已知实数x y满足x×x+y×y-xy+2x-y+1=0 试求x y的值

已知实数x y满足x×x+y×y-xy+2x-y+1=0 试求x y的值
已知实数x y满足x×x+y×y-xy+2x-y+1=0 试求x y的值

已知实数x y满足x×x+y×y-xy+2x-y+1=0 试求x y的值
相对方程,以元x为主.
方程变为
x^2+(2-y)x+y^2-y+1=0
其判别式=b^2-4ac=（y-2)^2-4y^2+4y-4
=y^2+4-4y+4y-4-4y^2=-3y^2大于等于0
显然y仅可取0,此时x取-1.

x×x+y×y-xy+2x-y+1
=x×x+2x+1+y×y-xy-y
=(x+1)²+y²-xy-y
所以x=-1
y=0
x=-1,y=0