初中數(shù)學(xué)++第三章第2節(jié)整式的加減第1課時(shí)課件+北師大版七年級數(shù)學(xué)上冊++

北師大版七年級數(shù)學(xué)上冊課件第3章第2節(jié)整式的加減第1課時(shí)課時(shí)目標(biāo)素養(yǎng)達(dá)成1.理解同類項(xiàng)及合并同類項(xiàng)的概念,會(huì)識別同類項(xiàng)模型觀念、抽象能力2.掌握合并同類項(xiàng)法則,能進(jìn)行同類項(xiàng)的合并模型觀念、運(yùn)算能力、抽象能力【課前預(yù)習(xí)】【要點(diǎn)歸納】【對點(diǎn)小練】1.下列各式中,與2x4y是同類項(xiàng)的為()

A.2x+4 B.2xyC.x4y

D.2x2y32.計(jì)算-2x2+3x2的結(jié)果為()A.-5x2

B.5x2

C.-x2

D.x23.合并同類項(xiàng):2xy2-4xy-3y2x+2xy=____________.

CD

-xy2-2xy

【典例微課】

【重點(diǎn)1】同類項(xiàng)(模型觀念、抽象能力)【典例1】(教材再開發(fā)·P88同類項(xiàng)定義的拓展)已知單項(xiàng)式2x2my7與單項(xiàng)式5x6yn+8是同類項(xiàng),求2m+n2的值.【自主解答】因?yàn)閱雾?xiàng)式2x2my7與單項(xiàng)式5x6yn+8是同類項(xiàng),所以2m=6,n+8=7,解得m=3,n=-1,所以2m+n2=6+1=7.【變式訓(xùn)練】1.(2024·東莞期中)已知-2xmy6與x3y2n是同類項(xiàng),則mn=_______.

【解析】因?yàn)?2xmy6與x3y2n是同類項(xiàng),所以m=3,2n=6,即m=3,n=3,所以mn=33=27.

27

2.如果單項(xiàng)式-xyb+2與2xa-2y4是同類項(xiàng),那么(a-b)2023=______.

【解析】因?yàn)閱雾?xiàng)式-xyb+2與2xa-2y4是同類項(xiàng),所以a-2=1,b+2=4,解得a=3,b=2,所以(a-b)2023=(3-2)2023=12023=1.

1

【重點(diǎn)2】合并同類項(xiàng)(模型觀念、運(yùn)算能力、抽象能力)【典例2】(教材再開發(fā)·P89例2強(qiáng)化)合并同類項(xiàng):(1)-6x-10x2+12x2-5x;(2)x2y-3xy2+2yx2-y2x.【自主解答】(1)原式=-5x-6x-10x2+12x2=-11x+2x2;(2)原式=x2y+2yx2-3xy2-y2x=3x2y-4xy2.

【課堂小測（8分鐘）】1.(2024·汕頭澄海期末)下面合并同類項(xiàng)正確的是()A.-3a2-2a2=-5a4B.-y2x+xy2=0C.4m-m=4D.-mn-mn=0【解析】A.-3a2-2a2=-5a2,原式計(jì)算錯(cuò)誤,不符合題意;B.-y2x+xy2=0,原式計(jì)算正確,符合題意;C.4m-m=3m,原式計(jì)算錯(cuò)誤,不符合題意;D.-mn-mn=-2mn,原式計(jì)算錯(cuò)誤,不符合題意.B

-1

3.若2x2yn-5xmy3是單項(xiàng)式,則(m-n)n=_______.

【解析】因?yàn)?x2yn-5xmy3是單項(xiàng)式,所以m=2,n=3,所以(m-n)n=(2-3)3=(-1)3=-1.

-1

4.合并同類項(xiàng):(1)3a2-2a+4a2-7a;(2)6x2y+xy2-x2y-2x2y.【解析】(1)原式=(3a2+4a2)+(-2a-7a)=7a2-9a;(2)原式=(6x2y-x2y-2x2y)+xy2=3x2y+xy2.【課后提升】【基礎(chǔ)練】1.(2024·肇慶懷集期末)與3a3b2是同類項(xiàng)的是()

A.-3b3a2 B.-2a3b2C.a2b2

D.a3b2c【解析】根據(jù)同類項(xiàng)的概念可得,與3a3b2是同類項(xiàng)的是-2a3b2.B

B

1

4.合并同類項(xiàng):(1)4x+5y-5x-4y;(2)7a+3a2+2a-a2+3.【解析】(1)原式=(4x-5x)+(5y-4y)=(4-5)x+(5-4)y=-x+y;(2)原式=(7a+2a)+(3a2-a2)+3=9a+2a2+3.【能力練】5.(2024·深圳期末)下列說法中,正確的是()A.-6πx2y3的系數(shù)是-6B.32x2y的次數(shù)是5C.-3和0是同類項(xiàng)D.-x3y+xy-7是三次三項(xiàng)式【解析】A.-6πx2y3的系數(shù)是-6π,原說法錯(cuò)誤,不符合題意;B.32x2y的次數(shù)是3,原說法錯(cuò)誤,不符合題意;C.-3和0是同類項(xiàng),正確,符合題意;D.-x3y+xy-7是四次三項(xiàng)式,原說法錯(cuò)誤,不符合題意.C6.(易錯(cuò)警示題·分類討論遺漏情況)(2024·永州期中)已知a,b為常數(shù),且三個(gè)單項(xiàng)式3xy2,axyb,-7xy相加得到的和仍然是單項(xiàng)式,那么a+b=__________.

【解析】因?yàn)閱雾?xiàng)式3xy2,axyb,-7xy相加得到的和仍然是單項(xiàng)式,所以3xy2,axyb是同類項(xiàng)相加等于零或axyb,-7xy是同類項(xiàng)相加等于零,所以b=2,a=-3或b=1,a=7所以a+b=-3+2=-1或a+b=7+1=8.

-1或8

7.(2024·汕頭潮南期末)(1)已知x=3時(shí),多項(xiàng)式ax3-bx+5的值是1,當(dāng)x=-3時(shí),求ax3-bx+5的值;(2)如果關(guān)于x的二次多項(xiàng)式-3x2+mx+nx2-x+3的值與x的取值無關(guān),求(m+n)(m-n)的值.【解析】(1)因?yàn)閤=3時(shí),多項(xiàng)式ax3-bx+5的值是1,所以27a-3b+5=1,所以27a-3b=-4,所以x=-3時(shí),-27a+3b+5=4+5=9;(2)-3x2+mx+nx2-x+3=(-3+n)x2+(m-1)x+3,因?yàn)殛P(guān)于x的二次多項(xiàng)式的值與x的取值無關(guān),所以-3+n=0,m-1=0,解得n=3,m=1,代入(m+n)(m-n)得,(1+3)×(1-3)=4×(-2)=-8.【培優(yōu)練】8.(模型觀念、運(yùn)算能力、應(yīng)用意識)小王購買了一套經(jīng)濟(jì)適用房,他準(zhǔn)備將地面鋪上地磚,地面結(jié)構(gòu)如圖所示.根據(jù)圖中的數(shù)據(jù)(單位:m),解答下列問題:(1)用含x,y的式子表示地面總面積.(2)當(dāng)x