\(\Leftrightarrow16x^4+5-6=6\sqrt[3]{4x^3+x}-6\)
\(\Leftrightarrow16x^4-1=6\left(\sqrt[3]{4x^3+x}-1\right)\)
\(\Leftrightarrow\left(4x^2-1\right)\left(4x^2+1\right)=6\left[\dfrac{4x^3+x-1}{\left(\sqrt[3]{4x^3+x}\right)^2+\sqrt[3]{4x^3+x}+1}\right]\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)-6\left(\dfrac{\left(2x-1\right)\left(2x^2+x+1\right)}{\left(\sqrt[3]{4x^3+x}\right)^2+\sqrt[3]{4x^3+x}+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left(2x+1\right)\left(4x^2+1\right)=\dfrac{6\left(2x^2+x+1\right)}{\left(\sqrt[3]{4x^3+x}\right)^2+\sqrt[3]{4x^3+x}+1}\left(l\right)\end{matrix}\right.\)
Vậy S=\(\left\{\dfrac{1}{2}\right\}\)
