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平方差公式
第十四章 整式的乘法与因式分解
多项式与多项式是如何相乘的?
(x + 3)( x+5)
=x2+5x +3x +15
=x2 +8x +15.
(a+b)(m+n) =am +an +bm +bn
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面积变了吗?
a米
5米
5米a米
(a-5)米
平方差公式
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(1)(x + 1)( x-1);
(2)(m + 2)( m-2);
(3)(2m+ 1)(2m-1);
(4)(5y + z)(5y-z).
计算下列多项式的积,你能发现什么规律?
算一算:看谁算得又快又准.
x2 - 12
m2-22
(2m)2 - 12
(5y)2 - z2
想一想:这些计算结果有什么特点?
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(a+b)(a−b)= a2−b2
也就是说,两个数的和与这两个数的差的
积,等于这两数的平方差.这个公式叫做(乘法
的)平方差公式.
1.(a – b ) ( a + b) = a2 - b2
2.(b + a )( -b + a ) = a2 - b2
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(a+b)(a-b)=(a)2-(b)2
相同为a
相反为b,-b
适当交换
合理加括号
平方差公式是
多项式乘法
(a+b)(p+q)
中,p=a,q=-b
的特殊形式.
a2-b2
a2-b2
b2-a2
b2-a2
计算:
(1) (3x+2 )( 3x-2 ) ; (2)(-x+2y)(-x-2y).
(2) 原式= (-x)2 - (2y)2 =x2 - 4y2.
解:(1)原式=(3x)2-22 =9x2-4.
应用平方差公式计算时,应注意以下几点:(1)左边是两个二
项式相乘,并且这两个二项式中一项完全相同,另一项互为相反数;(2)
右边是相同项的平方减去相反项的平方;(3)公式中的a和b可以是具体的
数,也可以是单项式或多项式.
例1
【练习】利用平方差公式计算:
(1)(3x-5)(3x+5);
(2)(-2a-b)(b-2a);
(3)(-7m+8n)(-8n-7m).
解:(1)原式=(3x)2-52=9x2-25.
(2)原式=(-2a)2-b2=4a2-b2.
(3)原式=(-7m)2-(8n)2=49m2-64n2.
计算:
(1) (y+2) (y-2) – (y-1) (y+5);
(2) 102×98 .解: (1) (y+2)(y-2)- (y-1)(y+5)
(2)102×98
=y2-4-y2-4y+5
=- 4y + 1.
=y2-22-(y2+4y-5)
=9996.
= (100+2)(100-2)
= 1002-22
= 10 000 – 4
通过合理变形,
利用平方差公式,
可以简化运算.
不符合平方差公
式运算条件的
乘法,按乘法
法则进行运算.
例2
【练习】计算:
(1) 51×49; (2)(3x+4)(3x-4)-(2x+3)(3x-2) .
解: (1) 原式=(50+1)(50-1)
= 502-12
=2500 – 1
=2499.
(2) 原式=(3x)2-42-(6x2+5x-6)
= 9x2-16-6x2-5x+6
= 3x2-5x-10.
先化简,再求值:(2x-y)(y+2x)-(2y+x)(2y-
x),其中x=1,y=2.
原式=5×12-5×22=-15.
解:原式=4x2-y2-(4y2-x2)
=4x2-y2-4y2+x2
=5x2-5y2.
当x=1,y=2时,
例3
对于任意的正整数n,整式(3n+1)(3n-1)-
(3-n)(3+n)的值一定是10的整数倍吗?
即(3n+1)(3n-1)-(3-n)(3+n)的值一定是10的
整数倍.
解:原式=9n2-1-(9-n2) =10n2-10.
∵(10n2-10)÷10=n2-1,n为正整数,∴n2-1为整数.
在探究整除性或倍数问题时,一般先将代数式化为最简,
然后根据结果的特征,判断其是否具有整除性或倍数关系.
例4
1.下列运算中,可用平方差公式计算的是( )
A.(x+y)(x+y) B.(-x+y)(x-y)
C.(-x-y)(y-x) D.(x+y)(-x-y)
C
2.计算(2x+1)(2x-1)等于( )
A.4x2-1 B.2x2-1 C.4x-1 D.4x2+1
A
3.两个正方形的边长之和为5,边长之差为2,那
么用较大的正方形的面积减去较小的正方形的
面积,差是________.
10
(1)(a+3b)(a- 3b);
=4a2-9.
=4x4-y2.
解:原式=(2a+3)(2a-3)
=a2-9b2 . =(2a)2-32
解:原式=(-2x2 )2-y2
解:原式=a2-(3b)2
(2)(3+2a)(-3+2a);
(3)(-2x2-y)(-2x2+y).
4.利用平方差公式计算:
5.计算: 20172 - 2016×2018.
解: 20172 - 2016×2018
= 20172 - (2017-1)×(2017+1)
= 20172- (20172-12 )
= 20172 - 20172+12
=1.
6.利用平方差公式计算:
(1)(a-2)(a+2)(a2 + 4) ;
解:原式=(a2-4)(a2+4)
=a4-16.
(2) (x-y)(x+y)(x2+y2)(x4+y4).
解:原式=(x2-y2)(x2+y2)(x4+y4)
=(x4-y4)(x4+y4)
=x8-y8.
7.先化简,再求值:(x+1)(x-1)+x2(1-x)+x3,
其中x=2.
解:原式=x2-1+x2-x3+x3
=2x2-1.
将x=2代入上式,
得原式=2×22-1=7.
8.已知x≠1,计算:(1+x)(1-x)=1-x2,(1-x)(1+
x+x2)=1-x3,(1-x)(1+x+x2+x3)=1-x4.
(1)观察以上各式并猜想:(1-x)(1+x+x2+…+xn)=
________;(n为正整数)(2)根据你的猜想计算:
①(1-2)(1+2+22+23+24+25)=________;
②2+22+23+…+2n=________(n为正整数);
③(x-1)(x99+x98+x97+…+x2+x+1)=________;
1-xn+1
-63
2n+1-2
x100-1
(3)通过以上规律请你进行下面的探索:
①(a-b)(a+b)=________;
②(a-b)(a2+ab+b2)=________;
③(a-b)(a3+a2b+ab2+b3)=________.
a2-b2
a3-b3
a4-b4
平方差
公 式
内 容
注 意
两个数的和与这两个数的差的积,
等于这两个数的平方差.
字母表示:(a+b)(a-b)=a2-b2
应用时,紧紧抓住 “一同一反”这一特征
,只有两个二项式的积才有可能应用平方
差公式;对于不能直接应用公式的,可能
要经过变形才可以应用