a) Ta có : \(f\left(x\right)+3f\left(\frac{1}{3}\right)=x^2\left(1\right)\Rightarrow f\left(\frac{1}{3}\right)+3f\left(\frac{1}{3}\right)=\left(\frac{1}{3}\right)^2\Leftrightarrow4f\left(\frac{1}{3}\right)=\frac{1}{9}\Leftrightarrow f\left(\frac{1}{3}\right)=\frac{1}{36}\)

Thay f(\(\frac{1}{3}\)) = \(\frac{1}{36}\) vào (1) được : \(f\left(x\right)=x^2-3f\left(\frac{1}{3}\right)=x^2-\frac{1}{12}\)

Vậy \(f\left(x\right)=x^2-\frac{1}{12}\)

b) \(f\left(x\right)+2f\left(\frac{1}{x}\right)=2x+\frac{1}{x}\)  (2) . Thay \(x=\frac{1}{x}\) vào \(f\left(x\right)\) và \(f\left(\frac{1}{x}\right)\) được :

\(f\left(\frac{1}{x}\right)+2f\left(x\right)=\frac{2}{x}+x\) \(\Leftrightarrow2f\left(\frac{1}{x}\right)+4f\left(x\right)=\frac{4}{x}+2x\) (3)

Lấy (3) trừ (2) theo vế được: \(\left[2f\left(\frac{1}{x}\right)+4f\left(x\right)\right]-\left[f\left(x\right)+2f\left(\frac{1}{x}\right)\right]=\left(2x+\frac{4}{x}\right)-\left(2x+\frac{1}{x}\right)\)

\(\Leftrightarrow3f\left(x\right)=\frac{3}{x}\Leftrightarrow f\left(x\right)=\frac{1}{x}\)

c) \(f\left(x\right)+2f\left(-x\right)=x+1\) (4)  . Thay x = -x vào f(x) và f(-x) được : 

\(f\left(-x\right)+2f\left(x\right)=-x+1\Leftrightarrow2f\left(-x\right)+4f\left(x\right)=-2x+2\) (5)

Lấy (5) trừ (4) theo vế được : 

\(\left[2f\left(-x\right)+4f\left(x\right)\right]-\left[f\left(x\right)+2f\left(-x\right)\right]=\left(-2x+2\right)-\left(x+1\right)\)

\(\Leftrightarrow3f\left(x\right)=-3x+1\Rightarrow f\left(x\right)=\frac{-3x+1}{3}\)

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