x的平方+4乘y的平方+2x+4y+2=0,求x-y的值
问题描述：
x的平方+4乘y的平方+2x+4y+2=0,求x-y的值
答
x^2+4*y^2+2x+4y+2=0
(x+1)^2+(2y+1)^2=0
x+1=0 x=-1
2y+1=0 y=-1/2
x-y=(-1)-(-1/2)=-1/2
答
x²+4y²+2x+4y+2=0
（x+1)²+（2y+1）²=0
x+1=0 2y+1=0
x=-1 ，y=-1/2
x-y=-1-（-1/2）=-1/2
答
整理一下,变成(x^2+2x+1)+(4y^2+4y+1)=0
即(x+1)^2+(2y+1)^2=0
由于(x+1)^2≥0,(2y+1)^2≥0
所以(x+1)^2=0,(2y+1)^2=0
x=-1,y=-1/2
x-y=-1/2
答
x^2+4y^2+2x+4y+2=0
x^2+2x+1+4y^2+4y+1=0
(x+1)^2+(2y+1)^2=0
x+1=0 2y+1=0
x=-1 y=-1/2
x-y
=-1-(-1/2)
=-1+1/2
=-1/2