北師大版七年級(jí)數(shù)學(xué)上冊(cè)第三章整式及其加減1代數(shù)式第2課時(shí)整式課件

第三章整式及其加減1代數(shù)式第2課時(shí)整式知識(shí)點(diǎn)1單項(xiàng)式基礎(chǔ)過(guò)關(guān)全練1.(2024山東濟(jì)寧梁山期中)代數(shù)式:2x2,-3,x-2y,t,

,

,m3+2m2-m,其中是單項(xiàng)式的有(

)A.5個(gè)

B.4個(gè)

C.3個(gè)

D.2個(gè)B解析單項(xiàng)式有2x2,-3,t,

,共4個(gè),故選B.2.(2024山西大同期中)若一個(gè)單項(xiàng)式的系數(shù)是2,次數(shù)是3,則

這個(gè)單項(xiàng)式可以是(

)A.2πy2

B.3x2

C.2xy3

D.2x3

D解析

A項(xiàng),2πy2的系數(shù)是2π;B項(xiàng),3x2的系數(shù)是3;C項(xiàng),2xy3的次

數(shù)是4;D項(xiàng),2x3的系數(shù)是2,次數(shù)是3.故選D.3.(2023安徽安慶期中)下列說(shuō)法正確的是(

)A.x+y是二次單項(xiàng)式

B.m2的次數(shù)是2,系數(shù)是0C.-2πab的系數(shù)是-2

D.32是單項(xiàng)式D解析

A.x+y不是單項(xiàng)式,原說(shuō)法錯(cuò)誤,故該選項(xiàng)不符合題意;B.m2的次數(shù)是2,系數(shù)是1,原說(shuō)法錯(cuò)誤,故該選項(xiàng)不符合題意;C.-2πab的系數(shù)是-2π,原說(shuō)法錯(cuò)誤,故該選項(xiàng)不符合題意;D.32是單項(xiàng)式,原說(shuō)法正確,故該選項(xiàng)符合題意.故選D.4.(2023江西中考)單項(xiàng)式-5ab的系數(shù)為

.-5解析-5ab的系數(shù)為-5.5.(2024廣東揭陽(yáng)榕城期中)單項(xiàng)式

的系數(shù)是

.解析單項(xiàng)式

的系數(shù)是

.6.(新獨(dú)家原創(chuàng))若某款圓柱形保溫杯的高度為hcm,底面圓

的直徑為dcm.(1)用代數(shù)式表示10個(gè)這樣的保溫杯的體積.(2)(1)中所寫(xiě)的代數(shù)式是單項(xiàng)式嗎?若是,請(qǐng)寫(xiě)出它的系數(shù)與

次數(shù).解析

(1)10·π

·h=

πd2h.所以10個(gè)保溫杯的體積為

πd2hcm3.(2)是單項(xiàng)式,系數(shù)是

π,次數(shù)是3.知識(shí)點(diǎn)2多項(xiàng)式7.(2024河南平頂山魯山期中)下列說(shuō)法中正確的是(

)A.

是單項(xiàng)式B.單項(xiàng)式-

ab3的次數(shù)是3C.單項(xiàng)式a2b的系數(shù)是0D.多項(xiàng)式

a-ab3+1是四次三項(xiàng)式D解析

A項(xiàng),

的分母中含有字母,不是單項(xiàng)式,故本選項(xiàng)錯(cuò)誤;B項(xiàng),單項(xiàng)式-

ab3的次數(shù)是4,故本選項(xiàng)錯(cuò)誤;C項(xiàng),單項(xiàng)式a2b的系數(shù)是1,故本選項(xiàng)錯(cuò)誤;D項(xiàng)正確.故選D.8.(2023河南周口期中)多項(xiàng)式-ax2-1的次數(shù)是m,項(xiàng)數(shù)是n,則n-

m=

(

)A.-1

B.0

C.1

D.2A解析因?yàn)槎囗?xiàng)式-ax2-1是三次二項(xiàng)式,所以m=3,n=2,所以n-m=2-3=-1.9.(2023遼寧葫蘆島南票期中)已知多項(xiàng)式-32m3n2+2mn2-

,它是

次

項(xiàng)式.五三解析多項(xiàng)式-32m3n2+2mn2-

是五次三項(xiàng)式.10.(新獨(dú)家原創(chuàng))多項(xiàng)式

的最高次項(xiàng)是

,常數(shù)項(xiàng)是

.-解析

=

x2-

x-

,因此最高次項(xiàng)是

,常數(shù)項(xiàng)是-

.易錯(cuò)警示多項(xiàng)式是幾個(gè)單項(xiàng)式的和,要找多項(xiàng)式的項(xiàng),首先應(yīng)把多

項(xiàng)式化為單項(xiàng)式的和的形式.11.下列多項(xiàng)式分別有幾項(xiàng)?每項(xiàng)的系數(shù)和次數(shù)分別是多少?(1)

a3-a;(2)-x2-

xy+2y2.解析

(1)有兩項(xiàng).

a3的系數(shù)是

,次數(shù)是3.-a的系數(shù)是-1,次數(shù)是1.(2)有三項(xiàng).-x2的系數(shù)是-1,次數(shù)是2.-

xy的系數(shù)是-

,次數(shù)是2.2y2的系數(shù)是2,次數(shù)是2.12.把多項(xiàng)式2x3y-4y2x+5x2按下列要求重新排列.(1)按x的降冪排列;(2)按y的降冪排列.解析

(1)2x3y+5x2-4y2x.(2)-4y2x+2x3y+5x2.知識(shí)點(diǎn)3整式13.(2024河南周口淮陽(yáng)期中)下列代數(shù)式中整式的個(gè)數(shù)有(

)2a,

,

+a,

,2(x2-4).A.4個(gè)

B.3個(gè)

C.2個(gè)

D.5個(gè)A解析

的分母中含有字母,不是整式,2a,

,

+a,2(x2-4)是整式,共4個(gè),故選A.14.(教材變式·P82隨堂練習(xí)T1)把下列各式填在相應(yīng)的括號(hào)

里.-

a2,

,

,ab2,x2-5x,

-y,0,π.單項(xiàng)式:{

…};多項(xiàng)式:{

…};整式:{

…}.解析單項(xiàng)式:

-

a2,ab2,0,π,…

;多項(xiàng)式:

x2-5x,

-y,…

;整式:

-

a2,ab2,x2-5x,

-y,0,π,…

.能力提升全練15.(2024河南平頂山寶豐期中,7,★☆☆)下列說(shuō)法中,正確的

是(

)A.單項(xiàng)式-

πr3的系數(shù)是-

,次數(shù)是4B.多項(xiàng)式ax2+bx+c是二次三項(xiàng)式C.-

ab2,-2x都是單項(xiàng)式,也都是整式D.多項(xiàng)式-2a2b+3ab-5的項(xiàng)是-2a2b,3ab,5C解析

A項(xiàng),單項(xiàng)式-

πr3的系數(shù)為-

π,次數(shù)為3,故該選項(xiàng)說(shuō)法錯(cuò)誤.B項(xiàng),多項(xiàng)式ax2+bx+c是三次三項(xiàng)式,故該選項(xiàng)說(shuō)法錯(cuò)誤.C項(xiàng)說(shuō)法正確.D項(xiàng),多項(xiàng)式-2a2b+3ab-5的項(xiàng)是-2a2b,3ab,-5,故該選項(xiàng)說(shuō)法錯(cuò)

誤.故選C.16.(2024遼寧阜新期末改編,13,★★☆)當(dāng)k=

時(shí),代

數(shù)式x2-8-3y2+(2k+1)xy中不含xy項(xiàng).-解析代數(shù)式x2-8-3y2+(2k+1)xy中不含xy項(xiàng),所以2k+1=0,解得k=-

.素養(yǎng)探究全練17.(推理能力)如果關(guān)于x的多項(xiàng)式mx4+4x2-

與多項(xiàng)式3xn+5x的次數(shù)相同,求

n3-2n2+3n-4的值.解析因?yàn)殛P(guān)于x的多項(xiàng)式mx4+4x2-

與多項(xiàng)式3xn+5x的次數(shù)相同,所以m≠0,n=4或m=0,n=2.所以當(dāng)m≠0,n=4時(shí),

n3-2n2+3n-4=8,當(dāng)m=0,n=2時(shí),

n3-2n2+3n-4=-2.綜上所述,

n3-2n2+3n-4的值為8或-2.微專題系數(shù)或次數(shù)中字母參數(shù)的取值(范圍)

方法指引1.系數(shù)中字母參數(shù)的取值范圍的求解方法根據(jù)單項(xiàng)式的系數(shù)不為0,或根據(jù)多項(xiàng)式某項(xiàng)系數(shù)不為0,求字

母參數(shù)的取值范圍.2.次數(shù)中字母參數(shù)的取值的求解方法單項(xiàng)式的次數(shù)是所有字母的指數(shù)和,多項(xiàng)式的次數(shù)是次數(shù)最

高的項(xiàng)的次數(shù),根據(jù)次數(shù)的定義列出關(guān)于某字母的方程,求字

母的值.1.(方程思想)(2024山東濟(jì)寧鄒城期末)若多項(xiàng)式2x|a-1|-(a-3)x+

7是關(guān)于x的二次三項(xiàng)式,則a的值為

.-1解析因?yàn)槎囗?xiàng)式2x|a-1|-(a-3)x+7是關(guān)于x的二次三項(xiàng)式,所以|a-1|=2,且-(a-3)≠0.解得a=-1.2.(易錯(cuò)題)已知多項(xiàng)式x2ym+1+xy2-3x3-6是六次四項(xiàng)式,單項(xiàng)式6x2ny5-m的次數(shù)與這個(gè)多項(xiàng)式的次數(shù)相同,求m+n的值.解析因?yàn)槎囗?xiàng)式x2ym+1+xy2-3x3-6是六次四項(xiàng)式,所以2+m+1=6,所以m=3.因?yàn)閱雾?xiàng)式6x2ny5-m的次數(shù)與多項(xiàng)式的次數(shù)