ta có pt
<=>\(16x^4+5=3\sqrt[3]{8\left(4x^3+x\right)}\)
Áp dụng bđt cô si, ta có \(3\sqrt[3]{8\left(4x^3+x\right)}=3\sqrt[3]{2.4x.\left(4x^2+1\right)}\le4x^2+1+2+4x\) =\(4x^2+4x+3\)
=>\(16x^4+5\le4x^2+4x+3\Leftrightarrow16x^4-4x^2-4x+2\le0\)
<=>\(8x^4-2x^2-2x+1\le0\Leftrightarrow\left(2x-1\right)^2\left(2x^2+2x+1\right)\le0\)
<=> x=1/2
^_^