[(x²+y²)-(x-y)²+2y(x-y)]/2y=1(x²+y²-x²-y²+2xy+2xy-2y²)/2y=1;2x-y=1;原式=1/(4x²-y²)-1/(2x+y)=(1-(2x-y))/(4x²-y²)=(1-1)/(4x²-y²)=0;...