已知2x-y=10,求式子[(x^2+y^2)-(x-y)^2+2y(x-y)]/4y
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[(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
=[(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
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[(x^2+y^2)-(x-y)^2+2y(x-y)]/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 (2x-y=10代入)
=10/2
=5
=[x^2+y^2-x^2+2xy-y^2+2xy-2y^2]/4y
=(4xy-2y^2)/4y
=2y(2x-y)/4y
=(2x-y)/2 (2x-y=10代入)
=10/2
=5
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[(x²+y²)-(x-y)²+2y(x-y)]/4y
=[2xy+2y(x-y)]/4y
=(2x-y)/2
=10/2
=5
=[2xy+2y(x-y)]/4y
=(2x-y)/2
=10/2
=5
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((x^2+y^2)-(x-y)^2+2y(x-y))/4y
=(x^2+y^2-x^2-y^2+2xy+2xy-2y^2)/4y
=(4xy-2y^2)/4y
=x-y/2
=(2x-y)/2
=10/2
=5
=(x^2+y^2-x^2-y^2+2xy+2xy-2y^2)/4y
=(4xy-2y^2)/4y
=x-y/2
=(2x-y)/2
=10/2
=5
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因为：2x-y=10
[(x²+y²)-(x-y)²+2y(x-y)]/4y
=[2xy+2y(x-y)]/4y
=(2x-y)/2
=10/2
=5
[(x²+y²)-(x-y)²+2y(x-y)]/4y
=[2xy+2y(x-y)]/4y
=(2x-y)/2
=10/2
=5
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[(x²+y²)-(x-y)²+2y(x-y)]/4y
=[2xy+2y(x-y)]/4y
=(2x-y)/2
=10/2
=5
=[2xy+2y(x-y)]/4y
=(2x-y)/2
=10/2
=5
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