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已知2x-y=10,求式子[(x^2+y^2)-(x-y)^2+2y(x-y)]/4y (4xy-2y²)/4y是怎么变成=2y(2x-y)÷4y 详细一点
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已知2x-y=10,
式子[(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
=2y*10/4y
=5
式子[(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
=2y*10/4y
=5
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