已知x²-x-2xy+y+y²,求代数式½（x²+y²）-xy的值
问题描述：
已知x²-x-2xy+y+y²,求代数式½（x²+y²）-xy的值
x²-x-2xy+y+y²=2，求代数式½（x²+y²）-xy的值
答
已知x²-x-2xy+y+y²=0x²-2xy+y²-x+y=0(x-y)²-(x-y)=0(x-y)(x-y-1)=0x-y=0 (x-y-1)=0(1)x-y=0 代数式½（x²+y²）-xy的值=1/2(x²+y²-2xy)=1/2(x-y)²=1/2* 0...发漏了不好意思啊是x²-x-2xy+y+y²=2x²-x-2xy+y+y²=2x²-2xy+y²-x+y-2=0(x-y)²-(x-y)-2=0(x-y-2)(x-y+1)=0x-y=2 x-y=-1(1)x-y=2代数式½（x²+y²）-xy的值=1/2(x²+y²-2xy)=1/2(x-y)²=1/2* 2²=2(2)x-y=-1代数式½（x²+y²）-xy的值=1/2(x²+y²-2xy)=1/2(x-y)²=1/2* (-1)²=1/2