2019年春七年級(jí)數(shù)學(xué)下冊(cè) 第一章 整式的乘除 4 整式的乘法同步課件（新版）北師大版.ppt

知識(shí)點(diǎn)一單項(xiàng)式與單項(xiàng)式的乘法,例1計(jì)算:(1)4xy2;(2)(0.3x3y4)2(-0.2x4y3)2;,(3)5x(-2.25axy)(-3x2y2).,解析(1)4xy2=x1+2y2+1z=-2x3y3z.(2)(0.3x3y4)2(-0.2x4y3)2=0.09x6y80.04x8y6=(0.090.04)x6+8y8+6=0.0036x14y14.(3)5x(-2.25axy)(-3x2y2)=5(-2.25)(-3)a1+1x1+1+1+2y1+2=a2x5y3.,知識(shí)點(diǎn)二單項(xiàng)式與多項(xiàng)式的乘法,例2計(jì)算:(1)a(3a-2);(2)(-2x2);(3)(-4a3+12a2b-7a3b3)(-4a2).,解析(1)a(3a-2)=3a2-2a.(2)(-2x2)=(-2x2)xy+(-2x2)y2=-x3y-2x2y2.(3)(-4a3+12a2b-7a3b3)(-4a2)=(-4a3)(-4a2)+12a2b(-4a2)-7a3b3(-4a2)=16a5-48a4b+28a5b3.,知識(shí)點(diǎn)三多項(xiàng)式與多項(xiàng)式的乘法,例3計(jì)算:(3x-2y)(2a+3b).,解析原式=3x2a+3x3b+(-2y)2a+(-2y)3b=6ax+9bx-4ay-6by.,題型將多項(xiàng)式看成整體進(jìn)行乘法計(jì)算例計(jì)算:(a+b)(a-b)4(a+b)4(b-a)5.,解析原式=(a+b)4(a-b)4(a+b)4(b-a)5=(a+b)4(b-a)4(a+b)4(b-a)5=(a+b)4(a+b)4(b-a)4(b-a)5=(a+b)8(b-a)9.,點(diǎn)撥將多項(xiàng)式看成整體進(jìn)行乘法運(yùn)算時(shí),一般會(huì)遇到互為相反數(shù)的乘方運(yùn)算,此時(shí)為了不出現(xiàn)符號(hào)運(yùn)算的錯(cuò)誤,通常遵循兩個(gè)原則:(1)少數(shù)服從多數(shù);(2)偶次方服從奇次方,如本題中將(a-b)4化為(b-a)4.,知識(shí)點(diǎn)一單項(xiàng)式與單項(xiàng)式的乘法1.第一宇宙速度是7.9103米/秒,衛(wèi)星以此速度運(yùn)行5103秒走過(guò)的路程為()A.3.95106米B.3.95105米C.39510米D.3.95107米,答案D走過(guò)的路程為7.91035103=39.5106=3.95107米,故選D.,2.計(jì)算:2x3(-2xy)=.,答案x7y4,解析2x3(-2xy)=2x3(-2xy)=x7y4.,3.計(jì)算:(1)2x35x2=;(2)3x2y5(-2xy2z)=.,答案(1)10 x5(2)-6x3y7z,解析(1)2x35x2=(25)(x3x2)=10 x5.(2)3x2y5(-2xy2z)=3(-2)(x2x)(y5y2)z=-6x3y7z.,4.計(jì)算:(1)(-6a2b);(2)(-x2y2)(3xy)2;(3)-a2b3abc.,解析(1)(-6a2b)=(-6)(a4a2)(b3b)=2a6b4.(2)(-x2y2)(3xy)2=(-x2y2)(9x2y2)=-9x4y4.(3)-a2b3abc=-a3b4c.,5.先化簡(jiǎn),再求值:(-3a3x)(-2a2x2)2+7(ax)3(a2x)2-a7x5,其中x=-2,a=-1.,解析原式=(-3a3x)4a4x4+7a3x3a4x2-a7x5=-12a7x5+7a7x5-a7x5=-6a7x5.當(dāng)a=-1,x=-2時(shí),原式=-6(-1)7(-2)5=-192.,知識(shí)點(diǎn)二單項(xiàng)式與多項(xiàng)式的乘法6.(2018河南鄭州金水月考)計(jì)算(-3xy2)(2y2-xyz+1)的結(jié)果是()A.-3xy4+3x2y3+3xy2B.-6xy4+3x2y3z-3xy2C.-6xy4-3x2y3z-3xy2D.-6xy4+3x2y2z,答案B(-3xy2)(2y2-xyz+1)=-6xy4+3x2y3z-3xy2,故選B.,7.(2017甘肅白銀五中月考)計(jì)算:-2a2-5a(a2b-ab2).,解析原式=-a3b-2a2b2-5a3b+5a2b2=-6a3b+3a2b2.,知識(shí)點(diǎn)三多項(xiàng)式與多項(xiàng)式的乘法8.(x+1)(2x-3)的計(jì)算結(jié)果是()A.2x2+x-3B.2x2-x-3C.2x2-x+3D.x2-2x-3,答案B(x+1)(2x-3)=2x2-3x+2x-3=2x2-x-3.故選B.,9.如果一個(gè)長(zhǎng)方形的長(zhǎng)為(4a2-2a+1),寬為(2a+1),則這個(gè)長(zhǎng)方形的面積為()A.8a3-4a2+2a-1B.8a3+4a2-2a-1C.8a3-1D.8a3+1,答案DS長(zhǎng)方形=(4a2-2a+1)(2a+1)=8a3+4a2-4a2-2a+2a+1=8a3+1.故選D.,10.計(jì)算:(1)(2x+1)(3x-1);(2)(3x+5a)(a-3x);(3)(2x-5y)(3x-y);(4)(x+y)(x2-2x-3).,解析(1)原式=6x2-2x+3x-1=6x2+x-1.(2)原式=3ax-9x2+5a2-15ax=3ax-15ax-9x2+5a2=-12ax-9x2+5a2.(3)原式=6x2-2xy-15xy+5y2=6x2-17xy+5y2.(4)原式=x3-2x2-3x+x2y-2xy-3y.,1.小明做了四道單項(xiàng)式乘法題,其中他做對(duì)的一道是()A.3x22x3=5x5B.3a34a3=12a9C.2m23m3=6m3D.3y36y3=18y6,答案D3x22x3=6x5;3a34a3=12a6;2m23m3=6m5;3y36y3=18y6.故選D.,2.計(jì)算:3x2(7x2-4x+2)-5x(2x-1)=.,答案21x4-12x3-4x2+5x,解析3x2(7x2-4x+2)-5x(2x-1)=21x4-12x3+6x2-10 x2+5x=21x4-12x3-4x2+5x.,3.計(jì)算圖1-4-1的面積.圖1-4-1,解析面積=bt+(a-t)t=at+bt-t2.,1.如果a2+a=1,那么(a-5)(a+6)的值為.,答案-29,解析(a-5)(a+6)=a2+a-30.因?yàn)閍2+a=1,所以(a-5)(a+6)=1-30=-29.,2.計(jì)算:(1)(m+2n)(m-2n);(2)(-a+b)(-a-b).,解析(1)原式=m2-2mn+2mn-4n2=m2-4n2.(2)原式=a2+ab-ab-b2=a2-b2.,3.計(jì)算:2m2(-2mn).,解析2m2(-2mn)=(m2mnm2n3)=2m5n4.,4.先化簡(jiǎn),再求值:(x+2y)(2x+y)-(3x-y)(x+2y),其中x=9,y=.,解析原式=2x2+xy+4xy+2y2-(3x2+6xy-xy-2y2)=-x2+4y2.當(dāng)x=9,y=時(shí),原式=-92+4=-80.,1.若(-x2ym)2(kxn+1y)=-2x6y3,則(km)n等于()A.-2B.2C.4D.-4,答案A因?yàn)?-x2ym)2(kxn+1y)=x4y2mkxn+1y=kx5+ny2m+1=-2x6y3,所以k=-2,5+n=6,2m+1=3,所以n=1,m=1.所以(km)n=(-2)11=-2,故選A.,2.定義為二階行列式,規(guī)定它的運(yùn)算法則為=ad-bc.例如:=14-23=-2.化簡(jiǎn)=.,答案4a+12,解析原式=(a+2)(a+3)-(a-2)(a+3)=a2+5a+6-a2-a+6=4a+12.,3.設(shè)x,y為任意實(shí)數(shù),定義運(yùn)算:x*y=(x+1)(y+1)-1,得到下列五個(gè)命題:x*y=y*x;x*(y+z)=x*y+x*z;(x+1)*(x-1)=(x*x)-1;x*0=0;(x+1)*(x+1)=x*x+2*x+1.其中正確的命題的序號(hào)是.,答案,解析x*y=xy+x+y=y*x,故正確;x*(y+z)=(x+1)(y+z+1)-1=xy+xz+x+y+z,x*y+x*z=(x+1)(y+1)-1+(x+1)(z+1)-1=xy+xz+2x+y+z,故錯(cuò)誤;(x+1)*(x-1)=(x+2)x-1=x2+2x-1,(x*x)-1=(x+1)(x+1)-1-1=x2+2x-1,故正確;x*0=x,故錯(cuò)誤;(x+1)*(x+1)=(x+2)(x+2)-1=x2+4x+3,x*x+2*x+1=(x+1)(x+1)-1+3(x+1)-1+1=x2+5x+3,故錯(cuò)誤.故答案為.,4.請(qǐng)說(shuō)明對(duì)任意正整數(shù)n,式子n(n+5)-(n+2)(n-3)的值必定能被6整除.,解析n(n+5)-(n+2)(n-3)=n2+5n-(n2-3n+2n-6)=n2+5n-n2+n+6=6n+6=6(n+1).n為正整數(shù),n+1為正整數(shù),n(n+5)-(n+2)(n-3)的值必定能被6整除.,一、選擇題1.(2018江西撫州臨川第一實(shí)驗(yàn)學(xué)校月考,2,)若(y+3)(y-2)=y2+my+n,則()A.m=5,n=6B.m=1,n=-6C.m=1,n=6D.m=5,n=-6,答案B(y+3)(y-2)=y2+y-6=y2+my+n,m=1,n=-6.,2.(2018江蘇揚(yáng)州樹(shù)人學(xué)校期中,7,)如果(x2-mx+1)(x+2)的積中不含x的二次項(xiàng),那么m的值是()A.1B.-1C.-2D.2,答案D原式=x3+2x2-mx2-2mx+x+2=x3+(2-m)x2-2mx+x+2,由題意得2-m=0,則m=2.,二、解答題3.2016江蘇昆山期中,19(2),計(jì)算:-2xy3x2y-x2y(-3xy+xy2).,解析原式=-6x3y2+3x3y2-x3y3=-3x3y2-x3y3.,1.(2018湖南郴州龍市中學(xué)月考,9,)化簡(jiǎn)x2y3的結(jié)果為()A.-x3y3B.x3y3C.-2x3y3D.2x3y3,答案C原式=x2xy3=-2x3y3.,2.(2016北京石景山期末,17,)計(jì)算:-6ab.,解析原式=-12a3b2+2a2b3.,3.(2016山東濟(jì)南商河期末,18,)計(jì)算:3x(x-y)-(2x-y)(x+y).,解析原式=3x2-3xy-(2x2+2xy-xy-y2)=x2-4xy+y2.,一、選擇題1.(2017臺(tái)灣省中考,3,)計(jì)算6x(3-2x)的結(jié)果,與下列哪一個(gè)式子相同()A.-12x2+18xB.-12x2+3C.16xD.6x,答案A6x(3-2x)=18x-12x2,故選A.,2.(2017湖北武漢中考,5,)計(jì)算(x+1)(x+2)的結(jié)果為()A.x2+2B.x2+3x+2C.x2+3x+3D.x2+2x+2,答案B(x+1)(x+2)=x2+2x+x+2=x2+3x+2.故選B.,二、填空題3.(2018天津中考,13,)計(jì)算2x4x3的結(jié)果等于.,答案2x7,解析原式=2x4+3=2x7.,4.(2015湖南常德中考,11,)計(jì)算:b(2a+5b)+a(3a-2b)=.,答案5b2+3a2,解析b(2a+5b)+a(3a-2b)=2ab+5b2+3a2-2ab=5b2+3a2.,1.(2016貴州畢節(jié)中考,3,)下列運(yùn)算正確的是()A.-2(a+b)=-2a+2bB.(a2)3=a5C.a3+4a=a3D.3a22a3=6a5,答案DA.原式=-2a-2b;B.原式=a6;C.a3與4a不是同類(lèi)項(xiàng),不能合并;D.原式=6a5.故選D.,2.(2015福建福州中考,12,)計(jì)算(x-1)(x+2)的結(jié)果是.,答案x2+x-2,解析(x-1)(x+2)=x2+2x-x-2=x2+x-2.,1.定義運(yùn)算:ab=a(1-b),下面給出了關(guān)于這種運(yùn)算的幾個(gè)結(jié)論:2(-2)=6;ab=ba;若a+b=0,則(aa)+(bb)=2ab;若ab=0,則a=0.其中正確結(jié)論的序號(hào)是.,答案,解析2(-2)=21-(-2)=6,正確.ba=b(1-a)ab,錯(cuò)誤.a+b=0,(aa)+(bb)=a(1-a)+b(1-b)=a-a2+b-b2=-a2-b2,由a+b=0,得b=-a,-a2-b2=-2a2,2ab=2a(-a)=-2a2,(aa)+(bb)=2ab,正確.當(dāng)ab=0時(shí),a=0或b=1,錯(cuò)誤,故填.,2.三角形表示運(yùn)算3abc,方框xywz表示運(yùn)算-4xywz,求n2m5的值.,解析由題意得,n2m5=(33mn)(-4n2m5)=9mn(-4m5n2)=9(-4)m1+5n1+2=-36m6n3.,1.觀察下列各式:(x-1)(x+1)=x2-1;(x-1)(x2+x+1)=x3-1;(x-1)(x3+x2+x+1)=x4-1;(1)根據(jù)以上規(guī)律,則(x-1)(x6+x5+x4+x3+x2+x+1)=;(2)你能否由此歸納出一般性規(guī)律:(x-1)(xn+xn-1+x+1)=;(3)根據(jù)(2)求出1