搜搜做题作业网已知2x-y=10,求（(x^2+y^2)-(x-y)^2+2y(x-y))/4y
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已知2x-y=10,求（(x^2+y^2)-(x-y)^2+2y(x-y))/4y
-(x-y)'2怎么变成)-(x^2-2xy+y^2)
-(x-y)'2怎么变成)-(x^2-2xy+y^2)
[(x^2+y^2)-(x-y)^2+2y(x-y)]/4y
=[(x^2+y^2)-(x^2-2xy+y^2)+2xy-2y^2]/4y
=[x^2+y^2-x^2+2xy-y^2+2xy-2y^2]/4y
=(4xy-2y^2)/4y
=2y(2x-y)/4y
=(2x-y)/2
=10/2
=5
再问： -(x-y)'2怎么变成)-(x^2-2xy+y^2)
再答： 不是完全平方差公式么
=[(x^2+y^2)-(x^2-2xy+y^2)+2xy-2y^2]/4y
=[x^2+y^2-x^2+2xy-y^2+2xy-2y^2]/4y
=(4xy-2y^2)/4y
=2y(2x-y)/4y
=(2x-y)/2
=10/2
=5
再问： -(x-y)'2怎么变成)-(x^2-2xy+y^2)
再答： 不是完全平方差公式么
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