\(\forall x\in R,x\ne0\) ta luôn có
\(f\left(x\right)+2f\left(\dfrac{1}{x}\right)=x\)
Thay \(x\) bởi \(\dfrac{1}{x}\) ta được: \(f\left(\dfrac{1}{x}\right)+2f\left(x\right)=\dfrac{1}{x}\)
Ta có hệ \(\left\{{}\begin{matrix}f\left(x\right)+2f\left(\dfrac{1}{x}\right)=x\\f\left(\dfrac{1}{x}\right)+2f\left(x\right)=\dfrac{1}{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(x\right)+2f\left(\dfrac{1}{x}\right)=x\left(1\right)\\2f\left(\dfrac{1}{x}\right)+4f\left(x\right)=\dfrac{2}{x}\left(2\right)\end{matrix}\right.\)
Lấy (2) - (1) ta được \(3f\left(x\right)=\dfrac{2}{x}-x=\dfrac{2-x^2}{x}\)
\(\Rightarrow f\left(x\right)=\dfrac{2-x^2}{3x}\)
