比如像(12a+3b+5c)+(1a-3c+2b)的
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/08 05:56:00
比如像(12a+3b+5c)+(1a-3c+2b)的
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,
(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,
比如像(12a+3b+5c)+(1a-3c+2b)的
比如像a×(b+c)这样的.
(a-b)(b-c)(c-a)/(b-a)(a-c)2(c-b)3
b/(a+c)=1/2,a/(b+c)=1/5 怎么得出a/(a+b+c)=1/6,b/(a+b+c)=1/3?
若|a|=3|b|=1|c|=5且|a+b|=a+b,|a+c|=-(a+c)求a+b+c的值
已知|a|=2,|b|=3,|c|=5,且|a+b|=a+b,|a+c|= -(a+c),求a-b-c的值
(a+b-c)^5(a-b-c)^2(c-a-b)^4(c+b-a)^3 计算
a-b=5,b-c=3,(c-a)(2b-a-c)=
a*1/5=b/6=c*2*3,比较a、b、c的大小
A=B+B+B+B+B+B+B+B+B+B+B+B+B+B+B A-B×3=C C+2×7+2=1
若a+b-2a-1-4b-2=3c-3-12c-5,则a+b+c的值为 ___ .
已知2/3/a=1/2/b=5/12/c,并且a、b、c都是大于0的自然数.求a、b、c各是多少?