(1) 令x=0.y=0,得f(0)=2f(0)-3,故f(0)=3
(2)设x1>x2>=0,f(x1)-f(x2)=f(x2+(x1-x2))-f(x2)=f(x2)+f(x1-x2)-f(x2)-3=f(x1-x2)-3
因为x1-x2>0 且当x>0时f(x)=0解得 (-无穷,0]+[1,8/5]
f(x²-x)+f(8-5x)=f(x^2-6x+8)+3>=0
f(x^2-6x+6)>=-3
因为f(1)=1 f(2)=1+1-3=-1,f(3)=f(1)+f(2)-3=-3
不等式可以化为f(x^2-6x+8)>=f(3)
有f(x)在定义域上单调递减,所以0