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
x(x-1)-(x²-y)=-2
x²-x-x²+y=-2
x-y=2
（x²+y²）/2-xy=(x²+y²-2xy)/2
=(x-y)²/2
=2

简化得
y-x=-2 两边平方得
x²+y²-2xy=4
两边同时除以2得
（x²+y²）/2-xy=2