x^2+ xy + y^2 =3 , 求函数 x^2 - xy + y^2 值域

x^2+y^2=3-xy
x^2+y^2>=2xy
所以3-xy>=2xy
xy<=1
-2xy>=-2

x^2-xy+y^2=x^2+xy+y^2-2xy=3-2xy>=3-2=1
即x^2-xy+y^2>=1

又(x+y)^2>=0，所以x^2y^2>=-2xy
所以3-xy>=-2xy
xy>=-3
-2xy<=6
x^2-xy+y^2=x^2+xy+y^2-2xy=3-2xy<=3+6=9

所以1<=x^2-xy+y^2<=9

x^2 - xy + y^2
=x^2+xy+y^2-2xy
=3-2xy

因此所求值域的问题转化为求xy的值域。

3=x^2+ xy + y^2
=(x^2+y^2)+xy
当xy>0时
原式
>=2xy+xy
=3xy
因此3xy<=3
xy<=1

当xy<0时，原式=3>=-2xy+xy
-xy<=3
xy>=-3

因此3-1*2<=3-2xy<=3-2*(-3)
1<=3-xy<=9

因此1<=x^2 - xy + y^2<=9

x^2+ xy + y^2 = 3
(x-y)^2=3-3xy>=0
xy<=1
(x+y)^2=3+xy>=0
xy>=-3
-3<=xy<=1
x^2 - xy + y^2
=3-2xy
x^2 - xy + y^2 >=3-2=1
x^2 - xy + y^2 <=3-(-6)=9
1<=(x^2 - xy + y^2 )<