3.5去括號(hào)基礎(chǔ)過(guò)關(guān)全練知識(shí)點(diǎn)去括號(hào)法則1.(2022江蘇徐州銅山期中)下列去括號(hào)正確的是()A.-3(b-1)=-3b+1B.-2(a-2)=-2a-4C.-3(b-1)=3b-3D.-2(a-2)=4-2a2.化簡(jiǎn)9x-{4x-[5x-(8x-2)]}的結(jié)果是()A.2x-2B.8x+2C.16x+2D.2x+23.(2022江蘇鹽城鹽都期中)下列運(yùn)算結(jié)果正確的是()A.2x+3y=5xyB.7a2b-4ab2=3a2bC.x-(3y-2)=x-3y-2D.-2(x+y)=-2x-2y4.若一個(gè)代數(shù)式加上-x2+x-2得x2-1,則這個(gè)代數(shù)式為.

5.若mn=m+3,則2mn+3m-5nm+10=.

6.已知A+B=C,且B=16(3x-6),C=12(x-4),能力提升全練7.(2020江蘇無(wú)錫中考,4,)若x+y=2,z-y=-3,則x+z的值等于()A.5B.1C.-1D.-58.(2022江蘇南京建鄴期中,5,)與a-b-c的值不相等的是()A.a-(b-c)B.a-(b+c)C.(a-b)+(-c)D.(-b)+(a-c)9.(2021江蘇常州中考,10,)計(jì)算:2a2-(a2+2)=.

10.(2022江蘇揚(yáng)州期末,13,)當(dāng)1<m<3時(shí),化簡(jiǎn)|m-1|-|m-3|=.

11.(2021江蘇無(wú)錫江陰期中,21,)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).

素養(yǎng)探究全練12.[運(yùn)算能力]有理數(shù)a,b,c在數(shù)軸上對(duì)應(yīng)的點(diǎn)分別為A,B,C,其位置如圖所示,試去掉絕對(duì)值符號(hào),并合并同類項(xiàng):|c|-|c+b|+|a-c|+|b+a|.

答案全解全析基礎(chǔ)過(guò)關(guān)全練1.D選項(xiàng)A的結(jié)果為-3b+3;選項(xiàng)B的結(jié)果為-2a+4;選項(xiàng)C的結(jié)果為-3b+3;選項(xiàng)D的結(jié)果為-2a+4=4-2a.2.D按去括號(hào)法則依次去括號(hào).3.D2x與3y不是同類項(xiàng),不能合并;7a2b與-4ab2不是同類項(xiàng),不能合并;x-(3y-2)=x-3y+2;-2(x+y)=-2x-2y.4.答案2x2-x+1解析設(shè)這個(gè)代數(shù)式為M,由題意,得M+(-x2+x-2)=x2-1,所以M=x2-1-(-x2+x-2)=x2-1+x2-x+2=2x2-x+1,所以所求代數(shù)式為2x2-x+1.5.答案1解析∵mn=m+3,∴2mn+3m-5nm+10=-3mn+3m+10=-3(m+3)+3m+10=-3m-9+3m+10=1.6.解析∵A+B=C,B=16(3x-6),C=1∴A=C-B=12(x-4)-16(3x-6)=12能力提升全練7.C∵x+y=2,z-y=-3,∴x+z=(x+y)+(z-y)=2+(-3)=-1,故選C.8.Aa-(b-c)=a-b+c;a-(b+c)=a-b-c;(a-b)+(-c)=a-b-c;(-b)+(a-c)=a-b-c.9.答案a2-2解析原式=2a2-a2-2=a2-2.10.答案2m-4解析∵1<m<3,∴m-1>0,m-3<0,∴|m-1|-|m-3|=(m-1)-(3-m)=2m-4.11.解析(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.素養(yǎng)探究全練12.解析由題意得