已知、5x²-2xy+y²+4x=1=0求[(x+y)(x-y)-(x-y)²+2y(x-3y)]÷(-4y)的值

已知、5x²-2xy+y²+4x=1=0求[(x+y)(x-y)-(x-y)²+2y(x-3y)]÷(-4y)的值
已知、5x²-2xy+y²+4x=1=0求[(x+y)(x-y)-(x-y)²+2y(x-3y)]÷(-4y)的值

已知、5x²-2xy+y²+4x=1=0求[(x+y)(x-y)-(x-y)²+2y(x-3y)]÷(-4y)的值
(x²-2xy+y²)+(4x²+4x+1)=0
(x-y)²+(2x+1)²=0
平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.
所以两个都等于0
所以x-y=0,2x+1=0
x=-1/2,y=x=-1/2
[(x+y)(x-y)-(x-y)²+2y(x-3y)]÷(-4y)
=[(x+y)*0-0²+2y(x-3y)]÷(-4y)
=-(x-3y)/2
=-(-1/2+3/2)/2
=-1/2

题目有问题：5x²-2xy+y²+4x=1=0，怎么会有1=0？？？？