2022年華東師大版數(shù)學(xué)八上《多項(xiàng)式與多項(xiàng)式相乘》課件

3.多項(xiàng)式與多項(xiàng)式相乘華東師大版八年級(jí)數(shù)學(xué)上冊(cè)3.多項(xiàng)式與多項(xiàng)式相乘華東師大版八年級(jí)數(shù)學(xué)上冊(cè)復(fù)習(xí)舊知(1)(-x)3·(-x)3·(-x)5=______;(2)(x2)4=_______;(3)(x3y5)4=______;(4)(xy)3·(xy)4·(xy)5=______;(5)(-3x3y)(-5x4y2z4)=___________;(6)-3ab2(-4a+3ab-2)=___________________.-x11x8x12y20x12y1215x7y3z412a2b2-9a2b3+6ab2復(fù)習(xí)舊知(1)(-x)3·(-x)3·(-x)5=_____情境導(dǎo)入某地區(qū)在退耕還林期間，將一塊長(zhǎng)m米、寬a米的長(zhǎng)方形林地的長(zhǎng)、寬分別增加n米和b米.用兩種方法表示這塊林地現(xiàn)在的面積，你知道下面的等式蘊(yùn)含著什么樣的運(yùn)算法那么嗎？mnbambmanbn(m+n)(a+b)=ma+mb+na+nba情境導(dǎo)入某地區(qū)在退耕還林期間，將一塊長(zhǎng)m米、(m+n)(a+b)=看成是一個(gè)整體(m+n)a+(m+n)b=ma+na+mb+nb(m+n)(a+b)=+na

+nbma+mb這個(gè)等式實(shí)際上給出了多項(xiàng)式乘以多項(xiàng)式的法那么:多項(xiàng)式與多項(xiàng)式相乘，先用一個(gè)多項(xiàng)式的每一項(xiàng)分別乘以另一個(gè)多項(xiàng)式的每一項(xiàng)，再把所得的積相加.(m+n)(a+b)=看成是一個(gè)整體(m+n)a+(m+n計(jì)算：〔1〕(x+2)(x-3)；〔2〕(2x+5y)(3x-2y).-3x=x2-x-6例3=x2+2x-6=6x2=6x2+11xy-10y2-10y2-4xy+15xy探究新知計(jì)算：〔1〕(x+2)(x-3)；〔2〕(2x+5y)(3x計(jì)算：〔1〕(m-2n)(m2+mn-3n2)〔2〕(3x2-2x+2)(2x+1)〔1〕(m-2n)(m2+mn-3n2)=m·m2+m·mn-m·3n2-2n·m2-2n·mn+2n·3n2=m3+m2n-3mn2-2m2n-2mn2+6n3=m3-m2n-5mn2+6n3例4計(jì)算：〔1〕(m-2n)(m2+mn-3n2)〔2〕(3x2計(jì)算：〔1〕(m-2n)(m2+mn-3n2)〔2〕(3x2-2x+2)(2x+1)〔2〕(3x2-2x+2)(2x+1)=6x3+3x2-4x2-2x+4x+2=6x3-x2+2x+2例4計(jì)算：〔1〕(m-2n)(m2+mn-3n2)〔2〕(3x2

(1)(x+2y)(5a+3b);解：(x+2y)(5a+3b)=x

·5a

+x

·3b

+2y

·5a

+2y

·3b=5ax+3bx+10ay+6by補(bǔ)充例題計(jì)算(1)(x+2y)(5a+3b);(2)(2x–3)(x+4);解：(2x–3)(x+4)2x2

+8x

–3x

–12=2x2

+5x=–12補(bǔ)充例題計(jì)算(2)(2x–3)(x+4);解：(2x–3)(x+補(bǔ)充例題計(jì)算(3)(3x+y)(x–2y)；解：(3x+y)(x–2y)=3x2

–6xy

+xy

–2y2=3x2

–5xy

–2y2

補(bǔ)充例題計(jì)算(3)(3x+y)(x–2y)；解：(3x〔1〕(x+5)(x-7)；計(jì)算：〔2〕(x+5y)(x-7y)；〔3〕(2m+3n)(2m-3n)；〔4〕(2a+3b)2.隨堂練習(xí)=x2-7x+5x-35=x2-2x-35=x2-7xy+5xy-35y2=x2-2xy-35y2=4m2+6mn-6mn-9n2=4m2-9n2=4a2+12ab+9b2〔1〕(x+5)(x-7)；計(jì)算：〔2〕(x+5y)(x-7算一算:

(1)(2x+1)(x+3)(2)(m+2n)(m+3n)(3)(a-1)2

(4)(a+3b)(a–3b)2x2+7x+3m2+5mn+6n2a2-2a+1a2-9b2算一算:2x2+7x+3m2+5mn+6n2a2-2a+1a(5)(x+2)(x+3)

(6)(x-4)(x+1)

(7)(y+4)(y-2)(8)(y-5)(y-3)

x2+5x+6x2-3x-4y2+2y-8y2-8y+15(5)(x+2)(x+3)課堂小結(jié)多項(xiàng)式乘以多項(xiàng)式，展開(kāi)后項(xiàng)數(shù)很有規(guī)律，在合并同類(lèi)項(xiàng)之前，展開(kāi)式的項(xiàng)數(shù)恰好等于兩個(gè)多項(xiàng)式的項(xiàng)數(shù)的積.

多項(xiàng)式與多項(xiàng)式相乘，先用一個(gè)多項(xiàng)式的每一項(xiàng)分別乘以另一個(gè)多項(xiàng)式的每一項(xiàng)，再把所得的積相加.

課堂小結(jié)多項(xiàng)式乘以多項(xiàng)式，展開(kāi)后項(xiàng)數(shù)很有規(guī)律，在合并復(fù)習(xí)舊知計(jì)算以下各式，說(shuō)說(shuō)你是怎么想的？〔1〕(am+bm)÷m；〔2〕(a2+ab)÷a.〔1〕(am+bm)÷m〔2〕(a2+ab)÷a=am÷m+bm÷m=a+b=a2÷a+ab÷a=a+b復(fù)習(xí)舊知計(jì)算以下各式，說(shuō)說(shuō)你是怎么想的？〔1〕(am+bm)新課導(dǎo)入試一試計(jì)算：〔1〕(ax+bx)÷x；解〔1〕·x(a+b)x=ax+bx所以(ax+bx)÷x=a+b新課導(dǎo)入試一試計(jì)算：〔1〕(ax+bx)÷x；解〔1〕試一試〔2〕(ma+mb+mc)÷m.·m(a+b+c)m=ma+mb+mc所以(ma+mb+mc)÷m=a+b+c試一試〔2〕(ma+mb+mc)÷m.·m(a+b+c)m=探究新知例2計(jì)算：〔1〕(9x4-15x2+6x)÷3x〔2〕(28a3b2c+a2b3-14a2b2)÷(-7a2b)探究新知例2計(jì)算：〔1〕(9x4-15x2+6x)÷3x〔2〔1〕(9x4-15x2+6x)÷3x解9x43x=9x4÷3x-15x2-15x2÷3x+6x+6x÷3x=(9÷3)x4-1-(15÷3)x2-1+(6÷3)x1-1=3x3-5x+2多項(xiàng)式除以單項(xiàng)式，先用這個(gè)多項(xiàng)式的每一項(xiàng)除以這個(gè)單項(xiàng)式，再把所得的商相加.〔1〕(9x4-15x2+6x)÷3x解9x43x=9x4÷〔2〕(28a3b2c+a2b3-14a2b2)÷(-7a2b)解28a3b2c-7a2b=28a3b2c÷(-7a2b)+a2b3+a2b3÷(-7a2b)-14a2b2-14a2b2÷(-7a2b)=-4a3-2b2-1c+(a2-2b3-1)-(-2a2-2b2-1)=-4abc

b2+2b〔2〕(28a3b2c+a2b3-14a2b2)÷(-7a2補(bǔ)充例題計(jì)算：(6xn+2+3xn+1-3xn-1)÷3xn-1解：(6xn+2+3xn+1-3xn-1)÷3xn-1=6xn+2÷3xn-1+3xn+1÷3xn-1-3xn-1÷3xn-1=2xn+2-n+1+xn+1-n+1-1=2x3+x2-1思路歸納

如果除式中字母的指數(shù)是多項(xiàng)式，計(jì)算時(shí)要把它看作一個(gè)整體，且要添括號(hào).【選自《狀元大課堂》】補(bǔ)充例題計(jì)算：(6xn+2+3xn+1-3xn-1)÷3xn補(bǔ)充例題化簡(jiǎn)：[4(xy-1)2+(xy+2)(xy-2)]÷xy解：[4(xy-1)2+(xy+2)(xy-2)]÷xy=(4x2y2-8xy+4+x2y2-4)÷xy=(5x2y2-8xy)÷xy=20xy-32【選自《狀元大課堂》】思維點(diǎn)撥進(jìn)行整式的混合運(yùn)算，應(yīng)按照運(yùn)算順序進(jìn)行化簡(jiǎn).補(bǔ)充例題化簡(jiǎn)：[4(xy-1)2+(xy+2)(xy-2)]隨堂練習(xí)〔1〕(3ab-2a)÷a1.計(jì)算：〔2〕(5ax2+15x)÷5x=3ab÷a-2a÷a=3b-2=5ax2÷5x+15x÷5x=ax+3隨堂練習(xí)〔1〕(3ab-2a)÷a1.計(jì)算：〔2〕(5ax2隨堂練習(xí)1.計(jì)算：〔3〕(12m2n-15mn2)÷6mn〔4〕(x3-2x2y)÷(-x2)=12m2n÷6mn-15mn2÷6mn=2m-2.5n=x3÷(-x2)-2x2y÷(-x2)=-x+2y隨堂練習(xí)1.計(jì)算：〔3〕(12m2n-15mn2)÷6mn〔〔1〕(4a3b3-6a2b3c-2ab5)÷(-2ab2)2.計(jì)算：解(4a3b3-6a2b3c-2ab5)÷(-2ab2)=4a3b3÷(-2ab2)-6a2b3c÷(-2ab2)-2ab5÷(-2ab2)=-2a2b+3abc+b3〔1〕(4a3b3-6a2b3c-2ab5)÷(-2ab2)（2）(x2y3-x3y2+2x2y2)÷xy2解（2）(x2y3-x3y2+2x2y2)÷xy2=x2y3÷xy2-x3y2÷xy2+2x2y2÷xy2=2xy-x2+4x（2）(x2y3-x3y2+2x2y2)÷〔1〕(9x4-15x2+6x)÷3x解9x43x=9x4÷3x-15x2-15x2÷3x+6x+6x÷3x=(9÷3)x4-1-(15÷3)x2-1+(6÷3)x1-1=3x3-5x+2多項(xiàng)式除以單項(xiàng)式，先用這個(gè)多項(xiàng)式的每一項(xiàng)除以這個(gè)單項(xiàng)式，再把所得的商相加.課堂小結(jié)〔1〕(9x4-15x2+6x)÷3x解9x43x=9x4÷3.多項(xiàng)式與多項(xiàng)式相乘華東師大版八年級(jí)數(shù)學(xué)上冊(cè)3.多項(xiàng)式與多項(xiàng)式相乘華東師大版八年級(jí)數(shù)學(xué)上冊(cè)復(fù)習(xí)舊知(1)(-x)3·(-x)3·(-x)5=______;(2)(x2)4=_______;(3)(x3y5)4=______;(4)(xy)3·(xy)4·(xy)5=______;(5)(-3x3y)(-5x4y2z4)=___________;(6)-3ab2(-4a+3ab-2)=___________________.-x11x8x12y20x12y1215x7y3z412a2b2-9a2b3+6ab2復(fù)習(xí)舊知(1)(-x)3·(-x)3·(-x)5=_____情境導(dǎo)入某地區(qū)在退耕還林期間，將一塊長(zhǎng)m米、寬a米的長(zhǎng)方形林地的長(zhǎng)、寬分別增加n米和b米.用兩種方法表示這塊林地現(xiàn)在的面積，你知道下面的等式蘊(yùn)含著什么樣的運(yùn)算法那么嗎？mnbambmanbn(m+n)(a+b)=ma+mb+na+nba情境導(dǎo)入某地區(qū)在退耕還林期間，將一塊長(zhǎng)m米、(m+n)(a+b)=看成是一個(gè)整體(m+n)a+(m+n)b=ma+na+mb+nb(m+n)(a+b)=+na

+nbma+mb這個(gè)等式實(shí)際上給出了多項(xiàng)式乘以多項(xiàng)式的法那么:多項(xiàng)式與多項(xiàng)式相乘，先用一個(gè)多項(xiàng)式的每一項(xiàng)分別乘以另一個(gè)多項(xiàng)式的每一項(xiàng)，再把所得的積相加.(m+n)(a+b)=看成是一個(gè)整體(m+n)a+(m+n計(jì)算：〔1〕(x+2)(x-3)；〔2〕(2x+5y)(3x-2y).-3x=x2-x-6例3=x2+2x-6=6x2=6x2+11xy-10y2-10y2-4xy+15xy探究新知計(jì)算：〔1〕(x+2)(x-3)；〔2〕(2x+5y)(3x計(jì)算：〔1〕(m-2n)(m2+mn-3n2)〔2〕(3x2-2x+2)(2x+1)〔1〕(m-2n)(m2+mn-3n2)=m·m2+m·mn-m·3n2-2n·m2-2n·mn+2n·3n2=m3+m2n-3mn2-2m2n-2mn2+6n3=m3-m2n-5mn2+6n3例4計(jì)算：〔1〕(m-2n)(m2+mn-3n2)〔2〕(3x2計(jì)算：〔1〕(m-2n)(m2+mn-3n2)〔2〕(3x2-2x+2)(2x+1)〔2〕(3x2-2x+2)(2x+1)=6x3+3x2-4x2-2x+4x+2=6x3-x2+2x+2例4計(jì)算：〔1〕(m-2n)(m2+mn-3n2)〔2〕(3x2

(1)(x+2y)(5a+3b);解：(x+2y)(5a+3b)=x

·5a

+x

·3b

+2y

·5a

+2y

·3b=5ax+3bx+10ay+6by補(bǔ)充例題計(jì)算(1)(x+2y)(5a+3b);(2)(2x–3)(x+4);解：(2x–3)(x+4)2x2

+8x

–3x

–12=2x2

+5x=–12補(bǔ)充例題計(jì)算(2)(2x–3)(x+4);解：(2x–3)(x+補(bǔ)充例題計(jì)算(3)(3x+y)(x–2y)；解：(3x+y)(x–2y)=3x2

–6xy

+xy

–2y2=3x2

–5xy

–2y2

補(bǔ)充例題計(jì)算(3)(3x+y)(x–2y)；解：(3x〔1〕(x+5)(x-7)；計(jì)算：〔2〕(x+5y)(x-7y)；〔3〕(2m+3n)(2m-3n)；〔4〕(2a+3b)2.隨堂練習(xí)=x2-7x+5x-35=x2-2x-35=x2-7xy+5xy-35y2=x2-2xy-35y2=4m2+6mn-6mn-9n2=4m2-9n2=4a2+12ab+9b2〔1〕(x+5)(x-7)；計(jì)算：〔2〕(x+5y)(x-7算一算:

(1)(2x+1)(x+3)(2)(m+2n)(m+3n)(3)(a-1)2

(4)(a+3b)(a–3b)2x2+7x+3m2+5mn+6n2a2-2a+1a2-9b2算一算:2x2+7x+3m2+5mn+6n2a2-2a+1a(5)(x+2)(x+3)

(6)(x-4)(x+1)

(7)(y+4)(y-2)(8)(y-5)(y-3)

x2+5x+6x2-3x-4y2+2y-8y2-8y+15(5)(x+2)(x+3)課堂小結(jié)多項(xiàng)式乘以多項(xiàng)式，展開(kāi)后項(xiàng)數(shù)很有規(guī)律，在合并同類(lèi)項(xiàng)之前，展開(kāi)式的項(xiàng)數(shù)恰好等于兩個(gè)多項(xiàng)式的項(xiàng)數(shù)的積.

多項(xiàng)式與多項(xiàng)式相乘，先用一個(gè)多項(xiàng)式的每一項(xiàng)分別乘以另一個(gè)多項(xiàng)式的每一項(xiàng)，再把所得的積相加.

課堂小結(jié)多項(xiàng)式乘以多項(xiàng)式，展開(kāi)后項(xiàng)數(shù)很有規(guī)律，在合并復(fù)習(xí)舊知計(jì)算以下各式，說(shuō)說(shuō)你是怎么想的？〔1〕(am+bm)÷m；〔2〕(a2+ab)÷a.〔1〕(am+bm)÷m〔2〕(a2+ab)÷a=am÷m+bm÷m=a+b=a2÷a+ab÷a=a+b復(fù)習(xí)舊知計(jì)算以下各式，說(shuō)說(shuō)你是怎么想的？〔1〕(am+bm)新課導(dǎo)入試一試計(jì)算：〔1〕(ax+bx)÷x；解〔1〕·x(a+b)x=ax+bx所以(ax+bx)÷x=a+b新課導(dǎo)入試一試計(jì)算：〔1〕(ax+bx)÷x；解〔1〕試一試〔2〕(ma+mb+mc)÷m.·m(a+b+c)m=ma+mb+mc所以(ma+mb+mc)÷m=a+b+c試一試〔2〕(ma+mb+mc)÷m.·m(a+b+c)m=探究新知例2計(jì)算：〔1〕(9x4-15x2+6x)÷3x〔2〕(28a3b2c+a2b3-14a2b2)÷(-7a2b)探究新知例2計(jì)算：〔1〕(9x4-15x2+6x)÷3x〔2〔1〕(9x4-15x2+6x)÷3x解9x43x=9x4÷3x-15x2-15x2÷3x+6x+6x÷3x=(9÷3)x4-1-(15÷3)x2-1+(6÷3)x1-1=3x3-5x+2多項(xiàng)式除以單項(xiàng)式，先用這個(gè)多項(xiàng)式的每一項(xiàng)除以這個(gè)單項(xiàng)式，再把所得的商相加.〔1〕(9x4-15x2+6x)÷3x解9x43x=9x4÷〔2〕(28a3b2c+a2b3-14a2b2)÷(-7a2b)解28a3b2c-7a2b=28a3b2c÷(-7a2b)+a2b3+a2b3÷(-7a2b)-14a2b2-14a2b2÷(-7a2b)=-4a3-2b2-1c+(a2-2b3-1)-(-2a2-2b2-1)=-4abc

b2+2b〔2〕(28a3b2c+a2b3-14a2b2)÷(-7a2補(bǔ)充例題計(jì)算：(6xn+2+3xn+1-3xn-1)÷3xn-1解：(6xn+2+3xn+1-3xn-1)÷3xn-1=6xn+2÷3xn-1+3xn+1÷3xn-1-3xn-1÷3xn-1=2xn+2-n+1+xn+1-n+1-1=2x3+x2-1思路歸納

如果除式中字母的指數(shù)是多項(xiàng)式，計(jì)算時(shí)要把它看作一個(gè)整體，且要添括號(hào).【選自《狀元大課堂》】補(bǔ)充例題計(jì)算：(6xn+2+3xn+1-3xn-1)÷3xn補(bǔ)充例題化簡(jiǎn)：[4(xy-1)2+(xy+2)(xy-2)]÷xy解：[4(xy-1)2+(xy+2)(xy-2)]÷xy=(4x2y2-8xy+4+x2y2-4)÷xy=(5x2y2-8xy)÷