3.4 合并同類項(xiàng) 蘇科版數(shù)學(xué)八年級(jí)上冊(cè)同步練習(xí)(解析版)

3.4合并同類項(xiàng)基礎(chǔ)過關(guān)全練知識(shí)點(diǎn)1同類項(xiàng)的概念1.下列說法正確的是()A.3x2與ax2y是同類項(xiàng)B.6與x是同類項(xiàng)C.3x3y2與-3x3y2是同類項(xiàng)D.2x2y3與-2x3y2是同類項(xiàng)2.若代數(shù)式3ax+7b4與代數(shù)式-a4b2y是同類項(xiàng),則xy的值是()A.9B.-9C.4D.-4知識(shí)點(diǎn)2合并同類項(xiàng)3.已知代數(shù)式ax與bx合并后的結(jié)果是零,則一定有()A.a=b=0B.a=b=x=0C.a+b=0D.a-b=04.化簡:a-2a+3a-4a+5a-6a+…+99a-100a=.

5.合并同類項(xiàng).(1)5a-3b-a+2b;(2)-3x2+7x-6+2x2-5x+1;(3)a2b-b2c+3a2b+2b2c;(4)-13a2b-12ab2+16a2

6.將a-b,x-y分別看成一個(gè)整體,合并同類項(xiàng):(1)4(a-b)2-2(a-b)+5(a-b)+3(a-b)2;(2)3(x-y)2-9(x-y)-8(x-y)2+6(x-y)-1.7.求代數(shù)式的值:12x2y-13xy-0.5x2y+0.5xy,8.關(guān)于x、y的多項(xiàng)式mx3+3nxy2+2x3-xy+y合并后不含三次項(xiàng),求2m+3n的值.能力提升全練9.(2022江蘇揚(yáng)州江都期中,3,)下列運(yùn)算正確的是()A.12xy-20xy=-8B.3xy2-4y2x=-xy2C.3x+4y=7xyD.3x2y-2xy2=xy10.(2022江蘇姜堰期末,3,)已知代數(shù)式-5xyn與3xmy3是同類項(xiàng),則m,n的值分別為()A.0,3B.1,3C.3,0D.3,111.(2020湖南湘潭中考,3,)已知2xn+1y3與13x4y3是同類項(xiàng),則n的值是()A.2B.3C.4D.512.(2021天津中考,13,)計(jì)算4a+2a-a的結(jié)果等于.

13.(2020江蘇蘇州中考,15,)若單項(xiàng)式2xm-1y2與單項(xiàng)式13x2yn+1是同類項(xiàng),則m+n=.

14.(2022江蘇江陰期中,22,)已知關(guān)于x,y的多項(xiàng)式my3+3nx2y+2y3-x2y+y2+x2+y合并后不含三次項(xiàng),求2m+3n的值.

素養(yǎng)探究全練15.[推理能力]觀察下面的三行單項(xiàng)式,2x2,4x3,8x4,16x5,32x6,….①-2x,4x2,-8x3,16x4,-32x5,64x6,….②2x2,-3x3,5x4,-9x5,17x6,-33x7,….③(1)根據(jù)你發(fā)現(xiàn)的規(guī)律,第①行第8個(gè)單項(xiàng)式為;

(2)第②行第8個(gè)單項(xiàng)式為,第③行第8個(gè)單項(xiàng)式為;

(3)取每行的第9個(gè)單項(xiàng)式,令這三個(gè)單項(xiàng)式的和為A.計(jì)算當(dāng)x=12時(shí),512A+

答案全解全析基礎(chǔ)過關(guān)全練1.C所含字母相同,并且相同字母的指數(shù)也相同的項(xiàng)叫做同類項(xiàng),故選C.2.A∵代數(shù)式3ax+7b4與代數(shù)式-a4b2y是同類項(xiàng),∴x+7=4,2y=4,∴x=-3,y=2,∴xy=(-3)2=9.故選A.3.C∵ax與bx合并后的結(jié)果是零,∴一定有a+b=0.4.答案-50a解析a-2a+3a-4a+5a-6a+…+99a-100a=(a-2a)+(3a-4a)+(5a-6a)+…+(99a-100a)=(-a5.解析(1)原式=(5-1)a+(-3+2)b=4a-b.(2)原式=(-3+2)x2+(7-5)x+(-6+1)=-x2+2x-5.(3)原式=(1+3)a2b+(-1+2)b2c=4a2b+b2c.(4)原式=-13+16a=-16a2b+12ab6.解析(1)4(a-b)2-2(a-b)+5(a-b)+3(a-b)2=4(a-b)2+3(a-b)2-2(a-b)+5(a-b)=7(a-b)2+3(a-b).(2)3(x-y)2-9(x-y)-8(x-y)2+6(x-y)-1=3(x-y)2-8(x-y)2+6(x-y)-9(x-y)-1=-5(x-y)2-3(x-y)-1.7.解析12x2y-13xy-0.5x=12-0.5=12-12x2y+當(dāng)x=3,y=-2時(shí),原式=168.解析mx3+3nxy2+2x3-xy+y=(m+2)x3+3nxy2-xy+y,∵合并后不含三次項(xiàng),∴m+2=0,3n=0,∴m=-2,n=0,∴2m+3n=2×(-2)+3×0=-4.能力提升全練9.B12xy-20xy=-8xy;3xy2-4y2x=-xy2;3x與4y不是同類項(xiàng),不能合并;3x2y與-2xy2不是同類項(xiàng),不能合并.10.B所含字母相同,并且相同字母的指數(shù)也相同的項(xiàng)叫做同類項(xiàng),所以m=1,n=3.11.B∵2xn+1y3與13x4y3是同類項(xiàng),∴n+1=4,解得n=3,故選12.答案5a解析4a+2a-a=(4+2-1)a=5a.13.答案4解析因?yàn)閱雾?xiàng)式2xm-1y2與單項(xiàng)式13x2yn+1是同類項(xiàng),所以m-1=2,n+1=2,所以m=3,n=1,所以14.解析my3+3nx2y+2y3-x2y+y2+x2+y=(m+2)y3+(3n-1)x2y+y2+x2+y,因?yàn)楹喜⒑蟛缓雾?xiàng),所以m+2=0,3n-1=0,即m=-2,n=13∴2m+3n=2×(-2)