已知x(x-1)-(x²-y)=3,求（2分之x²+y²）-xy的值

已知x(x-1)-(x²-y)=3,求（2分之x²+y²）-xy的值
已知x(x-1)-(x²-y)=3,求（2分之x²+y²）-xy的值

已知x(x-1)-(x²-y)=3,求（2分之x²+y²）-xy的值
即x²-x-x²+y=3
-x+y=3
x-y=-3
所以原式=(x²-2xy+y²)/2
=(x-y)²/2
=9/2