2024冀教版七年級(jí)下冊(cè)第八章 整式的乘法課堂練習(xí)題及答案第2課時(shí) 單項(xiàng)式乘多項(xiàng)式

第八章整式的乘法8.4

整式的乘法第2課時(shí)

單項(xiàng)式乘多項(xiàng)式第八章整式的乘法單項(xiàng)式乘多項(xiàng)式

1.計(jì)算:-x(x2+1)=(

)A.x3-1 B.-x3-x C.x3+x

D.x2-x基礎(chǔ)通關(guān)B65432179810

B65432179810A3.若

×xy=2x2y+xy2,則

內(nèi)應(yīng)填的式子是(

)A.2x+y

B.x+2y

C.2xy+2y

D.xy+24.今天數(shù)學(xué)課上,老師講了單項(xiàng)式乘多項(xiàng)式,放學(xué)回到家,嘉琪拿出課堂筆記復(fù)習(xí),發(fā)現(xiàn)一道題:-3xy·(4y-2x-1)=-12xy2+6x2y+

,的地方被鋼筆水弄污了,你認(rèn)為

內(nèi)應(yīng)填寫的是(

)A.3xy

B.-3xy

C.-1 D.1A654321798105.已知3x-4y+2=0,則代數(shù)式x(y-9)-y(x-12)的值為

.

6.若ab=1,則ab2(a2b-2a)+1=

.

60654321798107.計(jì)算:(1)(a2-a+1)·(-2a)2;(2)(3x2y-2x+1)(-2xy);解:(a2-a+1)·(-2a)2=(a2-a+1)·4a2=4a4-4a3+4a2.解:(3x2y-2x+1)(-2xy)=-6x3y2+4x2y-2xy.65432179810(3)x(x+2)+(x+1)·(-x2)-4x;(4)-3x2y3(x2-1)-(x2+1)·5x2y3.解:x(x+2)+(x+1)·(-x2)-4x=x2+2x-x3-x2-4x=-x3-2x.解:-3x2y3(x2-1)-(x2+1)·5x2y3=-3x4y3+3x2y3-5x4y3-5x2y3=-8x4y3-2x2y3.654321798108.先化簡(jiǎn),再求值:2xy(x-3y-1)-y(2x2-6xy-3x),其中x=-3,y=2.解:2xy(x-3y-1)-y(2x2-6xy-3x)=2x2y-6xy2-2xy-2x2y+6xy2+3xy=xy.當(dāng)x=-3,y=2時(shí),原式=(-3)×2=-6.65432179810單項(xiàng)式乘多項(xiàng)式的應(yīng)用9.一個(gè)長(zhǎng)方體的長(zhǎng)、寬、高分別為2x,2x-1,x2,則它的體積是(

)A.4x4-4x2 B.4x4-2x3C.4x3-2x2 D.4x4

10.張伯伯擁有一個(gè)長(zhǎng)方形魚塘,已知長(zhǎng)為2a-3b+3,寬為ab2,則魚塘的面積為

.

B2a2b2-3ab3+3ab2

6543217981011.要使-x3(x2+ax+1)+2x4中不含有x的四次項(xiàng),則a等于(

)A.1 B.2

C.3

D.412.已知x2-2=y,則x(x-2023y)-y(1-2023x)的值為(

)A.1 B.2

C.3

D.413.對(duì)于任意的有理數(shù)x,y,若存在a,b使得8x+(a-2b)y=ax-2b(x-2y)恒成立,則a+b=

.

能力突破BB14151413121114.解方程:x(x+1)-3x(x-2)+2x(x+1)=3.

151413121115.已知x(x-m)+n(x+m)=x2+5x-6對(duì)任意有理數(shù)x都成立,求m(n-1)+n(m+1)的值.

151413121116.閱讀下列文字,并解決問題.已知x2y=3,求2xy(x5y2-3x3y-4x)的值.解:2xy(x5y2-3x3y-4x)=2x6y3-6x4y2-8x2y=2(x2y)3-6(x2y)2-8x2y=2×33-6×32-8×3=-24.素養(yǎng)達(dá)標(biāo)16這樣的方法我們稱為“整體代入法”.請(qǐng)你用上述方法解決下面的問題:已知ab=3,求(2a3b2-3a2b-4a)·(-2b)的值.解:(2a3