已知实数x y 满足x^2+y^2-xy+2x-y+1=0,试求x y 的值?

网友回答

x^2+y^2-xy+2x-y+1=0
x^2+2x+1-y(x+1)+y^2=0
(x+1)^2-y(x+1)+y^2=0
(x+1-y/2)^2+3y^2/4=0
则y=0,x+1-y/2=0,x=-1
即x=-1,y=0
======以下答案可供参考======
供参考答案1:
x=-1,y=0
供参考答案2:
x=-1y=0供参考答案3:
x^2+y^2-xy+2x-y+1=0
x^2+2x+1-y(x+1)+y^2=0
(x+1)^2-y(x+1)+y^2=0
(x+1-y/2)^2+3y^2/4=0
所以y=0,x+1-y/2=0,
得x=-1最后为x=-1,y=0
供参考答案4:
x^2+y^2-xy+2x-y+1=0
x^2+2x+1-y(x+1)+y^2=0
(x+1)^2-y(x+1)+y^2=0
(x+1-y/2)^2+3y^2/4=0
则y=0,x+1-y/2=0,x=-1
即x=-1,y=0
(x+1-y/2)^2+3y^2/4=0
就是配方得出来的，配上而此项系数的平方