2017年秋人教版七年級(jí)上《2.2.2去括號(hào)》同步四維訓(xùn)練含答案.doc

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2.2.2　去括號(hào) 知識(shí)點(diǎn)一:去括號(hào)法則 1.下列去括號(hào)正確的是(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.先去括號(hào),再合并同類項(xiàng): (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. 知識(shí)點(diǎn)二:利用去括號(hào)法則化簡(jiǎn)求值的方法步驟 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. 當(dāng)a=1,b=-5時(shí),原式=91+(-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. 當(dāng)x=-時(shí),原式=-4. 拓展點(diǎn)一:結(jié)合去括號(hào)法則整體代入求值 1.已知3a-2b=2,則9a-6b=6　. 拓展點(diǎn)二:根據(jù)多項(xiàng)式的特點(diǎn)確定字母的值 2.學(xué)完整式的加減后,老師給出一道這樣的習(xí)題:“當(dāng)a=12,b=-7,c=0.5時(shí),求多項(xiàng)式4ab-{2a2b2c2-[ab-(5ab-2a2b2c2-3)]}的值”.聰聰同學(xué)經(jīng)過思考后指出,題目中給出的條件a=12,b=-7,c=0.5是多余的,你同意他的看法嗎?請(qǐng)說明理由. 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. 由此可見,該多項(xiàng)式的值與a,b,c的取值無關(guān),所以說題目中給出的條件a=12,b=-7,c=0.5是多余的. 拓展點(diǎn)三:去括號(hào)法則的應(yīng)用 3.導(dǎo)學(xué)號(hào)19054065已知a,b,c在數(shù)軸上對(duì)應(yīng)點(diǎn)的位置如圖所示,試化簡(jiǎn)|a+b|-|b-a|-|a-b-c|. 解由圖可知a<0,b>0,c>0,a0,b-a>0,a-b-c<0. 所以原式=(a+b)-(b-a)-[-(a-b-c)] =a+b-b+a+a-b-c=3a-b-c. 1.(2016湖北武漢模擬)下列式子正確的是(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)計(jì)算2-2(1-a)的結(jié)果是 (C　) A.a B.-a C.2a D.-2a 3.(2016內(nèi)蒙古寧城縣期末)已知a-b=-3,c+d=2,則(b+c)-(a-d)的值為(B　) A.1 B.5 C.-5 D.-1 4.(2016廣西欽州期末)-[x-(y-z)]去括號(hào)后應(yīng)得(A　) A.-x+y-z B.-x-y+z C.-x-y-z D.-x+y+z 5.(2015山東濟(jì)寧中考)化簡(jiǎn)-16(x-0.5)的結(jié)果是(D　) A.-16x-0.5 B.-16x+0.5 C.16x-8 D.-16x+8 6.(2016重慶銅梁縣期末)-x+y-z的相反數(shù)是 (C　) A.-x+y-z B.-z+x+y C.-y+z+x D.x+y+z 7.導(dǎo)學(xué)號(hào)19054066(2015吉林農(nóng)安縣期末)當(dāng)a=5時(shí),(a2-a)-(a2-2a+1)等于(A　) A.4 B.-4 C.-14 D.1 8.(2016廣西桂林秀峰區(qū)校級(jí)期中)化簡(jiǎn):-3x-(-x)=-2x　. 9.(2016山東濟(jì)寧期中)去括號(hào)、并合并同類項(xiàng):3x+1-2(4-x)=5x-7　. 10.(2015江蘇蘇州)若a-2b=3,則9-2a+4b的值為3　. 11.(2016廣西灌陽縣期中)在括號(hào)內(nèi)填入適當(dāng)?shù)捻?xiàng):a-2b+3c=-(-a+2b-3c　). 12.(2015陜西咸陽模擬)下列去括號(hào)正確嗎?如有錯(cuò)誤,請(qǐng)改正. (1)+(-a-b)=a-b; (2)5x-(2x-1)-xy=5x-2x+1+xy; (3)3xy-2(xy-y)=3xy-2xy-2y; (4)(a+b)-3(2a-3b)=a+b-6a+3b. 解(1)錯(cuò)誤,應(yīng)該是+(-a-b)=-a-b; (2)錯(cuò)誤,應(yīng)該是5x-(2x-1)-xy=5x-2x+1-xy; (3)錯(cuò)誤,應(yīng)該是3xy-2(xy-y)=3xy-2xy+2y; (4)錯(cuò)誤,應(yīng)該是(a+b)-3(2a-3b)=a+b-6a+9b. 13.(2015河南平頂山期中)先去括號(hào),再合并同類項(xiàng): (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山東新泰模擬)先化簡(jiǎn),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=222+62+1=21. (2)原式=-2ab+6a2-(2b2-5ab-a2+2ab) =-2ab+6a2-2b2+5ab+a2-2ab =7a2+ab-2b2. 把a(bǔ)=4,b=代入,得 原式=742+4-2=113. 15.導(dǎo)學(xué)號(hào)19054067(2015山東惠民期中)課堂上老師給大家出了這樣一道題,“當(dāng)x=2 016時(shí),求式子(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y+y3)的值”,小明一看,“x的值太大了,又沒有y的值,怎么算呢?”你能幫小明解決這個(gè)問題嗎?請(qǐng)寫出具體過程. 解(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y+y3)=2x3-3x2y-2xy2-x3+2xy2-y3-x3+3x2y+y3=0. 故不論x,y取什么值,代數(shù)式的值都為0. 16.導(dǎo)學(xué)號(hào)19054068如果當(dāng)x=3時(shí),式子px3+qx+1的值為2 016,那么當(dāng)x=-3時(shí),式子px3+qx+1的值是-2 014　. 17.已知關(guān)于x,y的多項(xiàng)式5x2-2xy2-[3xy+4y2+(9xy-2y2-2mxy2)+7x2]-1. (1)若該多項(xiàng)式不含三次項(xiàng),求m的值; (2)在(1)的條件下,當(dāng)x2+y2=13,xy=-6時(shí),求這個(gè)多項(xiàng)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?yàn)樵摱囗?xiàng)式不含三次項(xiàng), 所以-2+2m=0,故m的值為1; (2)原式=-2x2-2y2-12xy+(-2+2m)xy2-1=-2(x2+y2)-12xy-1 =-213-12(-6)-1=45. 4