\(f\left(x\right)+2f\left(\frac{1}{x}\right)=x^2\) (1)
\(\Rightarrow f\left(\frac{1}{x}\right)+2f\left(x\right)=\frac{1}{x^2}\Rightarrow2f\left(\frac{1}{x}\right)+4f\left(x\right)=\frac{2}{x^2}\) (2)
Trừ (2) cho (1): \(3f\left(x\right)=\frac{2}{x^2}-x^2\Rightarrow f\left(x\right)=\frac{2}{3x^2}-\frac{1}{3}x^2\)
\(\Rightarrow f\left(2019\right)=\frac{2}{3.2019^2}-\frac{1}{3}.2019^2\)