∵5x²+4x-2xy+y²+1=0
4x²+4x+1+x²-2xy+y²=0
(2x+1)²+(x-y)²=0
∵(2x+1)²≥0,(x-y)²≥0,
∴2x+1=0,x-y=0
∴x=y=-1/2,
∴（x+y)^2的平方根=土根号（x+y)^2=土根号(-1/2-1/2)²=土1