.2a2b C.6a2b2 D.5a2b2.計算3x3·(-2x7)的結(jié)果是(A)A.-6x10 B.-6x21C.-x10 D.x103.下列運(yùn)算正確的是(D)A.(-a)-1=a B.(-ab2)3=a3b5C.-a2b+2ba2=3a2b D.-2ab2·a3b=-2a4b34.(泰州中考)計算:12x·(-2x2)3=-4x7.5.計算:(1)(-4xy3)(-2x);解:原式=8x2y3.(2)(-2.4x2y3)(-0.5x4).解:原式=1.2x6y3.綜合實力提升練6.下列運(yùn)算正確的是(D)A.(-2ab)·(-3ab)3=-54a4b4B.5x2·(3x3)2=15x12C.(-0.1b)·(-10b2)3=-b7D.(3×10n)×13×107.(青島中考)計算(a2)3-5a3·a3的結(jié)果是(C)A.a5-5a6 B.a6-5a9C.-4a6 D.4a68.計算(2m2n-2)2·3m-2n3的結(jié)果是

12m2n9.已知A=3x2,B=-2xy2,C=-x2y2,則A·B2·C=-12x6y6.

10.計算:(1)(2x2y)·(3xy2)-4xy·(xy)2;解:原式=6x3y3-4x3y3=2x3y3.(2)25x2y3·516xyz·(-2x2y解:原式=18x3y4z·(-2x2y)=-14x5y511.已知x=4,y=18,求代數(shù)式17xy2·14(xy)2·14x解:17xy2·14(xy)2·14x5=12x8把x=4,y=18代入,得原式=12×48×1即該代數(shù)式的值是8.12.已知x3m=2,y2m=3,求(x2m)3+(ym)6-(x2y)3m·ym的值.解:∵x3m=2,y2m=3,∴(x2m)3+(ym)6-(x2y)3m·ym=(x3m)2+(y2m)3-(x6my4m)=(x3m)2+(y2m)3-(x3my2m)2=22+33-(2×3)2=-5.13.化簡:(-2x2y)·5xy3·-3解:原式=(-2)×5×-35×x2+1+3×y1+3+2=6x6y拓展探究突破練14.已知(-2xm+1y2n-1)·(5x2y)=-10x4y4,求-2m2n·-12解:由(-2xm+1y2n-1)·(5x2y)=-10xm+3·y2n=-10x4y4,可得m+3=4,則-2m2n·-12m3n22=-1第2課時單項式與多項式相乘學(xué)問要點(diǎn)基礎(chǔ)練學(xué)問點(diǎn)單項式與多項式相乘1.計算-2a(a2-1)的結(jié)果是(C)A.-2a3-2a B.-2a3+aC.-2a3+2a D.-a3+2a2.化簡a(a+1)-a(1-a)的結(jié)果是2a2.

3.計算:(1)3x2(-y-xy2+x2);解:原式=-3x2y-3x3y2+3x4.(2)(-4xy)·(xy+3x2y);解:原式=-4x2y2-12x3y2.(3)-1解:原式=-13x3y2+34x2y3-354.先化簡,再求值:x(x+1)-3x(x-2),其中x=3.解:x(x+1)-3x(x-2)=x2+x-3x2+6x=-2x2+7x,當(dāng)x=3時,原式=-2×32+7×3=-18+21=3.綜合實力提升練5.下列運(yùn)算正確的是(D)A.a(a+1)=a2+1 B.(a2)3=a5C.3a2+a=4a3 D.a5÷a2=a36.化簡x(2x-1)-x2(2-x)的結(jié)果是(B)A.-x3-x B.x3-xC.-x2-1 D.x3-17.計算:(2x2)3-6x3(x3+2x2+x)=(D)A.-12x5-6x4 B.2x6+12x5+6x4C.x2-6x-3 D.2x6-12x5-6x48.已知M,N分別表示不同的單項式,且3x(M-5x)=6x2y3+N,則(C)A.M=2xy3,N=-15xB.M=3xy3,N=-15x2C.M=2xy3,N=-15x2D.M=2xy3,N=15x29.已知ab2=-2,則-ab(a2b5-ab3+b)=(D)A.4 B.2 C.0 D.1410.一個長方體的長、寬、高分別為3a-4,2a,a,則它的體積等于(C)A.3a3-4a2 B.a2C.6a3-8a2 D.6a2-8a11.代數(shù)式y(tǒng)z(xz+2)-2y(3xz2+z+x)+5xyz2的值(A)A.只與x,y有關(guān) B.只與y,z有關(guān)C.與x,y,z都無關(guān) D.與x,y,z都有關(guān)12.已知3x·(xn+5)=3xn+1-8,那么x=-815.13.已知(-2x2)·(3x2-ax-6)-3x3+x2中不含x的三次項,求a的值.解:(-2x2)·(3x2-ax-6)-3x3+x2=-6x4+2ax3+12x2-3x3+x2=-6x4+(2a-3)x3+13x2,∵不含x的三次項,∴2a-3=0,解得a=3214.某中學(xué)擴(kuò)建教學(xué)樓,測量地基時,量得地基長為2am,寬為(2a-24)m.試用含a的代數(shù)式表示地基的面積,并計算當(dāng)a=25時地基的面積.解:依據(jù)題意得2a·(2a-24)=(4a2-48a)m2,當(dāng)a=25時,4a2-48a=4×252-48×25=1300(m2).拓展探究突破練15.當(dāng)m,n為何值時,12x[x(x+m)+nx(x+1)+m]的綻開式中不含有x2和x3解:12x[x(x+m)+nx(x+1)+m=12x(x2+mx+nx2+nx+m=n+12x3+m+n2x由結(jié)果中不含x2和x3的項,得1+n=0,m+n=0,解得m=1,n=-1.第3課時多項式與多項式相乘學(xué)問要點(diǎn)基礎(chǔ)練學(xué)問點(diǎn)多項式與多項式相乘1.(武漢中考)計算(a-2)(a+3)的結(jié)果是(B)A.a2-6 B.a2+a-6C.a2+6 D.a2-a+62.下列各式中,計算結(jié)果是x2+7x-18的是(D)A.(x-1)(x+18) B.(x+2)(x+9)C.(x-3)(x+6) D.(x-2)(x+9)3.如圖所示的長方形,有下列四種表示面積的方法:①(m+n)(a+b);②m(a+b)+n(a+b);③a(m+n)+b(m+n);④ma+mb+na+nb.其中正確的是(D)A.① B.④C.①④ D.①②③④4.(玉林中考)已知ab=a+b+1,則(a-1)(b-1)=2.

5.計算:(1)(2x+1)(x+5);解:原式=2x2+11x+5.(2)(x+2)(x-1)-3x(x+3).解:原式=x2-x+2x-2-3x2-9x=-2x2-8x-2.6.求(x-1)(2x+1)-2(x-5)(x+2)的值,其中x=-2.解:(x-1)(2x+1)-2(x-5)(x+2)=5x+19,把x=-2代入原式,原式=5×(-2)+19=-10+19=9.綜合實力提升練7.若(x-3)(x+2)=x2+ax+b,則a+b=(D)A.-1 B.3 C.5 D.-78.設(shè)M=(x-3)(x-7),N=(x-2)(x-8),則M與N的關(guān)系為(A)A.M>N B.M<NC.M=N D.不能確定9.假如關(guān)于x的多項式(2x-m)與(x+5)的乘積中,常數(shù)項為15,則m的值為(B)A.3 B.-3 C.10 D.-1010.已知a+b=4,ab=3,則代數(shù)式(a+2)(b+2)的值是(D)A.7 B.9 C.11 D.1511.若(x+m)與(x+3)的乘積中不含x的一次項,則m=-3.

12.三角形的一條邊長為4a+2,該邊上的高為2a-1,該三角形面積為S,試用含a的代數(shù)式表示S,并求當(dāng)a=2時S的值.解:S=12(4a+2)(2a-1)=4a2-當(dāng)a=2時,S=16-1=15.13.已知關(guān)于x的代數(shù)式(x2-3x-2)(ax+1),若運(yùn)算結(jié)果中不含有x的一次項,求代數(shù)式2a2-(2a+1)(a-1)的值.解:(x2-3x-2)(ax+1)=ax3-3ax2-2ax+x2-3x-2=ax3+(1-3a)x2-(2a+3)x-2,由題意得-(2a+3)=0,解