若xˆ2+yˆ2-xy-10x-10y+100≤0,则x-y=

若xˆ2+yˆ2-xy-10x-10y+100≤0,则x-y=
若xˆ2+yˆ2-xy-10x-10y+100≤0,则x-y=

若xˆ2+yˆ2-xy-10x-10y+100≤0,则x-y=
答：
xˆ2+yˆ2-xy-10x-10y+100≤0
2x^2+2y^2-2xy-20x-20y+200<=0
(x^2-2xy+y^2)+(x^2-20x+100)+(y^2-20y+100)=0
(x-y)^2+(x-10)^2+(y-10)^2<=0
根据平方数的非负性质有：
x-y=0
x-10=0
y-10=0
解得：x=y=10
所以：x-y=0

xˆ2+yˆ2-xy-10x-10y+100≤0
2xˆ2+2yˆ2-2xy-20x-20y+200≤0
(x-y)^2+(x-10)^2+(y-10)^2≤0
x-y=0且x=10,y=10
x-y=0