2021-2022学年人教版七年级数学上册《2.2整式的加减》期末复习自主提升训练(附答案)
1.下列运算结果正确的是( )
A.2x+3y=5xy B.﹣2(x+y)=﹣2x﹣2y
C.x﹣(3y﹣2)=x﹣3y﹣2 D.7a2b﹣4ab2=3a2b
2.如果﹣2x2﹣ay与x3yb﹣1是同类项,那么﹣a﹣b的值是( )
A.﹣3 B.﹣2 C.﹣1 D.1
3.若多项式2x3﹣8x2+x﹣1与多项式3x3+2mx2﹣5x+3的差不含二次项,则m等于( )
A.2 B.﹣2 C.4 D.﹣4
4.已知多项式M与多项式x3﹣2x2+3的和是2x3﹣6x2,则多项式M是 .
5.如果多项式4x3+2x2﹣(kx2+17x﹣6)中不含x2的项,则k的值为 .
6.已知5xm+2y3与是同类项,则(﹣m)3+n等于 .
7.化简:5a2b﹣2(a2b﹣2ab2)﹣3(2ab2﹣a2b).
8.先化简,再求值:2x2y﹣2[6xy﹣2(4xy﹣2)﹣2x2y]+8,其中x=﹣,y=2.
9.先化简,再求值:2(3xy﹣x2)﹣3(xy﹣2x2)﹣xy,其中x=﹣,y=3.
10.化简:
(1)5(3a2b﹣ab2)﹣4(﹣ab2+3a2b);
(2)﹣2(mn﹣3m2)﹣[m2﹣5(mn﹣m2)+2mn].
11.已知A=2x2﹣6ax+3,B=﹣7x2﹣8x﹣1,按要求完成下列各小题.
(1)当a=﹣2时,求A﹣3B的结果.
(2)若A+B的结果中不存在含x的一次项,求a的值.
12.先化简,再求值:2(x2y+xy2)﹣2(x2y﹣x)﹣2xy2﹣2y,其中x=2,y=﹣2.
13.先化简,再求值:3(x2﹣2xy)﹣(3x2﹣2xy+y2)+y2,其中x=,y=﹣2.
14.先化简,再求值:(4x2y﹣2xy2)﹣(5xy2﹣3x2y),其中x=﹣1,y=2.
15.已知代数式A=a4﹣3a2b2﹣ab3+5,B=2b4﹣2a2b2+ab3,C=a4﹣5a2b2+2b4﹣2.
小丽说:“代数式A+B﹣C的值与a,b的值无关.”她说得对吗?说说你的理由.
16.(1)化简:a2b﹣ab﹣(a2b﹣ab).
(2)先化简下式,再求值:5(3ba2﹣b2a)﹣(ab2+3a2b),其中a=,b=.
17.先化简,再求值:5ab﹣2[3ab﹣(4ab2+ab)]﹣5ab2,其中a=﹣,b=2.
18.先化简,再求值:4x2y﹣2[7xy﹣2(4xy﹣2)﹣2x2y]+8,其中x=﹣,y=2.
19.先化简,再求值:3y2﹣x2+2(2x2﹣3xy)﹣3(x2+y2)的值,其中x=1,y=﹣2.
20.先化简再求值:5a2+2b2﹣2(3b2﹣4a3)+(﹣b2﹣8a3+a2),其中a=﹣1,b=2.
21.已知代数式A=3x2﹣x+1,马小虎同学在做整式加减运算时,误将“A﹣B”看成“A+B”了,计算的结果是2x2﹣3x﹣2.
(1)请你帮马小虎同学求出正确的结果;
(2)x是最大的负整数,将x代入(1)问的结果求值.
22.先化简,再求值:2(a2b+ab2)﹣2(a2b﹣1)﹣ab2﹣2.其中a=1,b=﹣3.
23.先化简,再求值:2ab2﹣3a2b﹣2(a2b+ab2),其中a=1,b=﹣2.
24.计算:4xy+3y2﹣3x2+2xy﹣5xy﹣2x2﹣4y2.
25.化简:﹣2a2+7b+3a2﹣4b.
26.如果代数式3x4﹣2x3+5x2+kx3+mx2+4x+5﹣7x,合并同类项后不含x3和x2项,求mk的值.
27.﹣5yx2+4xy2﹣2xy+6x2y+2xy+5.
28.﹣5m2n+4mn2﹣2mn+6m2n+3mn
29.合并同类项
(1)3a﹣5a+6a. (2)x2y+4x2y﹣6x2y.
(3)﹣3mn2+8m2n﹣7mn2+m2n. (4)2x3﹣6x﹣6x3﹣2+9x+8.
30.数学老师给出这样一个题目:2A+B=﹣x2+2x,A=﹣3x2﹣2x+6.求:
(1)B的值;
(2)B﹣A的值.
参考答案
1.解:A、2x与3y不是同类项,不能合并,不符合题意;
B、根据去括号法则,括号前面是负号时,去掉括号后,里面的符号都要改变,﹣2(x+y)=﹣2x﹣2y,符合题意;
C、根据去括号法则,括号前面是负号时,去掉括号后,里面的符号都要改变,x﹣(3y﹣2)=x﹣3y+2,不符合题意;
D、7a2b和4ab2不是同类项,不能合并,不符合题意;
故选:B.
2.解:∵﹣2x2﹣ay与x3yb﹣1是同类项,
∴2﹣a=3,b﹣1=1,
解得:a=﹣1,b=2,
∴﹣a﹣b=﹣(﹣1)﹣2=1﹣2=﹣1.
故选:C.
3.解:∵多项式2x3﹣8x2+x﹣1与多项式3x3+2mx2﹣5x+3的差不含二次项,
∴2x3﹣8x2+x﹣1﹣(3x3+2mx2﹣5x+3)
=﹣x3﹣(8+2m)x2+6x﹣4,
∴8+2m=0,
解得:m=﹣4.
故选:D.
4.解:M=2x3﹣6x2﹣(x3﹣2x2+3)
=2x3﹣6x2﹣x3+2x2﹣3
=x3﹣4x2﹣3.
故答案为:x3﹣4x2﹣3.
5.解:合并得4x3+2x2﹣(kx2+17x﹣6)=4x3+(2﹣k)x2﹣17x+6,
根据题意得2﹣k=0,
解得k=2.
故答案是:2.
6.解:∵5xm+2y3与是同类项,
∴m+2=6,n+1=3,
解得m=4,n=2,
∴(﹣m)3+n=(﹣4)3+2=﹣64+2=﹣62.
故答案为:﹣62.
7.解:原式=5a2b﹣2a2b+4ab2﹣6ab2+3a2b.
=6a2b﹣2ab2.
8.解:原式=2x2y﹣2(6xy﹣8xy+4﹣2x2y)+8
=2x2y﹣12xy+16xy﹣8+4x2y+8
=6x2y+4xy,
当时,原式=6××2+4×(﹣)×2=﹣1.
9.解:原式=6xy﹣2x2﹣3xy+6x2﹣xy
=2xy+4x2,
当x=﹣,y=3时,
原式=2×(﹣)×3+4×
=﹣3+1
=﹣2.
10.解:(1)原式=15a2b﹣5ab2+4ab2﹣12a2b
=3a2b﹣ab2;
(2)原式=﹣2mn+6m2﹣m2+5mn﹣5m2﹣2mn
=mn.
11.解:(1)∵A=2x2﹣6ax+3,B=﹣7x2﹣8x﹣1,a=﹣2,
∴A﹣3B
=2x2﹣6ax+3+21x2+24x+3
=23x2+(24﹣6a)x+6
=23x2+36x+6;
(2)∵A=2x2﹣6ax+3,B=﹣7x2﹣8x﹣1,
∴A+B=2x2﹣6ax+3﹣7x2﹣8x﹣1=﹣5x2﹣(6a+8)x+2,
由A+B结果中不含x的一次项,得到6a+8=0,
解得:a=﹣.
12.解:原式=2x2y+2xy2﹣2x2y+2x﹣2xy2﹣2y
=2x﹣2y,
当x=2,y=﹣2时,
原式=2×2﹣2×(﹣2)
=4+4
=8.
13.解:原式=3x2﹣6xy﹣3x2+2xy﹣y2+y2
=﹣4xy;
当x=,y=﹣2时,
原式=﹣4××(﹣2)
=12.
14.解:原式=2x2y﹣xy2﹣5xy2+3x2y
=5x2y﹣6xy2,
当x=﹣1,y=2时.
原式=5×(﹣1)2×2﹣6×(﹣1)×22
=10+24
=34.
15.解:小丽的说法正确,理由如下:
∵A=a4﹣3a2b2﹣ab3+5,B=2b4﹣2a2b2+ab3,C=a4﹣5a2b2+2b4﹣2,
∴A+B﹣C=(a4﹣3a2b2﹣ab3+5)+(2b4﹣2a2b2+ab3)﹣(a4﹣5a2b2+2b4﹣2)
=a4﹣3a2b2﹣ab3+5+2b4﹣2a2b2+ab3﹣a4+5a2b2﹣2b4+2
=7,
则结果为常数,与a,b的值无关.
16.解:(1)原式=a2b﹣ab﹣a2b+ab
=ab;
(2)原式=15ba2﹣5b2a﹣ab2﹣3a2b
=12a2b﹣6ab2,
当a=,b=时,
原式=12××﹣6××=1﹣=.
17.解:原式=5ab﹣2(3ab﹣4ab2﹣ab)]﹣5ab2
=5ab﹣6ab+8ab2+ab﹣5ab2
=3ab2.
当a=﹣,b=2,
原式=3×(﹣)×22
=﹣4.
18.解:4x2y﹣2[7xy﹣2(4xy﹣2)﹣2x2y]+8
=4x2y﹣2[7xy﹣8xy+4﹣2x2y]+8
=4x2y﹣14xy+16xy﹣8+4x2y+8
=8x2y+2xy,
当x=﹣,y=2时,原式=8××2+2×(﹣)×2=0.
19.解:3y2﹣x2+2(2x2﹣3xy)﹣3(x2+y2)
=3y2﹣x2+4x2﹣6xy﹣3x2﹣3y2
=﹣6xy
当x=1,y=﹣2时,原式=﹣6×1×(﹣2)=12.
20.解:原式=5a2+2b2﹣6b2+8a3﹣b2﹣8a3+a2=6a2﹣5b2,
当a=﹣1,b=2时,原式=6﹣20=﹣14.
21.解:(1)根据题意知B=2x2﹣3x﹣2﹣(3x2﹣x+1)
=2x2﹣3x﹣2﹣3x2+x﹣1
=﹣x2﹣2x﹣3,
则A﹣B=(3x2﹣x+1)﹣(﹣x2﹣2x﹣3)
=3x2﹣x+1+x2+2x+3
=4x2+x+4;
(2)∵x是最大的负整数,
∴x=﹣1,
则原式=4×(﹣1)2﹣1+4
=4﹣1+4
=7.
22.解:原式=2a2b+2ab2﹣2a2b+2﹣ab2﹣2
=ab2,
当a=1,b=﹣3时,原式=1×(﹣3)2=9.
23.解:原式=2ab2﹣3a2b﹣2a2b﹣2ab2
=﹣5a2b;
当a=1,b=﹣2时,原式=﹣5×12×(﹣2)=10.
24.解:原式=xy﹣y2﹣5x2.
25.解:﹣2a2+7b+3a2﹣4b
=a2+3b.
26.解:由3x4﹣2x3+5x2+kx3+mx2+4x+5﹣7x,合并同类项后不含x3和x2项,得
﹣2+k=0,5+m=0.
解得k=2,m=﹣5.
mk=(﹣5)2=25.
27.解:原式=(﹣5+6)x2y+4xy2+5
=x2y+4xy2+5
28.解:﹣5m2n+4mn2﹣2mn+6m2n+3mn,
=(﹣5m2n+6m2n)+(﹣2mn+3mn)+4mn2,
=m2n+mn+4mn2.
29.解:(1)3a﹣5a+6a=(3﹣5+6)a=4a.
(2)x2y+4x2y﹣6x2y=(1+4﹣6)x2y=﹣x2y.
(3)﹣3mn2+8m2n﹣7mn2+m2n=(﹣3﹣7)mn2+(8+1)m2n=﹣10mn2+9m2n.
(4)2x3﹣6x﹣6x3﹣2+9x+8=(2﹣6)x3+(﹣6+9)x+(﹣2+8)=﹣4x3+3x+6.
30.解:(1)∵A=﹣3x2﹣2x+6,
∴2A=2(﹣3x2﹣2x+6)=﹣6x2﹣4x+12,
∵2A+B=﹣x2+2x,
∴B=﹣x2+2x﹣(﹣6x2﹣4x+12)
=﹣x2+2x+6x2+4x﹣12
=5x2+6x﹣12;
(2)由(1)知B=5x2+6x﹣12,A=﹣3x2﹣2x+6,
∴B﹣A=5x2+6x﹣12﹣(﹣3x2﹣2x+6)
=5x2+6x﹣12+3x2+2x﹣6
=8x2+8x﹣18.