若√(2x-y)+丨y+2丨=0,求代数式(x-y)²+(x+y)(x-y)的值
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若√(2x-y)+丨y+2丨=0,求代数式(x-y)²+(x+y)(x-y)的值
若√(2x-y)+丨y+2丨=0,求代数式(x-y)²+(x+y)(x-y)的值
若√(2x-y)+丨y+2丨=0,求代数式(x-y)²+(x+y)(x-y)的值
x-y)2+(x+y)(x-y)
=x²+y²+2xy+x²-y²
=2x²+2xy
又前式得:{①:2x-y=0
②:y+2=0
∴{y=-2
x=-1
带入得:
原式 =-2