设x(x-1)-(x2-y)=-2,求(x2+y2\2)-xy的值

设x(x-1)-(x2-y)=-2,求(x2+y2\2)-xy的值
设x(x-1)-(x2-y)=-2,求(x2+y2\2)-xy的值

设x(x-1)-(x2-y)=-2,求(x2+y2\2)-xy的值
x²-x-x²+y=-2
-x+y=-2
x-y=2
求的式子应该是(x²+y²)/2-xy吧?
(x²+y²)/2-xy
=(x²-2xy+y²)/2
=(x-y)²/2
=(2)²/2
=4/2
=2

x²-x-x²+y=-2
-x+y=-2
x-y=2
求的式子应该是(x²+y²)/2-xy吧？
(x²+y²)/2-xy
=(x²-2xy+y²)/2
=(x-y)²/2
=(2)²/2
=4/2
=2