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12.2 整式的乘法 华东师大版八年级数学上册同步练习题(含答案)
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这是一份12.2 整式的乘法 华东师大版八年级数学上册同步练习题(含答案),共14页。
12.2整式的乘法 知识点分类练习题
一.单项式乘单项式
1.计算2x2•(﹣3x)的结果是( )
A.﹣6x2 B.5x3 C.6x3 D.﹣6x3
2.下列计算正确的是( )
A.3x3•2x2y=6x5 B.2a2•3a3=6a5
C.(﹣2x)•(﹣5x2y)=﹣10x3y D.(﹣2xy)•(﹣3x2y)=6x3y
3.如果单项式﹣3ma﹣2bn2a+b与m3n8b是同类项,那么这两个单项式的积是( )
A.﹣3m6n16 B.﹣3m6n32 C.﹣3m3n8 D.﹣9m6n16
4.若单项式﹣8xay和x2yb的积为﹣2x5y6,则ab的值为( )
A.2 B.30 C.﹣15 D.15
5.已知单项式3x2y3与﹣5x2y2的积为mx4yn,那么m﹣n= .
6.计算(﹣8ab)•(a2b)= .
7.计算:(﹣2ab2)•(﹣3a2)= .
8.计算:
(1)(﹣3x)3•(5x2y);
(2)(﹣2)3+(﹣3)×(﹣4)2.
9.计算:
(1)3(x3)2x3﹣(3x3)3+(5x)2•x7;
(2)(﹣2ab)(3a2•2ab﹣b2)
10.(﹣2y3)2+(﹣4y2)3﹣(﹣2y)2•(﹣3y2)2.
二.单项式乘多项式
11.若A(m2﹣3n)=m3﹣3mn,则代数式A的值为( )
A.m B.mn C.mn2 D.m2n
12.计算:3a•(2a﹣5)= .
13.﹣2a(3a﹣4b)= .
14.计算:﹣6a•(﹣﹣a+2)
15.(﹣3y)(4x2y﹣2xy).
16.计算:
(1)(﹣5x)•(3x2﹣4x+5):
(2)﹣2a•(3ab2﹣5ab3):
(3)(﹣a2b)(2a﹣ab+3b);
(4)﹣2xn•(﹣3xn+1+4xn﹣1).
17.计算:6m•(3m2﹣m﹣1)
18.计算:(x2+x﹣1)•(﹣2x)= .
19.计算:
20.计算下列各题.
(1)3a2b(﹣4a2b+2ab2﹣ab);
(2).
21.[xy(x2﹣xy)﹣x2y(x﹣y)]•3xy2.
22..
23.已知a2﹣2a﹣3=0,则代数式3a(a﹣2)的值为 .
24.先化简,再求值:3a(2a2﹣4a+3)﹣2a2(3a+4),其中a=﹣2.
25.先化简,再求值:(x﹣2y)2﹣x(x+3y)﹣4y2,其中x=﹣4,y=.
26.今天数学课上,老师讲了单项式乘多项式,放学回到家,小明拿出课堂笔记复习,发现一道题:﹣7xy(2y﹣x﹣3)=﹣14xy2+7x2y□,□的地方被钢笔水弄污了,你认为□内应填写( )
A.+21xy B.﹣21xy C.﹣3 D.﹣10xy
27.要使﹣x3(x2+ax+1)+2x4中不含有x的四次项,则a等于( )
A.1 B.2 C.3 D.4
28.要使(﹣6x3)(x2+ax﹣3)的展开式中不含x4项,则a=( )
A.1 B.0 C.﹣1 D.
29.若要使x(x2+a)+3x﹣2b=x3+5x+4恒成立,则a,b的值分别是( )
A.﹣2,﹣2 B.2,2 C.2,﹣2 D.﹣2,2
30.某同学在计算﹣3x加上一个多项式时错将加法做成了乘法,得到的答案是3x3﹣3x2+3x,由此可以推断出正确的计算结果是( )
A.﹣x2﹣2x﹣1 B.x2+2x﹣1 C.﹣x2+4x﹣1 D.x2﹣4x+1
31.某同学在计算一个多项式乘以﹣3x2时,因抄错运算符号,算成了加上﹣3x2,得到的结果是x2﹣4x+1,那么正确的计算结果是多少?
三.多项式乘多项式
32.计算:(3m﹣1)(m+5).
33.计算:(x﹣2y)(2x+y)+x(﹣2x﹣y).
34.3(2x﹣1)(x﹣3)﹣2(3x﹣2)(2x﹣3)
35.计算:
(1)(a2+3)(a﹣2)﹣a(a2﹣2a﹣2);
(2)(2m+n)(2m﹣n)+(m+n)2﹣2(2m2﹣mn).
36.化简:
(1)(x3﹣1)(x6+x3+1)(x9+1);
(2)(x2﹣y2)(x2+xy+y2)(x2﹣xy+y2);
(3)(x+2y)2(x2﹣2xy+4y2)2.
37.化简:(3x﹣1)(2x2+3x﹣4)
38.化简:(x﹣y)(x+3y)﹣x(x+2y).
39.化简:x(x﹣3)﹣(x﹣1)(x+2).
40.计算:(x+3)(x﹣7)﹣x(x﹣1).
41.计算:(2a﹣b)(4a2+2ab+b2)= .
42.计算下列各题:
(1)(2m﹣3)(2m+5);
(2)(﹣a•a2)(﹣b)2+(﹣2a3b2)2+(﹣2a3b2).
43.计算:(a+b+c)(b+c+d)﹣(b+c)(a+b+c+d).
44.已知(x﹣1)(x+3)=x2+mx+n,则m﹣n的值是( )
A.﹣4 B.﹣1 C.1 D.5
45.(x﹣3)(x﹣5)=x2+px+15,则p的值是( )
A.﹣3 B.8 C.﹣8 D.﹣5
46.已知(x+my)(x+ny)=x2+2xy﹣8y2,求m2n+mn2的值.
47.解方程:(x﹣3)(x﹣2)=(x+9)(x+1)+4.
48.解方程:(2x+5)(3x﹣1)+(2x+3)(1﹣3x)=28.
49.(x+m)与(x+3)的乘积中不含x的一次项,则m的值是( )
A.0 B.±3 C.3 D.﹣3
50.若(3x+m)(3x+n)的结果中不含有x项,则m、n的关系是( )
A.mm=1 B.mm=0 C.m﹣n=0 D.m+n=0
51.若多项式mx+6y与x﹣3y的乘积中不含有xy项,则m的值为( )
A.﹣6 B.﹣3 C.0 D.2
52.若(x2+ax+2)(2x﹣2)的结果中不含x项,则a的值为( )
A.0 B.2 C. D.﹣2
53.已知(x﹣2)(x2﹣mx+n)的结果中不含x2项和x的项,求(m+n)(m2﹣mn+n2)的值.
54.已知(x2+mx+1)(x﹣n)的展开式中不含x项,x2项的系数为﹣2,求mn+m﹣n的值.
55.试说明:代数式(2x+3)(6x+2)﹣6x(2x+13)+8(7x+2)的值与x的取值无关.
56.在计算(x+a)(x+b)时,甲把b错看成了6,得到结果是:x2+8x+12.求出a的值.
57.在计算(x+a)(x+b)时,甲把b错看成了6,得到结果是:x2+8x+12.
(1)求出a的值;
(2)在(1)的条件下,且b=﹣3时,计算(x+a)(x+b)的结果.
58.小轩计算一道整式乘法的题:(2x+m)(5x﹣4),由于小轩将第一个多项式中的“+m”抄成“﹣m”,得到的结果为10x2﹣33x+20.
(1)求m的值;
(2)请计算出这道题的正确结果.
参考答案
一.单项式乘单项式
1.解:原式=2•(﹣3)x2•x=﹣6x3,
故选:D.
2.解:A、3x3×2x2y=6x5y,故此选项错误;
B、2a2×3a3=6a5,故此选项正确;
C、(﹣2x)×(﹣5x2y)=10x3y,故此选项错误;
D、(﹣2xy)×(﹣3x2y)=6x3y2,故此选错误.
故选:B.
3.解:∵单项式﹣3ma﹣2bn2a+b与m3n8b是同类项,
∴,
解得:,
故单项式﹣3ma﹣2bn2a+b与m3n8b是单项式﹣3m3n16与m3n16,
则这两个单项式的积是:﹣3m3n16•m3n16=﹣3m6n32.
故选:B.
4.解:﹣8xay×x2yb=﹣2xa+2yb+1=﹣2x5y6,
∴a+2=5,b+1=6,
解得a=3,b=5,
∴ab=3×5=15,
故选:D.
5.解:3x2y3×(﹣5x2y2)=﹣15x4y5,
∴mx4yn=﹣15x4y5,
∴m=﹣15,n=5
∴m﹣n=﹣15﹣5=﹣20
故答案为:﹣20
6.解:(﹣8ab)•(a2b)
=(﹣8×)•(a•a2)•(b•b)
=﹣6a3b2,
故答案为:﹣6a3b2.
7.解:(﹣2ab2)•(﹣3a2)
=6a3b2.
故答案为:6a3b2.
8.解:(1)(﹣3x)3•(5x2y)
=﹣27x3•5x2y
=﹣135x5y;
(2)(﹣2)3+(﹣3)×(﹣4)2
=﹣8+(﹣3)×16
=﹣8﹣48
=﹣56.
9.解:(1)原式=3x9﹣27x9+25x9
=x9;
(2)原式=﹣12a4b2+2ab3.
10.解:(﹣2y3)2+(﹣4y2)3﹣(﹣2y)2•(﹣3y2)2
=4y6﹣64y6﹣4y2•(9y4)
=4y6﹣64y6﹣36y6
=﹣96y6.
二.单项式乘多项式
11.解:∵A(m2﹣3n)=m3﹣3mn=m(m2﹣3n),
∴A=m.
故选:A.
12.解:3a•(2a﹣5)
=6a2﹣15a.
故答案为:6a2﹣15a.
13.解:﹣2a(3a﹣4b)=﹣6a2+8ab.
故答案为:﹣6a2+8ab.
14.解:﹣6a•(﹣﹣a+2)=3a3+2a2﹣12a.
15.解:(﹣3y)(4x2y﹣2xy)
=(﹣3y)(4x2y)+(﹣3y)(﹣2xy)
=﹣12x2y2+6xy2.
16.解:(1)原式=﹣15x3+20x2﹣25x;
(2)原式=﹣6a2b2+10a2b3;
(3)原式=﹣2a3b+a3b2﹣3a2b2;
(4)原式=6x2n+1﹣8x2n﹣1.
17.解:6m•(3m2﹣m﹣1)
=18m3﹣4m2﹣6m.
18.解:(x2+x﹣1)•(﹣2x)=﹣2x3﹣x2+2x.
19.解:原式=a2b2(﹣a2b﹣12ab+b2)
=a2b2•(﹣a2b)﹣a2b2•12ab+a2b2•b2
=﹣8a4b3﹣a3b3+a2b4.
20.解:(1)原式=﹣12a4b2+6a3b3﹣3a3b2;
(2)原式=﹣5x3y+5x2y2﹣x3y﹣2x2y2
=﹣6x3y+3x2y2.
21.解:[xy(x2﹣xy)﹣x2y(x﹣y)]•3xy2
=(x3y﹣x2y2﹣x3y+x2y2)•3xy2
=0.
22.解:原式=﹣8x3(2x3﹣x﹣1)﹣(4x4+8x3)
=﹣16x6+4x4+8x3﹣4x4﹣8x3
=﹣16x6.
23.解:∵a2﹣2a﹣3=0,
∴a2﹣2a=3,
∴3a(a﹣2)=3(a2﹣2a)=3×3=9.
故答案为:9.
24.解:3a(2a2﹣4a+3)﹣2a2(3a+4)
=6a3﹣12a2+9a﹣6a3﹣8a2
=﹣20a2+9a,
当a=﹣2时,原式=﹣20×4﹣9×2=﹣98.
25.解:原式=x2﹣4xy+4y2﹣x2﹣3xy﹣4y2
=﹣7xy,
当x=﹣4,y=时,原式=﹣7×(﹣4)×=14.
26.解:﹣7xy(2y﹣x﹣3)=﹣14xy2+7x2y+21xy.
故选:A.
27.解:原式=﹣x5﹣ax4﹣x3+2x4
=﹣x5+(2﹣a)x4﹣x3
∵﹣x3(x2+ax+1)+2x4中不含有x的四次项,
∴2﹣a=0,
解得,a=2.
故选:B.
28.解:原式=﹣6x5﹣6ax4+18x3,
由展开式不含x4项,得到a=0,
故选:B.
29.解:∵x(x2+a)+3x﹣2b=x3+5x+4恒成立,
∴x3+(a+3)x﹣2b=x3+5x+4,
∴,
解得.
故选:C.
30.解:由题意知,
这个多项式为=﹣x2+x﹣1,
∴正确的计算结果为﹣3x+(﹣x2+x﹣1)=﹣x2﹣2x﹣1.
故选:A.
31.解:这个多项式是(x2﹣4x+1)﹣(﹣3x2)=4x2﹣4x+1,(3分)
正确的计算结果是:(4x2﹣4x+1)•(﹣3x2)=﹣12x4+12x3﹣3x2.(3分)
三.多项式乘多项式
32.解:(3m﹣1)(m+5)
=3m2+15m﹣m﹣5
=3m2+14m﹣5.
33.解:原式=2x2+xy﹣4xy﹣2y2﹣2x2﹣xy
=﹣4xy﹣2y2.
34.解:原式=3(2x2﹣6x﹣x+3)﹣2(6x2﹣9x﹣4x+6)=6x2﹣21x+9﹣12x2+26x﹣12=﹣6x2+5x﹣3.
35.解:(1)原式=a3﹣2a2+3a﹣6﹣a3+2a2+2a
=5a﹣6;
(2)原式=4m2﹣n2+m2+2mn+n2﹣4m2+2mn
=m2+4mn.
36.解:(1)(x3﹣1)(x6+x3+1)(x9+1)
=(x9﹣1)(x9+1)
=x18﹣1;
(2)(x2﹣y2)(x2+xy+y2)(x2﹣xy+y2)
=(x﹣y)(x2+xy+y2)×(x+y)(x2﹣xy+y2)
=(x3﹣y3)(x3+y3)
=x6﹣y6;
(3)(x+2y)2(x2﹣2xy+4y2)2
=[(x+2y)(x2﹣2xy+4y2)]2
=(x3+8y3)2
=x6+16x3y3+64y6
37.解:原式=6x3+9x2﹣12﹣2x2﹣3x+4
=6x3+7x2﹣15x+4
38.解:原式=x2+3xy﹣xy﹣3y2﹣x2﹣2xy
=﹣3y2.
39.解:x(x﹣3)﹣(x﹣1)(x+2)
=x2﹣3x﹣(x2+2x﹣x﹣2)
=x2﹣3x﹣(x2+x﹣2)
=x2﹣3x﹣x2﹣x+2
=﹣4x+2.
故答案为:﹣4x+2.
40.解:(x+3)(x﹣7)﹣x(x﹣1)
=x2﹣7x+3x﹣21﹣x2+x
=﹣3x﹣21.
41.解:(2a﹣b)(4a2+2ab+b2)
=8a3+4a2b+2ab2﹣4a2b﹣2ab2﹣b3
=8a3﹣b3.
故答案为:8a3﹣b3.
42.解:(1)原式=4m2+10m﹣6m﹣15,
=4m2+4m﹣15;
(2)原式=﹣a3b2+4a6b4﹣2a3b2,
=﹣3a3b2+4a6b4.
43.解:(a+b+c)(b+c+d)﹣(b+c)(a+b+c+d)
=a(b+c+d)+(b+c)(b+c+d)﹣a(b+c)﹣(b+c)(b+c+d)
=a(b+c+d)﹣a(b+c)
=a(b+c)+ad﹣a(b+c)
=ad.
44.解:∵(x﹣1)(x+3)=x2+mx+n,
∴x2+2x﹣3=x2+mx+n,
∴m=2,n=﹣3,
∴m﹣n=2﹣(﹣3)=5,
故选:D.
45.解:∵(x﹣3)(x﹣5)=x2+px+15,
∴x2﹣8x+15=x2+px+15,
∴p=﹣8,
故选:C.
46.解:∵(x+my)(x+ny)=x2+2xy﹣8y2,
∴x2+nxy+mxy+mny2=x2+(m+n)xy+mny2=x2+2xy﹣8y2,
∴m+n=2,mn=﹣8,
∴m2n+mn2=mn(m+n)=﹣8×2=﹣16.
47.解:去括号后得:x2﹣5x+6=x2+10x+9+4,
移项合并得:﹣15x=7,
解得:x=﹣.
48.解:(2x+5)(3x﹣1)+(2x+3)(1﹣3x)=28
6x2+13x﹣5﹣6x2﹣9x+2x+3=28,
整理得:6x=30,
解得:x=5.
49.解:(x+m)(x+3)=x2+(m+3)x+3m,
∵乘积中不含x的一次项,
∴m+3=0,
∴m=﹣3,
故选:D.
50.解:(3x+m)(3x+n)=3x2+3mx+3nx+mn=3x2+3(m+n)x+mn,
∵展开式不含有x项,
∴m+n=0.
故选:D.
51.解:(mx+6y)(x﹣3y)
=mx2﹣3mxy+6xy﹣18y2
=mx2﹣(3m﹣6)xy﹣18y2,
∵乘积中不含有xy项,
∴3m﹣6=0,
∴m=2,
故选:D.
52.解:(x2+ax+2)(2x﹣2)
=2x3﹣2x2+2ax2﹣2ax+4x﹣4
=2x3+(﹣2+2a)x2+(﹣2a+4)x﹣4,
∵结果中不含x项,
∴﹣2a+4=0,
解得:a=2,
故选:B.
53.解:原式=x3﹣mx2+nx﹣2x2+2mx﹣2n=x3+(﹣m﹣2)x2+(n+2m)x﹣2n,
由结果不含x2项和x项,得到﹣m﹣2=0,n+2m=0,
解得:m=﹣2,n=4,
∴(m+n)(m2﹣mn+n2)=(﹣2+4)[(﹣2)2﹣(﹣2)×4+42]=2×28=56.
54.解:(x2+mx+1)(x﹣n)
=x3﹣nx2+mx2﹣mnx+x﹣n
=x3+(﹣n+m)x2+(﹣mn+1)x﹣n
∵展开式中不含x项,x2项的系数为﹣2,
∴﹣mn+1=0,﹣n+m=﹣2,
整理得:mn=1,m﹣n=﹣2,
∴mn+m﹣n
=1﹣2
=﹣1.
55.解:∵(2x+3)•(6x+2)﹣6x(2x+13)+8(7x+2)
=12x2+4x+18x+6﹣12x2﹣78x+56x+16
=22,
∴代数式的值与x的取值无关.
56.解:∵(x+a)(x+6)
=x2+6a+ax+6a
=x2+(6+a)x+6a,
∴6+a=8,6a=12,
∴a=2.
57.解:(1)∵(x+a)(x+6)
=x2+6x+ax+6a
=x2+(6+a)x+6a,
∴x2+(6+a)x+6a=x2+8x+12,
∴6+a=8,6a=12,
解得a=2;
(2)当a=2,b=﹣3时,
(x+a)(x+b)
=(x+2)(x﹣3)
=x2﹣3x+2x﹣6
=x2﹣x﹣6.
58.解:(1)由题知:(2x﹣m)(5x﹣4)
=10x2﹣8x﹣5mx+4m
=10x2﹣(8+5m)x+4m
=10x2﹣33x+20,
所以8+5m=33或4m=20,
解得:m=5.
故m的值为5;
(2)(2x+5)(5x﹣4)
=10x2﹣8x+25x﹣20
=10x2+17x﹣20.
12.2整式的乘法 知识点分类练习题
一.单项式乘单项式
1.计算2x2•(﹣3x)的结果是( )
A.﹣6x2 B.5x3 C.6x3 D.﹣6x3
2.下列计算正确的是( )
A.3x3•2x2y=6x5 B.2a2•3a3=6a5
C.(﹣2x)•(﹣5x2y)=﹣10x3y D.(﹣2xy)•(﹣3x2y)=6x3y
3.如果单项式﹣3ma﹣2bn2a+b与m3n8b是同类项,那么这两个单项式的积是( )
A.﹣3m6n16 B.﹣3m6n32 C.﹣3m3n8 D.﹣9m6n16
4.若单项式﹣8xay和x2yb的积为﹣2x5y6,则ab的值为( )
A.2 B.30 C.﹣15 D.15
5.已知单项式3x2y3与﹣5x2y2的积为mx4yn,那么m﹣n= .
6.计算(﹣8ab)•(a2b)= .
7.计算:(﹣2ab2)•(﹣3a2)= .
8.计算:
(1)(﹣3x)3•(5x2y);
(2)(﹣2)3+(﹣3)×(﹣4)2.
9.计算:
(1)3(x3)2x3﹣(3x3)3+(5x)2•x7;
(2)(﹣2ab)(3a2•2ab﹣b2)
10.(﹣2y3)2+(﹣4y2)3﹣(﹣2y)2•(﹣3y2)2.
二.单项式乘多项式
11.若A(m2﹣3n)=m3﹣3mn,则代数式A的值为( )
A.m B.mn C.mn2 D.m2n
12.计算:3a•(2a﹣5)= .
13.﹣2a(3a﹣4b)= .
14.计算:﹣6a•(﹣﹣a+2)
15.(﹣3y)(4x2y﹣2xy).
16.计算:
(1)(﹣5x)•(3x2﹣4x+5):
(2)﹣2a•(3ab2﹣5ab3):
(3)(﹣a2b)(2a﹣ab+3b);
(4)﹣2xn•(﹣3xn+1+4xn﹣1).
17.计算:6m•(3m2﹣m﹣1)
18.计算:(x2+x﹣1)•(﹣2x)= .
19.计算:
20.计算下列各题.
(1)3a2b(﹣4a2b+2ab2﹣ab);
(2).
21.[xy(x2﹣xy)﹣x2y(x﹣y)]•3xy2.
22..
23.已知a2﹣2a﹣3=0,则代数式3a(a﹣2)的值为 .
24.先化简,再求值:3a(2a2﹣4a+3)﹣2a2(3a+4),其中a=﹣2.
25.先化简,再求值:(x﹣2y)2﹣x(x+3y)﹣4y2,其中x=﹣4,y=.
26.今天数学课上,老师讲了单项式乘多项式,放学回到家,小明拿出课堂笔记复习,发现一道题:﹣7xy(2y﹣x﹣3)=﹣14xy2+7x2y□,□的地方被钢笔水弄污了,你认为□内应填写( )
A.+21xy B.﹣21xy C.﹣3 D.﹣10xy
27.要使﹣x3(x2+ax+1)+2x4中不含有x的四次项,则a等于( )
A.1 B.2 C.3 D.4
28.要使(﹣6x3)(x2+ax﹣3)的展开式中不含x4项,则a=( )
A.1 B.0 C.﹣1 D.
29.若要使x(x2+a)+3x﹣2b=x3+5x+4恒成立,则a,b的值分别是( )
A.﹣2,﹣2 B.2,2 C.2,﹣2 D.﹣2,2
30.某同学在计算﹣3x加上一个多项式时错将加法做成了乘法,得到的答案是3x3﹣3x2+3x,由此可以推断出正确的计算结果是( )
A.﹣x2﹣2x﹣1 B.x2+2x﹣1 C.﹣x2+4x﹣1 D.x2﹣4x+1
31.某同学在计算一个多项式乘以﹣3x2时,因抄错运算符号,算成了加上﹣3x2,得到的结果是x2﹣4x+1,那么正确的计算结果是多少?
三.多项式乘多项式
32.计算:(3m﹣1)(m+5).
33.计算:(x﹣2y)(2x+y)+x(﹣2x﹣y).
34.3(2x﹣1)(x﹣3)﹣2(3x﹣2)(2x﹣3)
35.计算:
(1)(a2+3)(a﹣2)﹣a(a2﹣2a﹣2);
(2)(2m+n)(2m﹣n)+(m+n)2﹣2(2m2﹣mn).
36.化简:
(1)(x3﹣1)(x6+x3+1)(x9+1);
(2)(x2﹣y2)(x2+xy+y2)(x2﹣xy+y2);
(3)(x+2y)2(x2﹣2xy+4y2)2.
37.化简:(3x﹣1)(2x2+3x﹣4)
38.化简:(x﹣y)(x+3y)﹣x(x+2y).
39.化简:x(x﹣3)﹣(x﹣1)(x+2).
40.计算:(x+3)(x﹣7)﹣x(x﹣1).
41.计算:(2a﹣b)(4a2+2ab+b2)= .
42.计算下列各题:
(1)(2m﹣3)(2m+5);
(2)(﹣a•a2)(﹣b)2+(﹣2a3b2)2+(﹣2a3b2).
43.计算:(a+b+c)(b+c+d)﹣(b+c)(a+b+c+d).
44.已知(x﹣1)(x+3)=x2+mx+n,则m﹣n的值是( )
A.﹣4 B.﹣1 C.1 D.5
45.(x﹣3)(x﹣5)=x2+px+15,则p的值是( )
A.﹣3 B.8 C.﹣8 D.﹣5
46.已知(x+my)(x+ny)=x2+2xy﹣8y2,求m2n+mn2的值.
47.解方程:(x﹣3)(x﹣2)=(x+9)(x+1)+4.
48.解方程:(2x+5)(3x﹣1)+(2x+3)(1﹣3x)=28.
49.(x+m)与(x+3)的乘积中不含x的一次项,则m的值是( )
A.0 B.±3 C.3 D.﹣3
50.若(3x+m)(3x+n)的结果中不含有x项,则m、n的关系是( )
A.mm=1 B.mm=0 C.m﹣n=0 D.m+n=0
51.若多项式mx+6y与x﹣3y的乘积中不含有xy项,则m的值为( )
A.﹣6 B.﹣3 C.0 D.2
52.若(x2+ax+2)(2x﹣2)的结果中不含x项,则a的值为( )
A.0 B.2 C. D.﹣2
53.已知(x﹣2)(x2﹣mx+n)的结果中不含x2项和x的项,求(m+n)(m2﹣mn+n2)的值.
54.已知(x2+mx+1)(x﹣n)的展开式中不含x项,x2项的系数为﹣2,求mn+m﹣n的值.
55.试说明:代数式(2x+3)(6x+2)﹣6x(2x+13)+8(7x+2)的值与x的取值无关.
56.在计算(x+a)(x+b)时,甲把b错看成了6,得到结果是:x2+8x+12.求出a的值.
57.在计算(x+a)(x+b)时,甲把b错看成了6,得到结果是:x2+8x+12.
(1)求出a的值;
(2)在(1)的条件下,且b=﹣3时,计算(x+a)(x+b)的结果.
58.小轩计算一道整式乘法的题:(2x+m)(5x﹣4),由于小轩将第一个多项式中的“+m”抄成“﹣m”,得到的结果为10x2﹣33x+20.
(1)求m的值;
(2)请计算出这道题的正确结果.
参考答案
一.单项式乘单项式
1.解:原式=2•(﹣3)x2•x=﹣6x3,
故选:D.
2.解:A、3x3×2x2y=6x5y,故此选项错误;
B、2a2×3a3=6a5,故此选项正确;
C、(﹣2x)×(﹣5x2y)=10x3y,故此选项错误;
D、(﹣2xy)×(﹣3x2y)=6x3y2,故此选错误.
故选:B.
3.解:∵单项式﹣3ma﹣2bn2a+b与m3n8b是同类项,
∴,
解得:,
故单项式﹣3ma﹣2bn2a+b与m3n8b是单项式﹣3m3n16与m3n16,
则这两个单项式的积是:﹣3m3n16•m3n16=﹣3m6n32.
故选:B.
4.解:﹣8xay×x2yb=﹣2xa+2yb+1=﹣2x5y6,
∴a+2=5,b+1=6,
解得a=3,b=5,
∴ab=3×5=15,
故选:D.
5.解:3x2y3×(﹣5x2y2)=﹣15x4y5,
∴mx4yn=﹣15x4y5,
∴m=﹣15,n=5
∴m﹣n=﹣15﹣5=﹣20
故答案为:﹣20
6.解:(﹣8ab)•(a2b)
=(﹣8×)•(a•a2)•(b•b)
=﹣6a3b2,
故答案为:﹣6a3b2.
7.解:(﹣2ab2)•(﹣3a2)
=6a3b2.
故答案为:6a3b2.
8.解:(1)(﹣3x)3•(5x2y)
=﹣27x3•5x2y
=﹣135x5y;
(2)(﹣2)3+(﹣3)×(﹣4)2
=﹣8+(﹣3)×16
=﹣8﹣48
=﹣56.
9.解:(1)原式=3x9﹣27x9+25x9
=x9;
(2)原式=﹣12a4b2+2ab3.
10.解:(﹣2y3)2+(﹣4y2)3﹣(﹣2y)2•(﹣3y2)2
=4y6﹣64y6﹣4y2•(9y4)
=4y6﹣64y6﹣36y6
=﹣96y6.
二.单项式乘多项式
11.解:∵A(m2﹣3n)=m3﹣3mn=m(m2﹣3n),
∴A=m.
故选:A.
12.解:3a•(2a﹣5)
=6a2﹣15a.
故答案为:6a2﹣15a.
13.解:﹣2a(3a﹣4b)=﹣6a2+8ab.
故答案为:﹣6a2+8ab.
14.解:﹣6a•(﹣﹣a+2)=3a3+2a2﹣12a.
15.解:(﹣3y)(4x2y﹣2xy)
=(﹣3y)(4x2y)+(﹣3y)(﹣2xy)
=﹣12x2y2+6xy2.
16.解:(1)原式=﹣15x3+20x2﹣25x;
(2)原式=﹣6a2b2+10a2b3;
(3)原式=﹣2a3b+a3b2﹣3a2b2;
(4)原式=6x2n+1﹣8x2n﹣1.
17.解:6m•(3m2﹣m﹣1)
=18m3﹣4m2﹣6m.
18.解:(x2+x﹣1)•(﹣2x)=﹣2x3﹣x2+2x.
19.解:原式=a2b2(﹣a2b﹣12ab+b2)
=a2b2•(﹣a2b)﹣a2b2•12ab+a2b2•b2
=﹣8a4b3﹣a3b3+a2b4.
20.解:(1)原式=﹣12a4b2+6a3b3﹣3a3b2;
(2)原式=﹣5x3y+5x2y2﹣x3y﹣2x2y2
=﹣6x3y+3x2y2.
21.解:[xy(x2﹣xy)﹣x2y(x﹣y)]•3xy2
=(x3y﹣x2y2﹣x3y+x2y2)•3xy2
=0.
22.解:原式=﹣8x3(2x3﹣x﹣1)﹣(4x4+8x3)
=﹣16x6+4x4+8x3﹣4x4﹣8x3
=﹣16x6.
23.解:∵a2﹣2a﹣3=0,
∴a2﹣2a=3,
∴3a(a﹣2)=3(a2﹣2a)=3×3=9.
故答案为:9.
24.解:3a(2a2﹣4a+3)﹣2a2(3a+4)
=6a3﹣12a2+9a﹣6a3﹣8a2
=﹣20a2+9a,
当a=﹣2时,原式=﹣20×4﹣9×2=﹣98.
25.解:原式=x2﹣4xy+4y2﹣x2﹣3xy﹣4y2
=﹣7xy,
当x=﹣4,y=时,原式=﹣7×(﹣4)×=14.
26.解:﹣7xy(2y﹣x﹣3)=﹣14xy2+7x2y+21xy.
故选:A.
27.解:原式=﹣x5﹣ax4﹣x3+2x4
=﹣x5+(2﹣a)x4﹣x3
∵﹣x3(x2+ax+1)+2x4中不含有x的四次项,
∴2﹣a=0,
解得,a=2.
故选:B.
28.解:原式=﹣6x5﹣6ax4+18x3,
由展开式不含x4项,得到a=0,
故选:B.
29.解:∵x(x2+a)+3x﹣2b=x3+5x+4恒成立,
∴x3+(a+3)x﹣2b=x3+5x+4,
∴,
解得.
故选:C.
30.解:由题意知,
这个多项式为=﹣x2+x﹣1,
∴正确的计算结果为﹣3x+(﹣x2+x﹣1)=﹣x2﹣2x﹣1.
故选:A.
31.解:这个多项式是(x2﹣4x+1)﹣(﹣3x2)=4x2﹣4x+1,(3分)
正确的计算结果是:(4x2﹣4x+1)•(﹣3x2)=﹣12x4+12x3﹣3x2.(3分)
三.多项式乘多项式
32.解:(3m﹣1)(m+5)
=3m2+15m﹣m﹣5
=3m2+14m﹣5.
33.解:原式=2x2+xy﹣4xy﹣2y2﹣2x2﹣xy
=﹣4xy﹣2y2.
34.解:原式=3(2x2﹣6x﹣x+3)﹣2(6x2﹣9x﹣4x+6)=6x2﹣21x+9﹣12x2+26x﹣12=﹣6x2+5x﹣3.
35.解:(1)原式=a3﹣2a2+3a﹣6﹣a3+2a2+2a
=5a﹣6;
(2)原式=4m2﹣n2+m2+2mn+n2﹣4m2+2mn
=m2+4mn.
36.解:(1)(x3﹣1)(x6+x3+1)(x9+1)
=(x9﹣1)(x9+1)
=x18﹣1;
(2)(x2﹣y2)(x2+xy+y2)(x2﹣xy+y2)
=(x﹣y)(x2+xy+y2)×(x+y)(x2﹣xy+y2)
=(x3﹣y3)(x3+y3)
=x6﹣y6;
(3)(x+2y)2(x2﹣2xy+4y2)2
=[(x+2y)(x2﹣2xy+4y2)]2
=(x3+8y3)2
=x6+16x3y3+64y6
37.解:原式=6x3+9x2﹣12﹣2x2﹣3x+4
=6x3+7x2﹣15x+4
38.解:原式=x2+3xy﹣xy﹣3y2﹣x2﹣2xy
=﹣3y2.
39.解:x(x﹣3)﹣(x﹣1)(x+2)
=x2﹣3x﹣(x2+2x﹣x﹣2)
=x2﹣3x﹣(x2+x﹣2)
=x2﹣3x﹣x2﹣x+2
=﹣4x+2.
故答案为:﹣4x+2.
40.解:(x+3)(x﹣7)﹣x(x﹣1)
=x2﹣7x+3x﹣21﹣x2+x
=﹣3x﹣21.
41.解:(2a﹣b)(4a2+2ab+b2)
=8a3+4a2b+2ab2﹣4a2b﹣2ab2﹣b3
=8a3﹣b3.
故答案为:8a3﹣b3.
42.解:(1)原式=4m2+10m﹣6m﹣15,
=4m2+4m﹣15;
(2)原式=﹣a3b2+4a6b4﹣2a3b2,
=﹣3a3b2+4a6b4.
43.解:(a+b+c)(b+c+d)﹣(b+c)(a+b+c+d)
=a(b+c+d)+(b+c)(b+c+d)﹣a(b+c)﹣(b+c)(b+c+d)
=a(b+c+d)﹣a(b+c)
=a(b+c)+ad﹣a(b+c)
=ad.
44.解:∵(x﹣1)(x+3)=x2+mx+n,
∴x2+2x﹣3=x2+mx+n,
∴m=2,n=﹣3,
∴m﹣n=2﹣(﹣3)=5,
故选:D.
45.解:∵(x﹣3)(x﹣5)=x2+px+15,
∴x2﹣8x+15=x2+px+15,
∴p=﹣8,
故选:C.
46.解:∵(x+my)(x+ny)=x2+2xy﹣8y2,
∴x2+nxy+mxy+mny2=x2+(m+n)xy+mny2=x2+2xy﹣8y2,
∴m+n=2,mn=﹣8,
∴m2n+mn2=mn(m+n)=﹣8×2=﹣16.
47.解:去括号后得:x2﹣5x+6=x2+10x+9+4,
移项合并得:﹣15x=7,
解得:x=﹣.
48.解:(2x+5)(3x﹣1)+(2x+3)(1﹣3x)=28
6x2+13x﹣5﹣6x2﹣9x+2x+3=28,
整理得:6x=30,
解得:x=5.
49.解:(x+m)(x+3)=x2+(m+3)x+3m,
∵乘积中不含x的一次项,
∴m+3=0,
∴m=﹣3,
故选:D.
50.解:(3x+m)(3x+n)=3x2+3mx+3nx+mn=3x2+3(m+n)x+mn,
∵展开式不含有x项,
∴m+n=0.
故选:D.
51.解:(mx+6y)(x﹣3y)
=mx2﹣3mxy+6xy﹣18y2
=mx2﹣(3m﹣6)xy﹣18y2,
∵乘积中不含有xy项,
∴3m﹣6=0,
∴m=2,
故选:D.
52.解:(x2+ax+2)(2x﹣2)
=2x3﹣2x2+2ax2﹣2ax+4x﹣4
=2x3+(﹣2+2a)x2+(﹣2a+4)x﹣4,
∵结果中不含x项,
∴﹣2a+4=0,
解得:a=2,
故选:B.
53.解:原式=x3﹣mx2+nx﹣2x2+2mx﹣2n=x3+(﹣m﹣2)x2+(n+2m)x﹣2n,
由结果不含x2项和x项,得到﹣m﹣2=0,n+2m=0,
解得:m=﹣2,n=4,
∴(m+n)(m2﹣mn+n2)=(﹣2+4)[(﹣2)2﹣(﹣2)×4+42]=2×28=56.
54.解:(x2+mx+1)(x﹣n)
=x3﹣nx2+mx2﹣mnx+x﹣n
=x3+(﹣n+m)x2+(﹣mn+1)x﹣n
∵展开式中不含x项,x2项的系数为﹣2,
∴﹣mn+1=0,﹣n+m=﹣2,
整理得:mn=1,m﹣n=﹣2,
∴mn+m﹣n
=1﹣2
=﹣1.
55.解:∵(2x+3)•(6x+2)﹣6x(2x+13)+8(7x+2)
=12x2+4x+18x+6﹣12x2﹣78x+56x+16
=22,
∴代数式的值与x的取值无关.
56.解:∵(x+a)(x+6)
=x2+6a+ax+6a
=x2+(6+a)x+6a,
∴6+a=8,6a=12,
∴a=2.
57.解:(1)∵(x+a)(x+6)
=x2+6x+ax+6a
=x2+(6+a)x+6a,
∴x2+(6+a)x+6a=x2+8x+12,
∴6+a=8,6a=12,
解得a=2;
(2)当a=2,b=﹣3时,
(x+a)(x+b)
=(x+2)(x﹣3)
=x2﹣3x+2x﹣6
=x2﹣x﹣6.
58.解:(1)由题知:(2x﹣m)(5x﹣4)
=10x2﹣8x﹣5mx+4m
=10x2﹣(8+5m)x+4m
=10x2﹣33x+20,
所以8+5m=33或4m=20,
解得:m=5.
故m的值为5;
(2)(2x+5)(5x﹣4)
=10x2﹣8x+25x﹣20
=10x2+17x﹣20.
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