解
√2x-y+y²+4y+4=0
√2x-y+(y+2)²=0
∵√2x-y≥0,(y+2)²≥0
∴2x-y=0,y+2=0
∴y=-2,x=-1
∴[(x-y)²+(x+y)(x-y)]÷2x
=(x²-2xy+y²+x²-y²)÷2x
=(2x²-2xy)÷2x
=x-y
=-1+2
=1
解
√2x-y+y²+4y+4=0
√2x-y+(y+2)²=0
∵√2x-y≥0,(y+2)²≥0
∴2x-y=0,y+2=0
∴y=-2,x=-1
∴[(x-y)²+(x+y)(x-y)]÷2x
=(x²-2xy+y²+x²-y²)÷2x
=(2x²-2xy)÷2x
=x-y
=-1+2
=1