\(\left(x+\dfrac{2}{3}\right)^2+\dfrac{2}{3}>=\dfrac{2}{3}\forall x\)
\(\dfrac{2}{\left(x^2-\dfrac{4}{9}\right)^6+3}< =\dfrac{2}{3}\forall x\)
mà \(\left(x+\dfrac{2}{3}\right)^2+\dfrac{2}{3}=\dfrac{2}{\left(x^2-\dfrac{4}{9}\right)^6+3}\)
nên \(\left\{{}\begin{matrix}x+\dfrac{2}{3}=0\\x^2-\dfrac{4}{9}=0\end{matrix}\right.\)
=>\(x=-\dfrac{2}{3}\)