若f(xy)=f(x)+f(y),f(2)=1求f(x)的解析式

f(x+y)=f(x)+f(y)+2xy f(x+y)+f(xy-1)=f(x)f(y)+2f(n)表达式 f(xy)=f(x)+f(y),证明f(x/y)=f(x)-f(y) f(xy)=f(x)+f(y),f(2)=1,f(3)=? f(xy)=f(x)+f(y)且f(8)=3则f(2)= f(xy)=f(x)+f(y),f(8)=3,则f(根2)等于 f(xy)=f(x)+f(y),f(8)=3,求f(√2), 若f(xy)=f(x)+f(y),f(2)=1求f(x)的解析式 若f(xy)=f(x)+f(y)且f(4)=1求f(1/16)若f（x+y)=f(xy) 求f(1） 已知f(x+y)=f(x)+f(y)+2xy,求f(0) 函数f(x)满足关系式f(xy)=f(x)+f(y)求f(1/3)+f(1/2)+f(1)+f(2)+f(3) f(x)满足F(x+y)=f(x)+F(y)+2xy,若f(1)=2,求f(-3)值 导数：f(x+y)=f(x)f(y),且f'(o)=1,求f'(x)f(x+y)=f(x)f(y),且f'(o)=1,求f'(x)f(x+y)=f(x)+f(y)+2xy,且f'(o)存在,求f'(x) f(1+x)=af(x),且f'(0)=b,求f'(1) 已知f(xy)=f(x)+f(y),x,y∈r,若f(3)=2,f(2)=5,则f(18)= f'(0)=1,f(x+y)=f(x)+f(y)+2xy,求f(x) f(x)满足f(x+y)+f(xy)=2f(x)f(y)求函数的奇偶性 f(xy)=f(x)+f(y)的单调性 若f(x)满足f(xy)= f(x) + f(y),据此能否推出f(x/y)= f(x) — f(y)?