把x^2-3xy+y^2=2变换成[(x-y)/根号2]^2+xy/2=1的形式,再令(x-y)/根号2=sina,根号下(xy/2)=cosa,再带入(x-y)^2 + 2xy 再算.以下就不写了,这样能求出
实数x,y满足x^2-3xy+y^2=2
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