1 ---- x(x-1)-(x^2-y)=-2,求(x^2+y^2)/2-xy的值

1.
∵x(x-1)-(x^2-y)=-2
∴x^2-x-x^2+y=-2
x-y=2
∵(x-y)^2=x^2+y^2-2xy=4
∴x^2+y^2=4+2xy
∴(x^2+y^2)/2-xy=(4+2xy)/2-xy=2+xy-xy=2
2.x^2+y^2-2x+2y+3=x^2-2x+1+y^2+2y++1+1
=(x-1)^2+(y+1)^2+1
∵(x-1)^2、(y+1)^2均大于等于0
∴x^2+y^2-2x+2y+3=(x-1)^2+(y+1)^2+1>0
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