把(a+b-c)(a-b+c)+(b-a-c)²因式分解

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把(a+b-c)(a-b+c)+(b-a-c)²因式分解
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把(a+b-c)(a-b+c)+(b-a-c)²因式分解
把(a+b-c)(a-b+c)+(b-a-c)²因式分解

把(a+b-c)(a-b+c)+(b-a-c)²因式分解
(a+b-c)(a-b+c)+(b-a-c)²
=[a+(b-c)][a-(b-c)]+[b-(a+c)]²
=a²-(b-c)²+b²-2b(a+c)+(a+c)²
=a²-b²+2bc+b²-c²-2ab-2bc+a²+2ac+c²
=2a²-2ab+2ac
明白请采纳,
有新问题请求助,

(a+b-c)(a-b+c)+(b-a-c)²
=(a+b-c+a+c-b)(a-b+c)
=2a(a-b+c)

(a+b-c)(a-b+c)+(b-a-c)^2
=(a+b-c)(a-b+c)+(a+c-b)^2
=(a+c-b)[(a+b-c)+(a+c-b)]
=(a+c-b)(2a)
=2a(a-b+c)
求采纳


(a+b-c)(a-b+c)+(b-a-c)²
=(a+b-c)(a-b+c)+(a-b+c)²
=(a+b-c+a-b+c)(a-b+c) (提取公因式(a-b+c) )
=2a(a-b+c)

=【a+(b-c)】【a-(b-c)】+【(b-c)-a】
=a^2-(b-c)^2+(b-c)^2-2a(b-c)+a^2
=2a^2-2ab-2ac
=2a(a-b-c)

(a+b-c)[a-(b-c)]+(b-a-c)²
=a²-(b-c)²+(b-c)²-2a(b-c)+a²
=a²-2a(b-c)+a²=2a(a-b+c)