Cho hàm số $f(x)$ xác định trên $R$ và thỏa mãn điều kiện
$f(xy) = \frac{(x) + f(y)}{x + y},\forall x,y \in R,x + y \ne 0           (1)$
Chứng minh rằng $f(x) = 0,\forall x \in R$
Trong $(1)$ cho $y = 1$, ta có:  $f(x) = \frac{{f(x) + f(1)}}{{x + 1}},\forall x \ne 1{\rm{  }} \Rightarrow xf(x) = f(1)$
Cho $x = 0$ ta được $f(1) = 0$
Từ đó với mọi $x \ne - 1;{\rm{ 0}}$ ta luôn có $f(x) = 0$
Trong $(1)$ lại cho $x = 2,y = 0$ ta có: 
$f(0) = \frac{{f(2) + f(0)}}{2} \Rightarrow f(0) = f(2) = 0$
Cuối cùng trong $(1)$ cho $x =  - 1;{\rm{ y = 0}}$ ta có:  $f(0) = \frac{{f( - 1) + f(0)}}{{ - 1}} \Rightarrow f( - 1) = - 2f(0) = 0$

Vậy với mọi $x \in R$ ta đều có $f(x) = 0$

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