标题: [数学] 2007-12-30 [打印本页]

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作者: jhfwin    时间: 2007-12-30 21:43     标题: 2007-12-30

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作者: ITmeansit    时间: 2008-1-13 01:44

x^2+y^2-a(x-y)≥0, (x-y)^2-a(x-y)+2≥0
a^2-8=0, a=2√2.(因x>y,x-y>0，所以a是正数).

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