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整式的乘法整式的乘法与因式分解章末小结与提升类型1幂的运算典例1如果(anbm)3=a9b12,那么m,n的值分别为()A.m=9,n=-4B.m=3,n=4C.m=4,n=3D.m=9,n=6【解析】直接利用积的乘方运算法则化简,进而求出即可.由已知得a3nb3m=a9b12,即3n=9,3m=12,解得n=3,m=4.【答案】 C【针对训练】1.(常州中考)下列运算正确的是(C)A.mm=2mB.(mn)3=mn3C.(m2)3=m6D.m6m2=m32.(南京中考)计算106(102)3104的结果是(C)A.103B.107C.108D.1093.已知:2m=3,32n=5,则22m+10n=225.类型2整式的乘除法典例2如果(x2+px)(x2-5x+7)的展开式中不含有x3项,则p=.【解析】(x2+px)(x2-5x+7)=x4+(p-5)x3+(7-5p)x2+7px,且展开式中不含x3项,p-5=0,解得p=5.【答案】 5【针对训练】1.若(x+1)(2x-3)=2x2+mx+n,则m=-1,n=-3.2.(海南中考)计算:(x+1)2+x(x-2)-(x+1)(x-1).解:原式=x2+2x+1+x2-2x-x2+1=x2+2.3.先化简,再求值:-2x3y4(-x2y2)(-x)-(x-2y)(3y+x)+x(x+2xy2),其中x=-1,y=-2.解:原式=2xy2(-x)-(x2-6y2+xy)+x2+2x2y2=-2x2y2-x2+6y2-xy+x2+2x2y2=6y2-xy,当x=-1,y=-2时,原式=6(-2)2-(-1)(-2)=22.类型3乘法公式典例3已知a+b=4,ab=1.(1)求a2+b2的值;(2)求a-b的值.【解析】(1)a+b=4,ab=1,a2+b2=(a+b)2-2ab=42-21=14.(2)a+b=4,ab=1,(a-b)2=(a+b)2-4ab=16-4=12,a-b=2.【针对训练】1.下列各式中可以运用平方差公式的有(B)(-1+2x)(-1-2x);(ab-2b)(-ab-2b);(-1-2x)(1+2x);(x2-y)(y2+x).A.1个B.2个C.3个D.4个2.对于任意的有理数a,b,现用“”定义一种运算:ab=a2-b2,根据这个定义,代数式(x+y)y可以化简为(C)A.xy+y2B.xy-y2C.x2+2xyD.x23.若(7x-a)2=49x2-bx+9,则|a+b|的值为45.4.计算:(1)(x+3)(x-3)(x2-9);解:(x+3)(x-3)(x2-9)=(x2-9)2=x4-18x2+81.(2)2(3+1)(32+1)(34+1)(38+1)(316+1).解:2(3+1)(32+1)(34+1)(38+1)(316+1)=(3-1)(3+1)(32+1)(34+1)(38+1)(316+1)=(32-1)(32+1)(34+1)(38+1)(316+1)=(34-1)(34+1)(38+1)(316+1)=332-1.5.先化简,再求值:(x+2y)(x-2y)-(x+4y)24y,其中x2-8x+y2-y+16=0.解:x2-8x+y2-y+16=0,即x2-8x+16+y2-y+=0,则(x-4)2+=0,则x-4=0且y-=0,解得x=4,y=.原式=x2-4y2-(x2+8xy+16y2)4y=(x2-4y2-x2-8xy-16y2)4y=-2x-5y,当x=4,y=时,原式=-8-.类型4因式分解典例4因式分解:9x2-1=.【解析】直接运用平方差公式即可求解.原式=(3x+1)(3x-1).【答案】 (3x+1)(3x-1)【针对训练】1.将下列多项式因式分解,结果中不含有因式(x-2)的是(C)A.x2-4B.x2-4x+4C.x2+2x+1D.x2-2x2.分解因式:(1)18axy-3ax2-27ay2;解:18axy-3ax2-27ay2=-3a(-6xy+x2+9y2)=-3a(x-3y)2.(2)(a2+4)2-16a2;解:(a2+4)2-16a2=(a2+4+4a)(a2+4-4a)=(a+2)2(a-2)2.(3)c(a-b)-2(a-b)2c+(a-b)3c.解:c(a-b)-2(a-b)2c+(a-b)3c=c(a-b)1-2(a-b)+(a-b)2=c(a-b)(a-b-1)2.
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