运用平方差计算下列各题(a-b+c-d)(a+b-c-d); (3a+b-2)(3a-b+2)
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运用平方差计算下列各题(a-b+c-d)(a+b-c-d); (3a+b-2)(3a-b+2)
运用平方差计算下列各题
(a-b+c-d)(a+b-c-d); (3a+b-2)(3a-b+2)
运用平方差计算下列各题(a-b+c-d)(a+b-c-d); (3a+b-2)(3a-b+2)
(a-b+c-d)(a+b-c-d)=[(a-d)-(b-c)][(a-d)+(b-c)]=(a-d)²-(b-c)²=a²+b²+c²+d²-2ad+2bc
(3a+b-2)(3a-b+2)=[3a+(b-2)][3a-(b-2)]=9a²-(b-2)²=9a²-b²+4b-4
(a-b+c-d)(a+b-c-d)
=(a-d-b+c)(a-d+b-c)
=(a-d)^2-(b-c)^2
=a^2-2ad+d^2-(b^2-2bc+c^2)
=a^2-2ad+d^2-b^2+2bc-c^2
(3a+b-2)(3a-b+2)
=(3a)^2-(b-2)^2
=9a^2-(b^2-4b+4)
=9a^2-b^2+4b-4
1。原式=[(a-d)-(b-c)][(a-d)+(b-c)]=(a-d)²-(b-c)²=a²+d²-2ad-b²-c²+2bc
2。原式=[3a+(b-2)][3a-(b-2)]=9a²-(b-2)²=9a²-b²-4+4b
第一个题可以写成:
[(a-d)-(b-c)][(a-d)+(b-c)] = (a-d)^2 - (b-c)^2
第二个题可以写成:
[3a+(b-2)][3a-(b-2)] = (3a)^2 - (b-2)^2
是这个意思吧
原式=[(a-d)-(b-c)][(a-d)+(b-c)]=(a-d)²-(b-c)²=a²+b²+c²+d²-2ad+2bc
原式=[3a+b-2][3a-(b-2)]=9a²-(b-2)²=9a²-b²+4b-4