
若(2x-y)的平方+|y+2丨=o,求代数式[(x-y)的平方+(x+y)(x-y)除以2x的值
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(2x-y)的平方+|y+2丨=o
2x-y=0 y+2=0
x=-1 y=-2
代数式[(x-y)的平方+(x+y)(x-y)除以2x
=[(x-y)^2+(x+y)(x-y)]/2x
=(x-y)(x-y+x+y)/2x
=(x-y)*2x/2x
=x-y
=-1-(-2)
=1
2x-y=0 y+2=0
x=-1 y=-2
代数式[(x-y)的平方+(x+y)(x-y)除以2x
=[(x-y)^2+(x+y)(x-y)]/2x
=(x-y)(x-y+x+y)/2x
=(x-y)*2x/2x
=x-y
=-1-(-2)
=1
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