Ta có:
\(\dfrac{x^2}{x^4+x^2+1}=\dfrac{x^2}{3x^2+x^4-2x^2+1}=\dfrac{x^2}{3x^2+\left(x^2-1\right)^2}\)
Mà \(\left(x^2-1\right)^2\ge0\forall x\)
\(\Rightarrow\dfrac{x^2}{3x^2+\left(x^2-1\right)^2}\le\dfrac{x^2}{3x^2}\)
\(\Rightarrow\dfrac{x^2}{3x^2+\left(x^2-1\right)^2}\le\dfrac{1}{3}\)
Dấu "=" xảy ra khi và chỉ khi \(x^2=1\Rightarrow x=\pm1\)
Vậy \(Max_B=\dfrac{1}{3}\) \(\Leftrightarrow x=\pm1\)