(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,若有一個(gè)大于0,則另一個(gè)小于0,不成立.
所以?xún)蓚€(gè)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