2.2.2 去括号
知识点一:去括号法则
1.下列去括号正确的是(B ) A.a-(b-c+d)=a-b+c+d B.a-(b-c+d)=a-b+c-d C.a+(b-c+d)=a-b+c-d D.a-(b-c+d)=a-b-c+d
2.先去括号,再合并同类项: (1)2(2b-3a)+3(2a-3b); (2)4a2+2(3ab-2a2)-(7ab-1). 解(1)2(2b-3a)+3(2a-3b)
=4b-6a+6a-9b=-5b;
(2)4a2+2(3ab-2a2)-(7ab-1) =4a2+6ab-4a2-7ab+1 =-ab+1.
知识点二:利用去括号法则化简求值的方法步骤
3.求下列整式的值.
(1)2a-3(a-2b)-[1-5(2a-b)],其中a=1,b=-5; (2)5x2-[(x2+5x2-2x)-2(x2-3x)],其中x=-. 解(1)2a-3(a-2b)-[1-5(2a-b)]
=2a-3a+6b-1+5(2a-b) =2a-3a+6b-1+10a-5b =9a+b-1.
当a=1,b=-5时,原式=9×1+(-5)-1=3. (2)5x2-[(x2+5x2-2x)-2(x2-3x)] =5x2-(x2+5x2-2x)+2(x2-3x) =5x2-x2-5x2+2x+2x2-6x=x2-4x. 当x=-时,原式=
-4×
.
拓展点一:结合去括号法则整体代入求值
1.已知3a-2b=2,则9a-6b=6 .
拓展点二:根据多项式的特点确定字母的值
2.学完整式的加减后,老师给出一道这样的习题:“当a=12,b=-7,c=0.5时,求多项式4ab-{2a2b2c2-[ab-(5ab-2a2b2c2-3)]}的值”.聪聪同学经过思考后指出,题目中给出的条件a=12,b=-7,c=0.5是多余的,你同意他的看法吗?请说明理由. 解同意她的看法.理由如下:
4ab-{2a2b2c2-[ab-(5ab-2a2b2c2-3)]} =4ab-2a2b2c2+[ab-(5ab-2a2b2c2-3)] =4ab-2a2b2c2+ab-(5ab-2a2b2c2-3) =4ab-2a2b2c2+ab-5ab+2a2b2c2+3
=(4ab+ab-5ab)+(-2a2b2c2+2a2b2c2)+3 =3.
由此可见,该多项式的值与a,b,c的取值无关,所以说题目中给出的条件a=12,b=-7,c=0.5是多余的.
拓展点三:去括号法则的应用
3.
导学号19054065已知
a,b,c在数轴上对应点的位置如图所示,试化简|a+b|-|b-a|-|a-b-c|.
解由图可知a<0,b>0,c>0,a
所以原式=(a+b)-(b-a)-[-(a-b-c)] =a+b-b+a+a-b-c=3a-b-c.
1.(2016·湖北武汉模拟)下列式子正确的是(D ) A.x-(y-z)=x-y-z B.-(x-y+z)=-x-y-z C.x+2y-2z=x-2(z+y)
D.-a+c+d+b=-(a-b)-(-c-d) 2.(2016·安徽宿州二模)计算2-2(1-a)的结果是 A.a B.-a C.2a D.-2a 3.(2016·内蒙古宁城县期末)已知a-b=-3,c+d=2,则(b+c)-(a-d)的值为(B ) A.1 B.5 C.-5 D.-1 4.(2016·广西钦州期末)-[x-(y-z)]去括号后应得(A ) A.-x+y-z B.-x-y+z C.-x-y-z D.-x+y+z 5.(2015·山东济宁中考)化简-16(x-0.5)的结果是(D ) A.-16x-0.5 B.-16x+0.5 C.16x-8 D.-16x+8 6.(2016·重庆铜梁县期末)-x+y-z的相反数是 A.-x+y-z B.-z+x+y C.-y+z+x D.x+y+z
7.导学号19054066(2015·吉林农安县期末)当a=5时,(a2-a)-(a2-2a+1)等于(A A.4 B.-4 C.-14 D.1 8.(2016·广西桂林秀峰区校级期中)化简:-3x-(-x)=-2x . 9.(2016·山东济宁期中)去括号、并合并同类项:3x+1-2(4-x)=5x-7 . 10.(2015·江苏苏州)若a-2b=3,则9-2a+4b的值为3 . 11.(2016·广西灌阳县期中)在括号内填入适当的项:a-2b+3c=-(-a+2b-3c ). 12.(2015·陕西咸阳模拟)下列去括号正确吗?如有错误,请改正. (1)+(-a-b)=a-b;
(2)5x-(2x-1)-xy=5x-2x+1+xy; (3)3xy-2(xy-y)=3xy-2xy-2y;
(C(C) )
)
(4)(a+b)-3(2a-3b)=a+b-6a+3b. 解(1)错误,应该是+(-a-b)=-a-b;
(2)错误,应该是5x-(2x-1)-xy=5x-2x+1-xy; (3)错误,应该是3xy-2(xy-y)=3xy-2xy+2y; (4)错误,应该是(a+b)-3(2a-3b)=a+b-6a+9b.
13.(2015·河南平顶山期中)先去括号,再合并同类项: (1)2(2b-3a)+3(2a-3b); (2)4a2+2(3ab-2a2)-(7ab-1).
解(1)2(2b-3a)+3(2a-3b)=4b-6a+6a-9b=-5b;
(2)4a2+2(3ab-2a2)-(7ab-1)=4a2+6ab-4a2-7ab+1=-ab+1. 14.(2015·山东新泰模拟)先化简,再求值: (1)3x2-3
+4,其中x=2;
(2)-2(ab-3a2)-[2b2-(5ab+a2)+2ab],其中a=4,b=.
解(1)原式=3x2-x2+6x-3+4=2x2+6x+1.
把x=2代入,得原式=2×22+6×2+1=21. (2)原式=-2ab+6a2-(2b2-5ab-a2+2ab) =-2ab+6a2-2b2+5ab+a2-2ab
=7a2+ab-2b2. 把a=4,b=代入,得
原式=7×42+4×-2×
=113.
15.导学号19054067(2015·山东惠民期中)课堂上老师给大家出了这样一道题,“当x=2 016时,求式子(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y+y3)的值”,小明一看,“x的值太大了,又没有y的值,怎么算呢?”你能帮小明解决这个问题吗?请写出具体过程.
解(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y+y3)=2x3-3x2y-2xy2-x3+2xy2-y3-x3+3x2y+y3=0.
故不论x,y取什么值,代数式的值都为0.
33
导学号19054068如果当x=3时,式子px+qx+1的值为2 016,那么当x=-3时,式子px+qx+1的值是-2
16.
014 .
17.已知关于x,y的多项式5x2-2xy2-[3xy+4y2+(9xy-2y2-2mxy2)+7x2]-1.
(1)若该多项式不含三次项,求m的值;
(2)在(1)的条件下,当x2+y2=13,xy=-6时,求这个多项式的值. 解(1)5x2-2xy2-[3xy+4y2+(9xy-2y2-2mxy2)+7x2]-1
=5x2-2xy2-(3xy+4y2+9xy-2y2-2mxy2+7x2)-1 =5x2-2xy2-(12xy+2y2-2mxy2+7x2)-1 =5x2-2xy2-12xy-2y2+2mxy2-7x2-1 =-2x2-2y2-12xy+(-2+2m)xy2-1,
因为该多项式不含三次项, 所以-2+2m=0,故m的值为1;
(2)原式=-2x2-2y2-12xy+(-2+2m)xy2-1=-2(x2+y2)-12xy-1 =-2×13-12×(-6)-1=45.