2.4　整　式同步訓(xùn)練 湘教版數(shù)學(xué)七年級(jí)上冊(cè)

第2章代數(shù)式2.4整式基礎(chǔ)過(guò)關(guān)知識(shí)點(diǎn)1單項(xiàng)式的有關(guān)概念1.單項(xiàng)式-32xy2A.-34C.942.下列單項(xiàng)式中,次數(shù)為6的是()A.3a5b2B.-2a4bC.4a5bD.-22a2b23.下列說(shuō)法正確的是()A.單項(xiàng)式3πr2的系數(shù)是3B.單項(xiàng)式2x2的系數(shù)是2C.單項(xiàng)式12ab2的系數(shù)是1D.單項(xiàng)式-5m2的系數(shù)是-104.判斷下列各式是不是單項(xiàng)式.如果不是,請(qǐng)簡(jiǎn)要說(shuō)明理由;如果是,請(qǐng)指出它們的系數(shù)和次數(shù).(1)x+2;(2)2x;(3)πr2知識(shí)點(diǎn)2多項(xiàng)式的有關(guān)概念5.多項(xiàng)式-5xy+xy2-1是()A.二次三項(xiàng)式B.三次三項(xiàng)式C.四次三項(xiàng)式D.五次三項(xiàng)式6.多項(xiàng)式3x2-2x-1的各項(xiàng)分別是()A.3x2,2x,1B.3x2,-2x,1C.-3x2,2x,-1D.3x2,-2x,-17.下列關(guān)于多項(xiàng)式2a2b+ab-1的說(shuō)法中,正確的是()A.次數(shù)是5B.二次項(xiàng)系數(shù)是0C.最高次項(xiàng)是2a2bD.常數(shù)項(xiàng)是18.多項(xiàng)式2x4-x2-6x3-2是次項(xiàng)式,其中二次項(xiàng)系數(shù)為,常數(shù)項(xiàng)為.9.已知單項(xiàng)式-12x4y3的次數(shù)與多項(xiàng)式a2+8am+1b+a2b2的次數(shù)相同,則m=10.已知關(guān)于x的二次三項(xiàng)式ax2+bx+c,常數(shù)項(xiàng)是一次項(xiàng)系數(shù)的13,一次項(xiàng)系數(shù)是二次項(xiàng)系數(shù)的13,若二次項(xiàng)系數(shù)是9,則a+b-6c=11.已知式子(a-1)x3-2x-(a+3).(1)若它是關(guān)于x的一次式,求a的值并寫(xiě)出常數(shù)項(xiàng);(2)若它是關(guān)于x的三次二項(xiàng)式,求a的值并寫(xiě)出最高次項(xiàng).知識(shí)點(diǎn)3整式12.代數(shù)式x2+5,-1,x2-3x+2,π,5x,x2+A.3個(gè)B.4個(gè)C.5個(gè)D.6個(gè)13.下列說(shuō)法中,正確的是()A.5不是單項(xiàng)式B.x+C.x2y的系數(shù)是0D.x-3214.下列式子:0,x-12,1y,?x+2π,?2m,1m+n,y能力提升15.下列整式中,是二次單項(xiàng)式的是()A.x2+1B.xyC.x2yD.-3x16.整式-3xy2的系數(shù)是()A.-3B.3C.-3xD.3x17.多項(xiàng)式2x2-5x2y-y2-3的次數(shù)和三次項(xiàng)分別是()A.2和5x2yB.3和5x2yC.4和-5x2yD.3和-5x2y18.按一定規(guī)律排列的單項(xiàng)式:a,-a2,a3,-a4,a5,-a6,…,第n個(gè)單項(xiàng)式是()A.anB.-anC.(-1)n+1anD.(-1)nan19.若多項(xiàng)式xy|m-n|+(n-2)x2y2+1是關(guān)于x,y的三次多項(xiàng)式,則mn=.20.已知多項(xiàng)式-13x2ym+1+12xy2-3x3+6是六次四項(xiàng)式,單項(xiàng)式3x2ny2素養(yǎng)探究21.按照規(guī)律填上所缺的單項(xiàng)式并回答問(wèn)題:(1)a,-2a2,3a3,-4a4,,;(2)試寫(xiě)出第2021個(gè)和第2022個(gè)單項(xiàng)式;(3)試寫(xiě)出第n個(gè)單項(xiàng)式;(4)試計(jì)算:當(dāng)a=-1時(shí),a+(-2a2)+3a3+(-4a4)+…+99a99+(-100a100)的值.22.歷史上的數(shù)學(xué)巨人歐拉最先把關(guān)于x的多項(xiàng)式用記號(hào)f(x)的形式來(lái)表示(f可用其他字母,但不同的字母表示不同的多項(xiàng)式),例如f(x)=x2+3x-5,把x=“某數(shù)”時(shí)的多項(xiàng)式的值用f(某數(shù))來(lái)表示.例如:x=-1時(shí),多項(xiàng)式x2+3x-5的值記為f(-1)=(-1)2+3×(-1)-5=-7.已知g(x)=-2x2-3x+1,h(x)=ax3+2x2-x.(1)求g(-2)的值;(2)若h(-2)=14,求g(a)的值.答案基礎(chǔ)過(guò)關(guān)1.B單項(xiàng)式的系數(shù)是-94,次數(shù)是2.C選項(xiàng)A的次數(shù)是7;選項(xiàng)B的次數(shù)是5;選項(xiàng)C的次數(shù)是6;選項(xiàng)D的次數(shù)是4.故選C.3.B單項(xiàng)式3πr2的系數(shù)是3π,故A說(shuō)法錯(cuò)誤;單項(xiàng)式12ab2的系數(shù)是12,故C說(shuō)法錯(cuò)誤;單項(xiàng)式-5m2的系數(shù)是-5,故D說(shuō)法錯(cuò)誤,4.解析(1)不是單項(xiàng)式,其為和的形式.(2)不是單項(xiàng)式,2x的字母x在分母中,不是數(shù)與字母的積的形式(3)是單項(xiàng)式,πr2的系數(shù)為π,次數(shù)為2.(4)是單項(xiàng)式,-35ab2的系數(shù)為-35,5.B多項(xiàng)式-5xy+xy2-1是三次三項(xiàng)式,故選B.6.D根據(jù)多項(xiàng)式項(xiàng)的定義求解,注意帶上前面的符號(hào).7.C多項(xiàng)式2a2b+ab-1的次數(shù)是3,故選項(xiàng)A不正確;多項(xiàng)式2a2b+ab-1的二次項(xiàng)系數(shù)是1,故選項(xiàng)B不正確;多項(xiàng)式2a2b+ab-1的最高次項(xiàng)是2a2b,故選項(xiàng)C正確;多項(xiàng)式2a2b+ab-1的常數(shù)項(xiàng)是-1,故選項(xiàng)D不正確.故選C.8.四;四;-1;-2解析根據(jù)多項(xiàng)式的相關(guān)概念進(jìn)行解答即可.9.5解析由題意可知4+3=m+1+1,所以m=5.10.6解析因?yàn)閍x2+bx+c的常數(shù)項(xiàng)是一次項(xiàng)系數(shù)的13,一次項(xiàng)系數(shù)是二次項(xiàng)系數(shù)的13,所以c=13b,b=13a.因?yàn)槎雾?xiàng)系數(shù)是9,所以a=9,所以b=13×9=3,所以11.解析(1)因?yàn)樵绞顷P(guān)于x的一次式,所以a-1=0,則a=1,常數(shù)項(xiàng)為-(a+3)=-4.(2)因?yàn)樵绞顷P(guān)于x的三次二項(xiàng)式,所以a+3=0,且a-1≠0,則a=-3,此時(shí)a-1=-4,所以最高次項(xiàng)為-4x3.12.Bx2+5,-1,x2-3x+2,π是整式,共4個(gè),故選B.13.D5是單項(xiàng)式,x+y2是多項(xiàng)式,x2y的系數(shù)是1,x-3214.3;3;6解析所給的式子中,單項(xiàng)式有0,-2m,-x23,共3個(gè);多項(xiàng)式有x-12,?x+2π整式有0,x-12,?x+2π能力提升15.B選項(xiàng)A是多項(xiàng)式;選項(xiàng)C是三次單項(xiàng)式;選項(xiàng)D是一次單項(xiàng)式,故選B.16.A單項(xiàng)式中的數(shù)字因數(shù)叫做單項(xiàng)式的系數(shù),依此即可求解.17.D2x2-5x2y-y2-3的次數(shù)是3,三次項(xiàng)是-5x2y,故選D.18.C觀察可知,奇數(shù)位置的單項(xiàng)式系數(shù)為1,偶數(shù)位置的單項(xiàng)式系數(shù)為-1,故第n個(gè)單項(xiàng)式為(-1)n+1·an.故選C.19.0或8解析∵多項(xiàng)式xy|m-n|+(n-2)x2y2+1是關(guān)于x,y的三次多項(xiàng)式,∴n-2=0,1+|m-n|=3,∴n=2,|m-n|=2,∴m-n=2或n-m=2,∴m=4或m=0,∴mn=0或8.解析∵多項(xiàng)式-13x2ym+1+1∴2+m+1=6,∴m=3,又∵單項(xiàng)式3x2ny2的次數(shù)與這個(gè)多項(xiàng)式的次數(shù)相同,∴2n+2=6,∴n=2,∴m2+n2=32+22=13.素養(yǎng)探究21.解析(1)5a5;-6a6.(2)第2021個(gè)單項(xiàng)式為2021a2021,第2022個(gè)單項(xiàng)式為-2022a2022.(3)第n個(gè)單項(xiàng)式為(-1)n+1·n·an.(4)原