第六章 整式的乘除 乘法公式的六种技巧 课件 2023-2024学年鲁教版六年级数学上册

　乘法公式的六种应用技巧 交换位置应用公式 1.计算: (1)(-3x-4)(3x-4); 解:(1)(-3x-4)(3x-4) =(-4-3x)(-4+3x) =(-4)2-(3x)2 =16-9x2. (2)(-2x+3y)2; 解:(2)(-2x+3y)2 =(3y-2x)2 =(3y)2-2×3y×2x+(2x)2 =9y2-12xy+4x2. (3)(3a+2b)(-3a+2b)(9a2+4b2). 解:(3)(3a+2b)(-3a+2b)(9a2+4b2) =(2b+3a)(2b-3a)(9a2+4b2) =(4b2-9a2)(4b2+9a2) =16b4-81a4. 改变系数应用公式 2.计算: (1)(x2+1)(2x2-2); 解:(1)(x2+1)(2x2-2) =2(x2+1)(x2-1) =2(x4-1) =2x4-2. (2)(2a-6b)(3a-9b); 解:(2)(2a-6b)(3a-9b) =2(a-3b)×3(a-3b) =6(a-3b)2 =6(a2-6ab+9b2) =6a2-36ab+54b2. 连续应用公式 3.计算: (1)(x+4)(x2+16)(x-4); 解:(1)(x+4)(x2+16)(x-4) =(x+4)(x-4)(x2+16) =(x2-16)(x2+16) =x4-256. (3)(a-2)2(a+2)2(a2+4)2. 解:(3)(a-2)2(a+2)2(a2+4)2 =[(a-2)(a+2)(a2+4)]2 =[(a2-4)(a2+4)]2 =(a4-16)2 =a8-32a4+256. 添括号后整体应用公式 4.计算: (1)(a-2b+3)(a+2b-3); 解:(1)(a-2b+3)(a+2b-3) =[a-(2b-3)][a+(2b-3)] =a2-(2b-3)2 =a2-(4b2-12b+9) =a2-4b2+12b-9. (2)(2x+y+1)(2x+y-1); 解:(2)(2x+y+1)(2x+y-1) =[(2x+y)+1][(2x+y)-1] =(2x+y)2-1 =4x2+y2+4xy-1. (3)(x-3y+2c)(x+3y+2c). 解:(3)(x-3y+2c)(x+3y+2c) =[(x+2c)-3y][(x+2c)+3y] =(x+2c)2-(3y)2 =x2+4xc+4c2-9y2. 变形后应用公式 (2)99.72; 解:(2)99.72 =(100-0.3)2 =1002-2×100×0.3+0.32 =10 000-60+0.09=9 940.09. (3)99×101×10 001; 解:(3)99×101×10 001 =(100-1)(100+1)×10 001 =(1002-1)(1002+1) =1004-1=99 999 999. (4)1.234 52+0.765 52+2.469×0.765 5. 解:(4)1.234 52+0.765 52+2.469×0.765 5 =1.234 52+2×1.234 5×0.765 5+0.765 52 =(1.234 5+0.765 5)2 =22=4. 6.已知x+y=-5,xy=-3. (1)求x2+y2的值; 解:(1)因为x+y=-5,xy=-3, 所以x2+y2 =(x+y)2-2xy =(-5)2-2×(-3) =31. (2)求(x-y)2的值. 解:(2)因为xy=-3,x2+y2=31, 所以(x-y)2 =x2+y2-2xy =31-2×(-3) =37. 巧添项,构造公式 8.计算: (1)(2+1)(22+1)(24+1)(28+1)(216+1)+1; 解:(1)(2+1)(22+1)(24+1)(28+1)(216+1)+1 =(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)+1 =(22-1)(22+1)(24+1)(28+1)(216+1)+1 =(24-1)(24+1)(28+1)(216+1)+1 =(28-1)(28+1)(216+1)+1 =(216-1)(216+1)+1 =232-1+1=232. 谢谢观赏！ 25 (3)16(2a+1)(2a-1)(a4+)(4a2+1). 解:(3)16(2a+1)(2a-1)(a4+)(4a2+1) =(4a2-1)(4a2+1)(16a4+1) =(16a4-1)(16a4+1) =256a8-1. (2)(x-)(x2+)(x+); 解:(2)(x-)(x2+)(x+) =(x-)(x+)(x2+) =(x2-)(x2+) =x4-. 5.利用公式进行简便运算: (1)40×39;(2)99.72; 解:(1)40×39 =(40+)(40-) =1 600-