2019年春七年级数学下册第一章整式的乘除4整式的乘法同步课件(新版)北师大版.pptx

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1、知识点一 单项式与单项式的乘法,例1 计算:(1)4xy2 ; (2)(0.3x3y4)2(-0.2x4y3)2;,(3)5x (-2.25axy)(-3x2y2).,解析 (1)4xy2 = x1+2y2+1z =-2x3y3z. (2)(0.3x3y4)2(-0.2x4y3)2=0.09x6y80.04x8y6 =(0.090.04)x6+8y8+6 =0.003 6x14y14. (3)5x (-2.25axy)(-3x2y2) =5 (-2.25)(-3)a1+1x1+1+1+2y1+2= a2x5y3.,知识点二 单项式与多项式的乘法,例2 计算: (1)a(3a-2); (2)(-

2、2x2) ; (3)(-4a3+12a2b-7a3b3)(-4a2).,解析 (1)a(3a-2)=3a2-2a. (2)(-2x2) =(-2x2) xy+(-2x2)y2 =-x3y-2x2y2. (3)(-4a3+12a2b-7a3b3)(-4a2) =(-4a3)(-4a2)+12a2b(-4a2)-7a3b3(-4a2) =16a5-48a4b+28a5b3.,知识点三 多项式与多项式的乘法,例3 计算:(3x-2y)(2a+3b).,解析 原式=3x2a+3x3b+(-2y)2a+(-2y)3b=6ax+9bx-4ay-6by.,题型 将多项式看成整体进行乘法计算 例 计算: (a

3、b)(a-b)4 (a+b)4(b-a)5.,解析 原式= (a+b)4(a-b)4 (a+b)4(b-a)5 = (a+b)4(b-a)4 (a+b)4(b-a)5 = (a+b)4(a+b)4(b-a)4(b-a)5= (a+b)8(b-a)9.,点拨 将多项式看成整体进行乘法运算时,一般会遇到互为相反数的乘 方运算,此时为了不出现符号运算的错误,通常遵循两个原则:(1)少数服 从多数;(2)偶次方服从奇次方,如本题中将(a-b)4化为(b-a)4.,知识点一 单项式与单项式的乘法 1.第一宇宙速度是7.9103米/秒,卫星以此速度运行5103秒走过的路程 为 ( ) A.3.95106

4、米 B.3.95105米 C.39510米 D.3.95107米,答案 D 走过的路程为7.91035103=39.5106=3.95107米,故选D.,2.计算:2x3(-2xy) = .,答案 x7y4,解析 2x3(-2xy) =2x3(-2xy) = x7y4.,3.计算: (1)2x35x2= ; (2)3x2y5(-2xy2z)= .,答案 (1)10x5 (2)-6x3y7z,解析 (1)2x35x2=(25)(x3x2)=10x5. (2)3x2y5(-2xy2z)=3(-2)(x2x)(y5y2)z=-6x3y7z.,4.计算: (1) (-6a2b); (2)(-x2y2)

5、3xy)2; (3)- a2b3 abc.,解析 (1) (-6a2b)= (-6)(a4a2)(b3b)=2a6b4. (2)(-x2y2)(3xy)2=(-x2y2)(9x2y2)=-9x4y4. (3)- a2b3 abc=- a3b4c.,5.先化简,再求值:(-3a3x)(-2a2x2)2+7(ax)3(a2x)2-a7x5,其中x=-2,a=-1.,解析 原式=(-3a3x)4a4x4+7a3x3a4x2-a7x5 =-12a7x5+7a7x5-a7x5 =-6a7x5. 当a=-1,x=-2时, 原式=-6(-1)7(-2)5=-192.,知识点二 单项式与多项式的乘法 6.(

6、2018河南郑州金水月考)计算(-3xy2)(2y2-xyz+1)的结果是 ( ) A.-3xy4+3x2y3+3xy2 B.-6xy4+3x2y3z-3xy2 C.-6xy4-3x2y3z-3xy2 D.-6xy4+3x2y2z,答案 B (-3xy2)(2y2-xyz+1)=-6xy4+3x2y3z-3xy2,故选B.,7.(2017甘肃白银五中月考)计算: -2a2 -5a(a2b-ab2).,解析 原式=-a3b-2a2b2-5a3b+5a2b2=-6a3b+3a2b2.,知识点三 多项式与多项式的乘法 8.(x+1)(2x-3)的计算结果是 ( ) A.2x2+x-3 B.2x2-x

7、3 C.2x2-x+3 D.x2-2x-3,答案 B (x+1)(2x-3)=2x2-3x+2x-3=2x2-x-3.故选B.,9.如果一个长方形的长为(4a2-2a+1),宽为(2a+1),则这个长方形的面积 为 ( ) A.8a3-4a2+2a-1 B.8a3+4a2-2a-1 C.8a3-1 D.8a3+1,答案 D S长方形=(4a2-2a+1)(2a+1)=8a3+4a2-4a2-2a+2a+1=8a3+1.故选D.,10.计算:(1)(2x+1)(3x-1); (2)(3x+5a)(a-3x); (3)(2x-5y)(3x-y); (4)(x+y)(x2-2x-3).,解析 (1

8、)原式=6x2-2x+3x-1=6x2+x-1. (2)原式=3ax-9x2+5a2-15ax=3ax-15ax-9x2+5a2=-12ax-9x2+5a2. (3)原式=6x2-2xy-15xy+5y2=6x2-17xy+5y2. (4)原式=x3-2x2-3x+x2y-2xy-3y.,1.小明做了四道单项式乘法题,其中他做对的一道是 ( ) A.3x22x3=5x5 B.3a34a3=12a9 C.2m23m3=6m3 D.3y36y3=18y6,答案 D 3x22x3=6x5;3a34a3=12a6;2m23m3=6m5;3y36y3=18y6.故选D.,2.计算:3x2(7x2-4x+

9、2)-5x(2x-1)= .,答案 21x4-12x3-4x2+5x,解析 3x2(7x2-4x+2)-5x(2x-1) =21x4-12x3+6x2-10x2+5x =21x4-12x3-4x2+5x.,3.计算图1-4-1的面积.图1-4-1,解析 面积=bt+(a-t)t=at+bt-t2.,1.如果a2+a=1,那么(a-5)(a+6)的值为 .,答案 -29,解析 (a-5)(a+6)=a2+a-30.因为a2+a=1,所以(a-5)(a+6)=1-30=-29.,2.计算: (1)(m+2n)(m-2n);(2)(-a+b)(-a-b).,解析 (1)原式=m2-2mn+2mn-4

10、n2=m2-4n2. (2)原式=a2+ab-ab-b2=a2-b2.,3.计算:2m2(-2mn) .,解析 2m2(-2mn) = (m2mnm2n3)=2m5n4.,4.先化简,再求值:(x+2y)(2x+y)-(3x-y)(x+2y),其中x=9,y= .,解析 原式=2x2+xy+4xy+2y2-(3x2+6xy-xy-2y2) =-x2+4y2. 当x=9,y= 时,原式=-92+4 =-80.,1.若(-x2ym)2(kxn+1y)=-2x6y3,则(km)n等于 ( ) A.-2 B.2 C.4 D.-4,答案 A 因为(-x2ym)2(kxn+1y)=x4y2mkxn+1y=

11、kx5+ny2m+1=-2x6y3,所以k=-2,5+n= 6,2m+1=3, 所以n=1,m=1. 所以(km)n=(-2)11=-2,故选A.,2.定义 为二阶行列式,规定它的运算法则为 =ad-bc.例如: =1 4-23=-2.化简 = .,答案 4a+12,解析 原式=(a+2)(a+3)-(a-2)(a+3)=a2+5a+6-a2-a+6=4a+12.,3.设x,y为任意实数,定义运算:x*y=(x+1)(y+1)-1,得到下列五个命题: x*y=y*x;x*(y+z)=x*y+x*z;(x+1)*(x-1)=(x*x)-1;x*0=0;(x+1)*(x +1)=x*x+2*x+1

12、 其中正确的命题的序号是 .,答案 ,解析 x*y=xy+x+y=y*x,故正确; x*(y+z)=(x+1)(y+z+1)-1=xy+xz+x+y+z,x*y+x*z=(x+1)(y+1)-1+(x+1)(z+ 1)-1=xy+xz+2x+y+z,故错误; (x+1)*(x-1)=(x+2)x-1=x2+2x-1, (x*x)-1=(x+1)(x+1)-1-1=x2+2x-1,故正确; x*0=x,故错误; (x+1)*(x+1)=(x+2)(x+2)-1=x2+4x+3, x*x+2*x+1=(x+1)(x+1)-1+3(x+1)-1+1=x2+5x+3,故错误.故答案为.,4.请说明对

13、任意正整数n,式子n(n+5)-(n+2)(n-3)的值必定能被6整除.,解析 n(n+5)-(n+2)(n-3) =n2+5n-(n2-3n+2n-6) =n2+5n-n2+n+6 =6n+6=6(n+1). n为正整数,n+1为正整数, n(n+5)-(n+2)(n-3)的值必定能被6整除.,一、选择题 1.(2018江西抚州临川第一实验学校月考,2,)若(y+3)(y-2)=y2+my+ n,则 ( ) A.m=5,n=6 B.m=1,n=-6 C.m=1,n=6 D.m=5,n=-6,答案 B (y+3)(y-2)=y2+y-6=y2+my+n,m=1,n=-6.,2.(2018江苏扬

14、州树人学校期中,7,)如果(x2-mx+1)(x+2)的积中不 含x的二次项,那么m的值是 ( ) A.1 B.-1 C.-2 D.2,答案 D 原式=x3+2x2-mx2-2mx+x+2=x3+(2-m)x2-2mx+x+2,由题意得2-m =0,则m=2.,二、解答题 3.2016江苏昆山期中,19(2),计算:-2xy3x2y-x2y(-3xy+xy2).,解析 原式=-6x3y2+3x3y2-x3y3 =-3x3y2-x3y3.,1.(2018湖南郴州龙市中学月考,9,)化简 x2y3 的结果为 ( ) A.-x3y3 B.x3y3 C.-2x3y3 D.2x3y3,答案 C 原式=

15、x2xy3=-2x3y3.,2.(2016北京石景山期末,17,)计算:-6ab .,解析 原式=-12a3b2+2a2b3.,3.(2016山东济南商河期末,18,)计算:3x(x-y)-(2x-y)(x+y).,解析 原式=3x2-3xy-(2x2+2xy-xy-y2)=x2-4xy+y2.,一、选择题 1.(2017台湾省中考,3,)计算6x(3-2x)的结果,与下列哪一个式子 相同 ( ) A.-12x2+18x B.-12x2+3 C.16x D.6x,答案 A 6x(3-2x)=18x-12x2,故选A.,2.(2017湖北武汉中考,5,)计算(x+1)(x+2)的结果为 ( )

16、A.x2+2 B.x2+3x+2 C.x2+3x+3 D.x2+2x+2,答案 B (x+1)(x+2)=x2+2x+x+2=x2+3x+2.故选B.,二、填空题 3.(2018天津中考,13,)计算2x4x3的结果等于 .,答案 2x7,解析 原式=2x4+3=2x7.,4.(2015湖南常德中考,11,)计算:b(2a+5b)+a(3a-2b)= .,答案 5b2+3a2,解析 b(2a+5b)+a(3a-2b)=2ab+5b2+3a2-2ab=5b2+3a2.,1.(2016贵州毕节中考,3,)下列运算正确的是 ( ) A.-2(a+b)=-2a+2b B.(a2)3=a5 C.a3+4

17、a= a3 D.3a22a3=6a5,答案 D A.原式=-2a-2b;B.原式=a6;C.a3与4a不是同类项,不能合并;D. 原式=6a5.故选D.,2.(2015福建福州中考,12,)计算(x-1)(x+2)的结果是 .,答案 x2+x-2,解析 (x-1)(x+2)=x2+2x-x-2=x2+x-2.,1.定义运算:ab=a(1-b),下面给出了关于这种运算的几个结论: 2(-2)=6; ab=ba; 若a+b=0,则(aa)+(bb)=2ab; 若ab=0,则a=0. 其中正确结论的序号是 .,答案 ,解析 2(-2)=21-(-2)=6,正确. ba=b(1-a)ab,错误. a+

18、b=0,(aa)+(bb)=a(1-a)+b(1-b)=a-a2+b-b2=-a2-b2, 由a+b=0,得b=-a,-a2-b2=-2a2,2ab=2a(-a)=-2a2, (aa)+(bb)=2ab,正确. 当ab=0时,a=0或b=1,错误,故填.,2.三角形 表示运算3abc,方框xy wz表示运算-4xywz,求n2 m5的值.,解析 由题意得, n2 m5=(33mn)(-4n2m5)=9mn(-4m5n2)=9(-4)m 1+5n1+2=-36m6n3.,1.观察下列各式: (x-1)(x+1)=x2-1; (x-1)(x2+x+1)=x3-1; (x-1)(x3+x2+x+1)

19、x4-1; (1)根据以上规律,则(x-1)(x6+x5+x4+x3+x2+x+1)= ; (2)你能否由此归纳出一般性规律:(x-1)(xn+xn-1+x+1)= ; (3)根据(2)求出1+2+22+234+235的结果.,解析 (1)根据题意得(x-1)(x6+x5+x4+x3+x2+x+1)=x7-1. (2)根据题意得(x-1)(xn+xn-1+x+1)=xn+1-1. (3)原式=(2-1)(1+2+22+234+235)=236-1.,2.化简:2(m-1)m+m(m+1)(m-1)m-m(m+1).若m是任意整数,请观察化简 后的结果,你发现原式表示一个什么数?,解析 原式=2(m2-m+m2+m)(m2-m-m2-m) =22m2(-2m) =-8m3. 发现:原式=(-2m)3,即无论m取什么整数,原式表示一个偶数的立方.,

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