设x,y是实数,且 x2+xy+y2=3.那么,x2-xy+y2的取值范围是????

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设x^2-xy+y^2=A

x^2-xy+y^2=A与x2+xy+y2=3相加可喊罩以郑神闹得到2(x^2+y^2)=3+A (1)

x^2-xy+y^2=A与x2+xy+y2=3相减得到2xy=3-A (2)

(1)+(2)*2得:2(x^2+2xy+y^2)=2(x+y)^2=9-A>=0

A>=0,所瞎世以A<=9

(1)-(2)*2得:2(x-y)^2=A-3>=0,所以A>=3

综上:3<=A<=9

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