4a^2-b^2-4c^2+4bc 如何分解；谢谢指导
作者&投稿：佛沾 （若有异议请与网页底部的电邮联系）
2(2a-b-2c)(4a^2+b^2+4c^2)在实数范围里肯定不能再分解了。
要证明它另当别论。
4a²-4ab+5b²-4bc=4c-2c²-4
4a²-4ab+5b²-4bc-4c+2c²+4=0
(4a²-4ab+b²)+(4b²-4bc+c²)+(c²-4c+4)=0
(2a-b)²+(2b-c)²+(c-2)²=0
因为(2a-b)²≥0，(2b-c)²≥0，(c-2)²≥0
所以要使它们的和0，则它们都必须等于0
所以2a-b=0，2b-c=0，c-2=0
解得a=0.5，b=1，c=2
a+b+c=0.5+1+2=3.5
=(2a)²-(b-2c)²
=(2a+b-2c)(2a-b+2c)
解：原式=4a²-(b²-4bc+4c²)
=(2a)²-(b-2c)²
=(2a+b-2c)(2a-b+2c)
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