Đáp án: $f\left(2\right)=-\dfrac{13}{32}$
Giải thích các bước giải:
Ta có :
$f\left(x\right)+3f\left(\dfrac1x\right)=x^2$
$\to f\left(\dfrac1x\right)+3f\left(\dfrac1{\dfrac1x}\right)=\left(\dfrac1x\right)^2$
$\to f\left(\dfrac1x\right)+3f\left(x\right)=\dfrac1{x^2}$
$\to f\left(\dfrac1x\right)=\dfrac1{x^2}-3f\left(x\right)$
$\to f\left(x\right)+3\cdot \left(\dfrac1{x^2}-3f\left(x\right)\right)=x^2$
$\to -8f\left(x\right)+\dfrac3{x^2}=x^2$
$\to 8f\left(x\right)=\dfrac3{x^2}-x^2$
$\to f\left(x\right)=\dfrac18\left(\dfrac3{x^2}-x^2\right)$
$\to f\left(2\right)=\dfrac18\left(\dfrac3{2^2}-2^2\right)$
$\to f\left(2\right)=-\dfrac{13}{32}$
