文档介绍：(1)(3x-5y)-(6x+7y)+(9x-2y)(2)2a-[3b-5a-(3a-5b)](3)(6m2n-5mn2)-6(m2n-mn2)解:(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-:A=3x2-4xy+2y2,B=x2+2xy-5y2求:(1)A+B(2)A-B(3)若2A-B+C=0,求C。解:(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):(1)m2+(-mn)-n2+(-m2)-(-)(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]解:(1)m2+(-mn)-n2+(-m2)-(-)=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。解: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=-1y5和-x5y2n+1是同类项,求3m+2n的值。解:∵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。+y=6,xy=-4,求:(5x-4y-3xy)-(8x-y+2xy)的值。解:(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****过程中,注意使用。三、练****一)计算:(1)a-(a-3b+4c)+3(-c+2b)(2)(3x2-2xy+7)-(-4x2+5xy