2x-y=2, 求[(x²+y²)-(x-y)²+2y(x-y)]÷4y
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先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y
[(x²+y²)-(x-y)²+2y(x-y)]/4y
=[x²+y²-(x²+y²-2xy)+2xy-2y²]/4y
=[x²+y²-x²-y²+2xy+2xy-2y²]/4y
=[4xy-2y²]/4y
=[4x-2y]/4
=2*(2x-y)/4
=(2x-y)/2
=1
[(x²+y²)-(x-y)²+2y(x-y)]/4y
=[x²+y²-(x²+y²-2xy)+2xy-2y²]/4y
=[x²+y²-x²-y²+2xy+2xy-2y²]/4y
=[4xy-2y²]/4y
=[4x-2y]/4
=2*(2x-y)/4
=(2x-y)/2
=1
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