已知x^2+xy=3,xy+y^2=-2,则2x^2-xy-3y^2=

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已知x^2+xy=3,xy+y^2=-2,则2x^2-xy-3y^2=
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已知x^2+xy=3,xy+y^2=-2,则2x^2-xy-3y^2=
已知x^2+xy=3,xy+y^2=-2,则2x^2-xy-3y^2=

已知x^2+xy=3,xy+y^2=-2,则2x^2-xy-3y^2=
x^2+xy=3可化为x（x+y)=3
xy+y^2=-2可化为y(x+y)=-2
两式相加减得（x+y）^2=1
(x+y)(x-y)=5
得{x+y=1 {x+y=-1
x-y=5 x-y=-5
这两个二元一次方程组
解得两组解{x1=3 {x2=-3
y1=-2 y2=2
代入2x^2-xy-3y^2可得唯一答案12

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