原式=(x²+2xy+y²-x²+y²)÷2y=(2xy+2y²)÷2y=x+y
[(x+y)²-(x+y)(x-y)]÷2y =[(x²+2xy+y²)-(x²-y²)]÷2y=(2y²+2xy)÷2y=y+x
解：[(x+y)²-(x+y)(x-y)]÷2y=[(x+y)(x+y-x+y)]÷2y=[(x+y)×2y]÷2y=x+y
