函数f(x)对一切实数x,y都有f(x+y)-f(y)=(x+2y+1)x成立,且f(1)=0

f(x+y)-f(y)=(x+2y+1)x设x＞0,y＞0,有x+y＞y,(x+2y+1)x＞0即f(x+y)-f(y)=(x+2y+1)x＞0则f(x)在(0,+∞)上递增f(1+0)-f(0)=f(1)-f(0)=(1+1)·1-f(0)=2f(0)=-2f(1/2+1/2)-f(1/2)=f(1)-f(1/2)=(1/2+2·1/2+1)/2-f(1/2)=5...