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已知a=2000x+1999,b=2000x+2000,c=2000x+2001,a^2+b^2+c^2-ab-bc-ac的值
人气:418 ℃ 时间:2020-03-28 14:18:04
解答
a^2+b^2+c^2-ab-bc-ac
=1/2(2a^2+2b^2+2c^2-2ab-2bc-2ac)
=1/2[(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)]
=1/2[(a-b)^2+(a-c)^2+(b-c)^2]
=1/2[1+4+1]
=3
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