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由x2+xy+y2=1知,-xy=-1+(x2+y2)……(1),又由(x+y)2≥0知,x2+y2≥-2xy=-2+2(x2+y2),即x2+y2≥-2+2(x2+y2),所以-(x2+y2)≥-2,所以x2+y2≤2,即x2+y2的最大值為2.同樣1=x2+xy+y2=x2+y2+xy≤x2+y2+(x2+y2)/2=3...
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