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因为:
a^2 + b^2 + c^2 = ab + bc + ac
2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0
(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = (a - b)^2 + (a - c)^2 + (b - c)^2 = 0
为了让上式成立,则必须有:
(a - b)^2 = (a - c)^2 = (b - c)^2 = 0
即 a = b = c
那么,:(a+b-2c-1)^2 + (b+c-2a-2)^2 + (c+a-2b-3)^2
= 1^2 + 2^2 + 3^2
= 14
1年前
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