Cho $a,b$ là các số thực cho trước. Xác định tất cả các hàm số $f(x)$ thỏa mãn mỗi một tính chất sau đây:
a) $f(a-x)=f(x)$, với mọi $x\in R$
b) $f(a-x)+f(x)=b$, với mọi $x\in R$
Giải
Đặt $x=\frac{a}{2}-t, t\in R$ suy ra $t=\frac{a}{2}-x$ và $a-x=\frac{a}{2}+t$. Ta có
a) $f(a-x)=f(x), \forall x\in R \Rightarrow f(\frac{a}{2}+1)=f(\frac{a}{2}-t), \forall t\in R        (*)$
Đặt $g(t)=f(\frac{a}{2}+t)$ khi đó (*) $\Leftrightarrow g(t)=g(-t), \forall t\in R \Leftrightarrow g(t)$ là hàm số chẵn trên $R$
b) $f(a-x)+f(x)=b, \forall x\in R \Leftrightarrow f(\frac{a}{2}+t)-\frac{b}{2}=\frac{b}{2}-f(\frac{a}{2}-1), \forall t\in R       (**)$
Đặt $h(t)=f(\frac{a}{2}+t)-\frac{b}{2}$ khi đó (**) $\Leftrightarrow h(t)=-h(t), \forall t\in R \Leftrightarrow h(t)$ là hàm số lẻ trên $R$.
Vậy $f(x)=h(x-\frac{a}{2})+\frac{b}{2}$ với $g(x)$ là một hàm số chẵn tùy ý xác định trên $R$

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