Đáp án: $x=\dfrac{1}{2}$
Giải thích các bước giải:
$16x^4+5=6\sqrt[3]{4x^3+x}$
$\rightarrow 16x^4+5=16x^4-8x^2+1+8x^2+4=(4x^2-1)^2+8x^2+4\ge 8x^2+4$
$\rightarrow 16x^4+5\ge \dfrac{1}{2}(8x^2+4x^2+1+4x^2+1+2+2+2)$
$\rightarrow 16x^4+5\ge \dfrac{1}{2}6\sqrt[6]{8x^2.(4x^2+1).(4x^2+1).2.2.2}$
$\rightarrow 16x^4+5\ge 3\sqrt[6]{64x^2.(4x^2+1)^2}$
$\rightarrow 16x^4+5\ge 6\sqrt[3]{x(4x^2+1)}$
$\rightarrow 16x^4+5\ge 6\sqrt[3]{4x^3+x}$
$\rightarrow VT\ge VP\rightarrow\text{Dấu = xảy ra khi x=$\dfrac{1}{2}$}$
