+ Nếu x < 0 ta có \(\left\{{}\begin{matrix}16x^4+5>0\\6\sqrt[3]{4x^3+x}< 0\end{matrix}\right.\Rightarrow16x^4+5\ne6\sqrt[3]{4x^3+x}\) ( loại )
=> \(x\ge0\)
Theo bđt AM-GM ta có: \(6\sqrt[3]{4x^3+x}=3\sqrt[3]{4x\left(4x^2+1\right)\cdot2}\le4x^2+4x+3\)
Dấu "=" \(\Leftrightarrow x=\frac{1}{2}\)
+ Ta lại có : \(16x^4+5=16x^4-8x^2+1+4x^2-4x+1+4x^2+4x+3\)
\(=\left(4x^2-1\right)^2+\left(2x-1\right)^2+4x^2+4x+3\ge4x^2+4x+3\forall x\ge0\)
Dấu "=" \(\Leftrightarrow x=\frac{1}{2}\)
Do đó \(pt\Leftrightarrow x=\frac{1}{2}\)
