RT,若（√2x-y）+y^2+4y+4=0,求[(x-y)^2+(x+y)(x-y)]÷2x的值

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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======以下答案可供参考======
供参考答案1:
（√2x-y）+y^2+4y+4=0
（√2x-y）+（y+2）²=0
∴﹛2x-y=0
y+2=0∴x=-1, y=-2
[(x-y)^2+(x+y)(x-y)]÷2x
=[(x-y)(x-y+x+y)]÷2x
=(x+y)×2x÷2x
=x+y=-1-2=-3供参考答案2:
（√2x-y）+y^2+4y+4=0
（√2x-y）+(y+2)²=0
∴√2x-y=0
y+2=0∴x=-√2
y=-2[(x-y)^2+(x+y)(x-y)]÷2x
=(x-y)(x-y+x+y)÷2x
=x-y=-√2+2