分式有意義,y≠0
[(x²+y²)-(x-y)²+2y(x-y)]÷(4y)=1
x²+y²-(x²-2xy+y²)+(2xy-2y²)=4y
4xy-2y²=4y
2y²-4xy+4y=0
2y(y-2x+2)=0
y≠0,要等式成立,只有y-2x+2=0
2x-y=2
4x/(4x²-y²) - 1/(2x+y)
=4x/[(2x+y)(2x-y)]- 1/(2x+y)
=[4x-(2x-y)]/[(2x+y)(2x-y)]
=(2x+y)/[(2x+y)(2x-y)]
=1/(2x-y)
=1/2