2023八年級(jí)數(shù)學(xué)上冊(cè)第十四章整式的乘法與因式分解14.3因式分解課時(shí)1提公因式法作業(yè)課件新版新人教版

14.3因式分解課時(shí)1提公因式法1.[2022唐山路北區(qū)期末]下列各式從左到右的變形中,是因式分解的為(

)A.ax+bx+c=x(a+b)+cB.x2-1=(x+1)(x-1)C.x(a-b)=ax+bxD.x2-1+y2=(x+1)(x-1)+y2知識(shí)點(diǎn)1因式分解的概念答案1.B2.[2021天津河北區(qū)期末]多項(xiàng)式12ab3+8a3b各項(xiàng)的公因式是(

)A.ab B.2ab C.4ab D.4ab2知識(shí)點(diǎn)2公因式的概念答案2.C3.[2022長(zhǎng)沙開(kāi)福區(qū)期中]多項(xiàng)式3x2y2-12x2y4-6x3y3各項(xiàng)的公因式是(

)A.3xy B.3x2y2C.3x2y3 D.3x3y4知識(shí)點(diǎn)2公因式的概念答案3.B

多項(xiàng)式3x2y2-12x2y4-6x3y3中,各項(xiàng)系數(shù)的最大公約數(shù)是3,各項(xiàng)都含有字母x,y,字母x的最低次數(shù)是2,字母y的最低次數(shù)是2,所以3x2y2-12x2y4-6x3y3各項(xiàng)的公因式是3x2y2.4.

多項(xiàng)式2(a+b)2-8(a+b)(a-b)各項(xiàng)的公因式是(

)A.4(a+b) B.2(a+b)C.2(a+b)2(a-b) D.2(a+b)(a-b)知識(shí)點(diǎn)2公因式的概念答案4.B5.[2022滄州期末]將多項(xiàng)式a2-16a進(jìn)行因式分解的結(jié)果是(

)A.a(a+4)(a-4) B.(a-4)2C.a(a-16) D.(a+4)(a-4)知識(shí)點(diǎn)3用提公因式法分解因式答案5.C6.

將-3a2b-6ab2的各項(xiàng)提公因式后,另一個(gè)因式為(

)A.a+2b B.-a+2bC.-a-b D.a-2b知識(shí)點(diǎn)3用提公因式法分解因式答案6.A

多項(xiàng)式-3a2b-6ab2各項(xiàng)的公因式為-3ab,提出公因式-3ab后,另一個(gè)因式就是用原多項(xiàng)式-3a2b-6ab2除以公因式-3ab所得的商,顯然第一項(xiàng)為a,第二項(xiàng)為2b,所以另一個(gè)因式為a+2b.7.

分解因式-4x2y+2xy2-2xy的結(jié)果是(

)A.-2xy(2x-y+1) B.2xy(-2x+y)C.2xy(-2xy+y-1) D.-2xy(2x+y-1)知識(shí)點(diǎn)3用提公因式法分解因式答案7.A

-4x2y+2xy2-2xy=-(4x2y-2xy2+2xy)=-(2xy·2x-2xy·y+2xy)=-2xy(2x-y+1).8.

把多項(xiàng)式3(x-y)2+2(y-x)3分解因式結(jié)果正確的是(

)A.(x-y)2(3-2x-2y)B.(x-y)2(3-2x+2y)C.(x-y)2(3+2x-2y)D.(y-x)2(3+2x+2y)知識(shí)點(diǎn)3用提公因式法分解因式答案8.B

3(x-y)2+2(y-x)3=3(x-y)2-2(x-y)3=(x-y)2·[3-2(x-y)]=(x-y)2(3-2x+2y).9.

分解因式2m(m-n)2-8m2(n-m)=

.

知識(shí)點(diǎn)3用提公因式法分解因式答案9.2m(m-n)(5m-n)

2m(m-n)2-8m2(n-m)=2m(m-n)2+8m2(m-n)=2m(m-n)(m-n+4m)=2m(m-n)(5m-n).10.

用提公因式法將下列各式分解因式:(1)4x2y3+8x2y2z-12xy2z;(2)5x(x-2y)3-20y(2y-x)3.知識(shí)點(diǎn)3用提公因式法分解因式答案10.解:(1)4x2y3+8x2y2z-12xy2z=4xy2(xy+2xz-3z).(2)5x(x-2y)3-20y(2y-x)3=5x(x-2y)3+20y(x-2y)3=5(x-2y)3(x+4y).11.

若長(zhǎng)和寬分別是a,b的長(zhǎng)方形的周長(zhǎng)為10,面積為4,則a2b+ab2的值為(

)A.14 B.16 C.20 D.40知識(shí)點(diǎn)4用提公因式法分解因式的應(yīng)用答案11.C

根據(jù)題意,得2(a+b)=10,ab=4,∴a+b=5,∴a2b+ab2=ab(a+b)=20.12.利用因式分解簡(jiǎn)化運(yùn)算:(1)2021+20212-2021×2022;(2)56.2×1999-462×199.9.知識(shí)點(diǎn)4用提公因式法分解因式的應(yīng)用答案12.解:(1)2021+20212-2021×2022=2021×(1+2021-2022)=0.(2)56.2×1999-462×199.9=562×199.9-462×199.9=199.9×(562-462)=199.9×100=19990.1.

把-a(x-y)-b(y-x)+c(x-y)分解因式,結(jié)果正確的是(

)A.(x-y)(-a-b-c) B.(y-x)(a-b-c)C.-(x-y)(a+b+c) D.-(y-x)(a+b-c)答案1.B

-a(x-y)-b(y-x)+c(x-y)=-a(x-y)+b(x-y)+c(x-y)=(x-y)(-a+b+c)=(y-x)(a-b-c).2.[2022長(zhǎng)春外國(guó)語(yǔ)學(xué)校期中](-2)2021+(-2)2022等于(

)A.-22021 B.-22022 C.22021 D.-2答案2.C

(-2)2021+(-2)2022=(-2)2021×(1-2)=22021.3.

多項(xiàng)式(x+2)(2x-1)-(x+2)可以因式分解成2(x+m)(x+n),則m-n的值是(

)A.0 B.4 C.3或-3 D.1答案3.C

∵(x+2)(2x-1)-(x+2)=(x+2)(2x-2)=2(x+2)(x-1),(x+2)(2x-1)-(x+2)可以因式分解成2(x+m)(x+n),∴2(x+2)(x-1)=2(x+m)(x+n),∴m=2,n=-1或m=-1,n=2,∴m-n=3或m-n=-3.4.

分解因式:(a-b)2-6b+6a=

.

答案4.(a-b)(a-b+6)

(a-b)2-6b+6a=(a-b)2+6(a-b)=(a-b)(a-b+6).5.[2021婁底期末]已知ab=2,a-3b=-5,則代數(shù)式a2b-3ab2+ab的值為

.

答案5.-8

∵ab=2,a-3b=-5,∴a2b-3ab2+ab=ab(a-3b+1)=2×(-5+1)=-8.6.[2021內(nèi)江中考]若實(shí)數(shù)x滿足x2-x-1=0,則x3-2x2+2021=

.

答案6.2020

∵x2-x-1=0,∴x2-x=1,∴x3-2x2+2021=x3-x2-x2+2021=x(x2-x)-x2+2021=x-x2+2021=-(x2-x)+2021=-1+2021=2020.7.

用提公因式法將下列各式分解因式:(1)-4a3b2+12a2b-4ab;(2)(a2-ab)+c(a-b);(3)(3a-4b)(7a-8b)+(11a-12b)(7a-8b).答案7.解:(1)-4a3b2+12a2b-4ab=-(4ab·a2b-4ab·3a+4ab)=-4ab(a2b-3a+1).(2)(a2-ab)+c(a-b)=a(a-b)+c(a-b)=(a-b)(a+c).(3)(3a-4b)(7a-8b)+(11a-12b)(7a-8b)=(7a-8b)(3a-4b+11a-12b)=(7a-8b)(14a-16b)=2(7a-8b)(7a-8b)=2(7a-8b)2.

答案8.解:7y(x-3y)2-2(3y-x)3=7y(x-3y)2+2(x-3y)3=(x-3y)2[7y+2(x-3y)]=(x-3y)2(7y+2x-6y)=(x-3y)2(2x+y),∵2x+y=6,x-3y=1,∴原式=6.9.

已知△ABC的三邊長(zhǎng)a,b,c滿足a2-bc-ab+ac=0.求證:△ABC是等腰三角形.答案9.解:因?yàn)閍2-bc-ab+ac=0,所以a2-ab-bc+ac=0,所以(a2-ab)+(ac-bc)=0,所以a(a-b)+c(a-b)=0,則(a-b)(a+c)=0,因?yàn)閍+c≠0,所以a-b=0,所以a=b,所以△ABC是等腰三角形.10.

先閱讀下列因式分解的過(guò)程,再回答所提出的問(wèn)題:

1+x+x(x+1)+x(x+1)2=(1+x)[1+x+x(x+1)]=(1+x)(1+x)2=(1+x)3.(1)上述分解因式的方法是

,共應(yīng)用了

次;

(2)若分解1+x+x(x+1)+x(x+1)2+…+x(x+1)2022,則需應(yīng)用上述方法

次,結(jié)果是

;

(3)分解因式:1+x+x(x+1)+x(x