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合并同类项的100道练习题

合并同类项的100道练习题

例1、合并同类项

(1)(3x-5y)-(6x+7y)+(9x-2y)

(2)2a-[3b-5a-(3a-5b)]

(3)(6m2n-5mn2)-6(m2n-mn2)

(1)(3x-5y)-(6x+7y)+(9x-2y)

=3x-5y-6x-7y+9x-2y (正确去掉括号)

=(3-6+9)x+(-5-7-2)y (合并同类项)

=6x-14y

(2)2a-[3b-5a-(3a-5b)] (应按小括号,中括号,大括号的顺序逐层去括号)

=2a-[3b-5a-3a+5b] (先去小括号)

=2a-[-8a+8b] (及时合并同类项)

=2a+8a-8b (去中括号)

=10a-8b

(3)(6m2n-5mn2)-6(m2n-mn2) (注意第二个括号前有因数6)

=6m2n-5mn2-2m2n+3mn2 (去括号与分配律同时进行)

=(6-2)m2n+(-5+3)mn2 (合并同类项)

=4m2n-2mn2

例2.已知:A=3x2-4xy+2y2,B=x2+2xy-5y2

求:(1)A+B (2)A-B (3)若2A-B+C=0,求C.

(1)A+B=(3x2-4xy+2y2)+(x2+2xy-5y2)

=3x2-4xy+2y2+x2+2xy-5y2(去括号)

=(3+1)x2+(-4+2)xy+(2-5)y2(合并同类项)

=4x2-2xy-3y2(按x的降幂排列)

(2)A-B=(3x2-4xy+2y2)-(x2+2xy-5y2)

=3x2-4xy+2y2-x2-2xy+5y2 (去括号)

=(3-1)x2+(-4-2)xy+(2+5)y2 (合并同类项)

=2x2-6xy+7y2 (按x的降幂排列)

(3)∵2A-B+C=0

∴C=-2A+B

=-2(3x2-4xy+2y2)+(x2+2xy-5y2)

=-6x2+8xy-4y2+x2+2xy-5y2 (去括号,注意使用分配律)

=(-6+1)x2+(8+2)xy+(-4-5)y2 (合并同类项)

=-5x2+10xy-9y2 (按x的降幂排列)

例3.计算:

(1)m2+(-mn)-n2+(-m2)-(-0.5n2)

(2)2(4an+2-an)-3an+(an+1-2an+1)-(8an+2+3an)

(3)化简:(x-y)2-(x-y)2-[(x-y)2-(x-y)2]

(1)m2+(-mn)-n2+(-m2)-(-0.5n2)

=m2-mn-n2-m2+n2 (去括号)

=(-)m2-mn+(-+)n2 (合并同类项)

=-m2-mn-n2 (按m的降幂排列)

(2)2(4an+2-an)-3an+(an+1-2an+1)-(8an+2+3an)

=8an+2-2an-3an-an+1-8an+2-3an (去括号)

=0+(-2-3-3)an-an+1 (合并同类项)

=-an+1-8an

(3)(x-y)2-(x-y)2-[(x-y)2-(x-y)2] [把(x-y)2看作一个整体]

=(x-y)2-(x-y)2-(x-y)2+(x-y)2 (去掉中括号)

=(1--+)(x-y)2 (“合并同类项”)

=(x-y)2

例4求3x2-2{x-5[x-3(x-2x2)-3(x2-2x)]-(x-1)}的值,其中x=2.

分析:由于已知所给的式子比较复杂,一般情况都应先化简整式,然后再代入所给数值x=-2,去括号时要注意符号,并且及时合并同类项,使运算简便.

原式=3x2-2{x-5[x-3x+6x2-3x2+6x]-x+1} (去小括号)

=3x2-2{x-5[3x2+4x]-x+1} (及时合并同类项)

=3x2-2{x-15x2-20x-x+1} (去中括号)

=3x2-2{-15x2-20x+1} (化简大括号里的式子)

=3x2+30x2+40x-2 (去掉大括号)

=33x2+40x-2

当x=-2时,原式=33×(-2)2+40×(-2)-2=132-80-2=50

例5.若16x3m-1y5和-x5y2n+1是同类项,求3m+2n的值.

∵16x3m-1y5和-x5y2n+1是同类项

∴对应x,y的次数应分别相等

∴3m-1=5且2n+1=5

∴m=2且n=2

∴3m+2n=6+4=10

本题考察我们对同类项的概念的理解.

例6.已知x+y=6,xy=-4,求:(5x-4y-3xy)-(8x-y+2xy)的值.

(5x-4y-3xy)-(8x-y+2xy)

=5x-4y-3xy-8x+y-2xy

=-3x-3y-5xy

=-3(x+y)-5xy

∵x+y=6,xy=-4

∴原式=-3×6-5×(-4)=-18+20=2

说明:本题化简后,发现结果可以写成-3(x+y)-5xy的形式,因而可以把x+y,xy的值代入原式即可求得最后结果,而没有必要求出x,y的值,这种思考问题的思想方法叫做整体代换,希望同学们在学习过程中,注意使用.

三、练习

(一)计算:

(1)a-(a-3b+4c)+3(-c+2b)

(2)(3x2-2xy+7)-(-4x2+5xy+6)

(3)2x2-{-3x+6+[4x2-(2x2-3x+2)]}

(二)化简

(1)a>0,