2020版七年級(jí)數(shù)學(xué)下冊(cè)第一章整式的乘除1.4整式的乘法第2課時(shí)課件新版北師大版

1、4整式的乘法第2課時(shí)【知識(shí)再現(xiàn)知識(shí)再現(xiàn)】乘法對(duì)加法的分配律乘法對(duì)加法的分配律:a(b+c)=_.:a(b+c)=_. ab+acab+ac【新知預(yù)習(xí)新知預(yù)習(xí)】閱讀教材閱讀教材p16p16【想一想想一想】, ,解決以下問(wèn)題解決以下問(wèn)題: :仿照有理數(shù)的乘法分配律進(jìn)行計(jì)算仿照有理數(shù)的乘法分配律進(jìn)行計(jì)算: :(1)(3x-1)(-2)=_.(1)(3x-1)(-2)=_.(2)(-3x-1)3x=_.(2)(-3x-1)3x=_.(3)ax(cx-b+1)=_.(3)ax(cx-b+1)=_.-6x+2-6x+2-9x-9x2 2-3x-3xacxacx2 2-abx+ax-abx+ax你發(fā)現(xiàn)的規(guī)律

2、是你發(fā)現(xiàn)的規(guī)律是: :(1)(1)語(yǔ)言敘述語(yǔ)言敘述: :單項(xiàng)式與多項(xiàng)式相乘單項(xiàng)式與多項(xiàng)式相乘, ,就是根據(jù)就是根據(jù)_用單項(xiàng)式去乘多項(xiàng)式的用單項(xiàng)式去乘多項(xiàng)式的_,_,再把所得的再把所得的積積_._.(2)(2)字母表示字母表示:m(a+b+c)=_.:m(a+b+c)=_.分配分配律律每一項(xiàng)每一項(xiàng)相加相加ma+mb+mcma+mb+mc【基礎(chǔ)小練基礎(chǔ)小練】請(qǐng)自我檢測(cè)一下預(yù)習(xí)的效果吧請(qǐng)自我檢測(cè)一下預(yù)習(xí)的效果吧! !1.m(a1.m(a2 2-b-b2 2+c)+c)等于等于( )( )a.maa.ma2 2-mb-mb2 2+m+mb.mab.ma2 2+mb+mb2 2+mc+mcc.mac.m

3、a2 2-mb-mb2 2+mc+mcd.mad.ma2 2-b-b2 2+c+cc c2.2.計(jì)算計(jì)算(3x(3x2 2+1)2x+1)2x的結(jié)果是的結(jié)果是 ( )( )a.5xa.5x3 3+2x+2xb.6xb.6x3 3+1+1c.6xc.6x3 3+2x+2xd.6xd.6x2 2+2x+2xc c3.3.計(jì)算計(jì)算-4a(2a-4a(2a2 2+3a-1)+3a-1)的結(jié)果是的結(jié)果是( )( )a.-8aa.-8a3 3+12a+12a2 2-4a-4ab.-8ab.-8a3 3-12a-12a2 2+1+1c.-8ac.-8a3 3-12a-12a2 2+4a+4ad.8ad.8a

4、3 3+12a+12a2 2+4a+4ac c4.4.計(jì)算計(jì)算:(2x-5)3x=_.:(2x-5)3x=_.6x6x2 2-15x-15x知識(shí)點(diǎn)一知識(shí)點(diǎn)一 單項(xiàng)式乘多項(xiàng)式單項(xiàng)式乘多項(xiàng)式(p16(p16例例2 2補(bǔ)充補(bǔ)充) )【典例典例1 1】(2019(2019南通崇川區(qū)月考南通崇川區(qū)月考) )計(jì)算計(jì)算: :(1)(-6xy)(1)(-6xy) (2)(-2mn(2)(-2mn2 2) )2 2-4mn-4mn3 3(mn+1).(mn+1).(3)3a(3)3a2 2(a(a3 3b b2 2-2a)-4a(-a-2a)-4a(-a2 2b)b)2 2. .2321(2xyx y ).3【

5、自主解答自主解答】(1)(1)原式原式=(-6xy)=(-6xy)2xy2xy2 2+ +(-6xy) (-6xy) =-12x=-12x2 2y y3 3+2x+2x4 4y y3 3. .(2)(2)原式原式=4m=4m2 2n n4 4-4m-4m2 2n n4 4-4mn-4mn3 3=-4mn=-4mn3 3. .(3)3a(3)3a5 5b b2 2-6a-6a3 3-4a-4a(a(a4 4b b2 2) )=3a=3a5 5b b2 2-6a-6a3 3-4a-4a5 5b b2 2=-a=-a5 5b b2 2-6a-6a3 3. .321(x y )3【學(xué)霸提醒學(xué)霸提醒】單

6、項(xiàng)式乘以多項(xiàng)式的三點(diǎn)注意單項(xiàng)式乘以多項(xiàng)式的三點(diǎn)注意1.1.單項(xiàng)式與多項(xiàng)式相乘單項(xiàng)式與多項(xiàng)式相乘, ,根據(jù)分配律根據(jù)分配律, ,用單項(xiàng)式乘多項(xiàng)用單項(xiàng)式乘多項(xiàng)式的各項(xiàng)式的各項(xiàng), ,就將其轉(zhuǎn)化為單項(xiàng)式的乘法就將其轉(zhuǎn)化為單項(xiàng)式的乘法, ,不可漏乘項(xiàng)不可漏乘項(xiàng). .2.2.在確定積的每一項(xiàng)符號(hào)時(shí)在確定積的每一項(xiàng)符號(hào)時(shí), ,既要看多項(xiàng)式中每一項(xiàng)既要看多項(xiàng)式中每一項(xiàng)的符號(hào)的符號(hào), ,又要看單項(xiàng)式的符號(hào)又要看單項(xiàng)式的符號(hào), ,才能正確確定積的每一才能正確確定積的每一項(xiàng)的符號(hào)項(xiàng)的符號(hào). .3.3.對(duì)于含有乘方、乘法、加減法的混合運(yùn)算的題目對(duì)于含有乘方、乘法、加減法的混合運(yùn)算的題目, ,要注意運(yùn)算順序要注意運(yùn)算順

7、序, ,也要注意合并同類項(xiàng)也要注意合并同類項(xiàng), ,得出最簡(jiǎn)結(jié)果得出最簡(jiǎn)結(jié)果. .【題組訓(xùn)練題組訓(xùn)練】1.(20191.(2019寧波海曙區(qū)期中寧波海曙區(qū)期中) )把把2a(ab-b+c)2a(ab-b+c)化簡(jiǎn)后得化簡(jiǎn)后得( )( )a.2aa.2a2 2b-ab+acb-ab+acb.2ab.2a2 2-2ab+2ac-2ab+2acc.2ac.2a2 2b+2ab+2acb+2ab+2acd.2ad.2a2 2b-2ab+2acb-2ab+2acd d2.(20192.(2019青島中考青島中考) )計(jì)算計(jì)算(-2m)(-2m)2 2(-mm(-mm2 2+3m+3m3 3) )的的結(jié)果是

8、結(jié)果是( )( )a.8ma.8m5 5b.-8mb.-8m5 5c.8mc.8m6 6d.-4md.-4m4 4+12m+12m5 5a a3.(20193.(2019德惠月考德惠月考) )計(jì)算計(jì)算(x(x3 3) )2 2(x(x2 2+2x+1)+2x+1)的結(jié)果是的結(jié)果是( )( )a.xa.x4 4+2x+2x3 3+x+x2 2b.xb.x5 5+2x+2x4 4+x+x3 3c.xc.x8 8+2x+2x7 7+x+x6 6d.xd.x8 8+2x+2x4 4+x+x3 3c c4.4.計(jì)算計(jì)算:-2a:-2a2 2(a-3ab)=_.(a-3ab)=_.-2a-2a3 3+6a

9、+6a3 3b b5.5.計(jì)算計(jì)算:(1)3x:(1)3x2 2(-y-xy(-y-xy2 2+x+x2 2).).(2)(2019(2)(2019淮安中考淮安中考)ab(3a-2b)+2ab)ab(3a-2b)+2ab2 2. .解解: :(1)3x(1)3x2 2(-y-xy(-y-xy2 2+x+x2 2) )=3x=3x2 2(-y)-3x(-y)-3x2 2(xy(xy2 2)+3x)+3x2 2x x2 2=-3x=-3x2 2y-3xy-3x3 3y y2 2+3x+3x4 4. .(2)ab(3a-2b)+2ab(2)ab(3a-2b)+2ab2 2=3a=3a2 2b-2ab

10、b-2ab2 2+2ab+2ab2 2=3a=3a2 2b.b.知識(shí)點(diǎn)二知識(shí)點(diǎn)二 單項(xiàng)式與多項(xiàng)式相乘的應(yīng)用單項(xiàng)式與多項(xiàng)式相乘的應(yīng)用(p17(p17習(xí)題習(xí)題t2t2補(bǔ)充補(bǔ)充) )【典例典例2 2】如圖如圖, ,一長(zhǎng)方形地塊用來(lái)建造學(xué)校、小區(qū)、一長(zhǎng)方形地塊用來(lái)建造學(xué)校、小區(qū)、噴泉噴泉, ,求這塊地的面積求這塊地的面積. .【規(guī)范解答規(guī)范解答】這塊地的面積為這塊地的面積為: :3x(2x-y)+(3x+2y)(3x+x)+x(2x-y)3x(2x-y)+(3x+2y)(3x+x)+x(2x-y)列代數(shù)式列代數(shù)式=6x=6x2 2-3xy+12x-3xy+12x2 2+8xy+2x+8xy+2x2 2

11、-xy-xy單項(xiàng)式乘以多項(xiàng)式單項(xiàng)式乘以多項(xiàng)式=20 x=20 x2 2+4xy.+4xy.合并同類項(xiàng)合并同類項(xiàng)【學(xué)霸提醒學(xué)霸提醒】單項(xiàng)式與多項(xiàng)式相乘的單項(xiàng)式與多項(xiàng)式相乘的“三種題型三種題型”1.1.化簡(jiǎn)求值務(wù)必是先化簡(jiǎn)化簡(jiǎn)求值務(wù)必是先化簡(jiǎn), ,再求值再求值. .2.2.探究規(guī)律常見(jiàn)的有探究規(guī)律常見(jiàn)的有: :探究數(shù)字的變化規(guī)律探究數(shù)字的變化規(guī)律, ,數(shù)形結(jié)合數(shù)形結(jié)合探究規(guī)律探究規(guī)律. .3.3.列式計(jì)算常與面積等問(wèn)題結(jié)合出題列式計(jì)算常與面積等問(wèn)題結(jié)合出題. .【題組訓(xùn)練題組訓(xùn)練】1.1.一個(gè)三角形的底為一個(gè)三角形的底為2m,2m,高為高為m+2n,m+2n,它的面積是它的面積是( )( )a.2

12、ma.2m2 2+4mn+4mnb.mb.m2 2+2mn+2mnc.mc.m2 2+4mn+4mnd.2md.2m2 2+2mn+2mnb b2.2.要使要使(x(x3 3+ax+ax2 2-x)(-8x-x)(-8x4 4) )的運(yùn)算結(jié)果中不含的運(yùn)算結(jié)果中不含x x6 6的的項(xiàng)項(xiàng), ,則則a a的值應(yīng)為的值應(yīng)為 ( )( )a.8a.8b.-8b.-8c.c. d.0d.018d d3.(3.(易錯(cuò)提醒題易錯(cuò)提醒題) )已知已知x(x-2)=3,x(x-2)=3,則代數(shù)式則代數(shù)式2x2x2 2-4x-7-4x-7的值為的值為( )( )a.6a.6b.-4b.-4c.13c.13d.-1d

13、.-1d d4.4.若若2x(x-2)=ax2x(x-2)=ax2 2+bx,+bx,則則a,ba,b的值為的值為( )( )a.a=1,b=2a.a=1,b=2b.a=2,b=-2b.a=2,b=-2c.a=2,b=4c.a=2,b=4d.a=2,b=-4d.a=2,b=-4d d5.5.若若abab2 2=-1,=-1,求求-ab(a-ab(a2 2b b5 5-ab-ab3 3-2b)-2b)的值的值. .世紀(jì)金榜導(dǎo)學(xué)號(hào)世紀(jì)金榜導(dǎo)學(xué)號(hào)解解: :原式原式=-a=-a3 3b b6 6+a+a2 2b b4 4+2ab+2ab2 2=-(ab=-(ab2 2) )3 3+(ab+(ab2 2

14、) )2 2+2(ab+2(ab2 2),),因?yàn)橐驗(yàn)閍bab2 2=-1,=-1,所以原式所以原式=1+1-2=0.=1+1-2=0.【火眼金睛火眼金睛】計(jì)算計(jì)算:-9xy:-9xy2 2(-2x+4y-1).(-2x+4y-1).【正解正解】-9xy-9xy2 2(-2x+4y-1)(-2x+4y-1)=-9xy=-9xy2 2(-2x)+(-9xy(-2x)+(-9xy2 2)4y+(-9xy)4y+(-9xy2 2)(-1)(-1)=18x=18x2 2y y2 2-36xy-36xy3 3+9xy+9xy2 2. . 【一題多變一題多變】某同學(xué)在計(jì)算一個(gè)多項(xiàng)式乘以某同學(xué)在計(jì)算一個(gè)多項(xiàng)

15、式乘以-2a-2a時(shí)時(shí), ,因抄錯(cuò)運(yùn)算符因抄錯(cuò)運(yùn)算符號(hào)號(hào), ,算成了加上算成了加上-2a,-2a,得到的結(jié)果是得到的結(jié)果是a a2 2+2a-1,+2a-1,那么正確那么正確的計(jì)算結(jié)果是多少的計(jì)算結(jié)果是多少? ? 解解: :因?yàn)橛?jì)算一個(gè)多項(xiàng)式乘以因?yàn)橛?jì)算一個(gè)多項(xiàng)式乘以-2a-2a時(shí)時(shí), ,因抄錯(cuò)運(yùn)算符號(hào)因抄錯(cuò)運(yùn)算符號(hào), ,算成了加上算成了加上-2a,-2a,得到的結(jié)果是得到的結(jié)果是a a2 2+2a-1, +2a-1, 所以這個(gè)多項(xiàng)式為所以這個(gè)多項(xiàng)式為:a:a2 2+2a-1+2a=a+2a-1+2a=a2 2+4a-1, +4a-1, 所以正確的計(jì)算結(jié)果是所以正確的計(jì)算結(jié)果是:-2a(a:-

16、2a(a2 2+4a-1)=-2a+4a-1)=-2a3 3-8a-8a2 2+2a.+2a. 【母題變式母題變式】【變式一變式一】若要使若要使x(xx(x2 2+a+3)=x(x+a+3)=x(x2 2+5)+2(b+2)+5)+2(b+2)成立成立, ,則則a,ba,b的值分別為的值分別為_(kāi)._.2,-22,-2【變式二變式二】閱讀閱讀: :已知已知x x2 2y=3,y=3,求求2xy(x2xy(x5 5y y2 2-3x-3x3 3y-4x)y-4x)的的值值. .分析分析: :考慮到考慮到x,yx,y的可能值較多的可能值較多, ,不能逐一代入求解不能逐一代入求解, ,故故考慮整體思想考慮整體思想, ,將將x x2 2y=3y=3整體代入整體代入. .解解: :2xy(x2xy(x5 5y y2 2-3x-3x3 3y-4x)y-4x)=2x=2x6 6y y3 3-6x-6x4 4y y2 2-8x-8x2 2y y=2(x=2(x2 2y)y)3 3-6(x-6(x2 2y)y)2 2-8x-8x2 2y y=2=23 33 3-6-63 32 2-8-83 3=-24.=-24.你能用上述方法解決以下問(wèn)題嗎你能用上述方法解決以下問(wèn)題嗎? ?試一試試一試! !已知已知ab=3,ab=3,求求(2a(2a3 3b