设f(x)是R上的函数,且满足f(0)=1,并且对任意实数x,y.有f(x-y)=f(x)-y(2x-y+1),求f(x

恒等式f(x-y)=f(x)-y(2x-y+1),可化为f(x-y)-(x-y)²-(x-y)=f(x)-x²-x
设g(x)=f(x)-x²-x,则g(x-y)=g(x)对于任意实数x,y成立.
∴g(x)为常数函数,
又g(0)=f(0)=1,
∴g(x)必定恒为1,即f(x)=x²+x+1.