Question

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

求f（x）的表达式。

Answer 1

令 y = x得：
f(x - x) = f(x) - x(2x - x + 1)
f(0) = f(x) - x(x + 1)
1 = f(x) - x(x + 1)
f(x) = x(x + 1) + 1
f(x) = x² + x + 1

Answer 2

f(x-y)=f(x)-y(2x-y+1),

put y= x
f(0) = f(x) -x(2x-x+1)
f(x) = 1+ x(x+1)
= x^2+x+1

Answer 3

令y=x得
f(0)=f(x)-x(2x-x+1)
f(x)=f(0)+x^2+x
=1+x^2+x

Answer 4

令 y = x得：
f(x - x) = f(x) - x(2x - x + 1)
f(0) = f(x) - x(x + 1)
1 = f(x) - x(x + 1)
f(x) = x(x + 1) + 1
f(x) = x² + x + 1