原式=(x²-2xy+y²)+(y²+2y+1)
=(x-y)²+(y+1)²=0
∵(x-y)²≥0,(y+1)²≥0
∴只有当(x-y)²=0且(y+1)²=0时,(x-y)²+(y+1)²=0
则x-y=0,y+1=0,即y=-1,x=-1
则x+2y=-3