Đáp án:
$f(\dfrac{1}{3})=\dfrac{161}{27}$
Giải thích các bước giải:
Ta có:
$f(x)+2f(\dfrac{1}{x})=x^2$
$\rightarrow f(\dfrac{1}{x})+2f(\dfrac{1}{\frac{1}{x}})=(\dfrac{1}{x})^2$
$\rightarrow f(\dfrac{1}{x})+2f(x)=\dfrac{1}{x^2}$
$\rightarrow 2(f(\dfrac{1}{x})+2f(x))-f(x)+2f(\dfrac{1}{x})=2.\dfrac{1}{x^2}-x^2$
$\rightarrow 3f(x)=2.\dfrac{1}{x^2}-x^2$
$\rightarrow f(x)=\dfrac{2}{3x^2}-\dfrac{x^2}{3}$
$\rightarrow f(\dfrac{1}{3})=\dfrac{2}{3(\dfrac{1}{3})^2}-\dfrac{(\dfrac{1}{3})^2}{3}=\dfrac{161}{27}$
