襄阳市中考 已知[（x²+y²)-(x-y)²+2y(x-y)]÷4y=1 求4x/4x²-y² - 1/2x+y的值
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[（x²+y²)-(x-y)²+2y(x-y)]÷4y
=[（x²+y²)-(x^2-2xy+y^2)+2xy-2y^2]÷4y
=(4xy-2y^2)÷4y
=x-y/2
=1
2x-y=2
4x/(4x²-y²) - 1/(2x+y)
=4x/(4x²-y²) - (2x-y)/(4x²-y²)
=(2x+y)/(4x²-y²)
=1/(2x-y)
=1/2
=[（x²+y²)-(x^2-2xy+y^2)+2xy-2y^2]÷4y
=(4xy-2y^2)÷4y
=x-y/2
=1
2x-y=2
4x/(4x²-y²) - 1/(2x+y)
=4x/(4x²-y²) - (2x-y)/(4x²-y²)
=(2x+y)/(4x²-y²)
=1/(2x-y)
=1/2
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