解析:
由题意可知y≠0
已知[(x²+y²)-(x-y)²+2y(x-y)]÷4y=1,那么:
(x²+y²)-(x²-2xy+y²)+2xy-2y²=4y
即4xy-2y²=4y
两边同除以2y可得:2x-y=2
所以:4x/(4x²-y²)-1/(2x+y)
=[4x-(2x-y)]/(4x²-y²)
=(2x+y)/(4x²-y²)
=1/(2x-y)
=1/2
解析:
由题意可知y≠0
已知[(x²+y²)-(x-y)²+2y(x-y)]÷4y=1,那么:
(x²+y²)-(x²-2xy+y²)+2xy-2y²=4y
即4xy-2y²=4y
两边同除以2y可得:2x-y=2
所以:4x/(4x²-y²)-1/(2x+y)
=[4x-(2x-y)]/(4x²-y²)
=(2x+y)/(4x²-y²)
=1/(2x-y)
=1/2