Lời giải:
PT \(\Leftrightarrow \sqrt[3]{x-\frac{1}{2}}+1=16x^3+2\\ \Leftrightarrow \frac{x-\frac{1}{2}+1}{\sqrt[3]{(x-\frac{1}{2})^2}-\sqrt[3]{x-\frac{1}{2}}+1}=2(8x^3+1)\\ \Leftrightarrow \frac{2x+1}{2(\sqrt[3]{(x-\frac{1}{2})^2}-\sqrt[3]{x-\frac{1}{2}}+1)}=2(2x+1)(4x^2-2x+1)\)
\(\Leftrightarrow (2x+1)\left[\frac{1}{2(\sqrt[3]{(x-\frac{1}{2})^2}-\sqrt[3]{x-\frac{1}{2}}+1)}-2(4x^2-2x+1)\right]=0\)
Nếu $2x+1=0\Leftrightarrow x=\frac{-1}{2}$ (tm)
Nếu $\frac{1}{2(\sqrt[3]{(x-\frac{1}{2})^2}-\sqrt[3]{x-\frac{1}{2}}+1)}-2(4x^2-2x+1)=0$
\(\Leftrightarrow 1=4(\sqrt[3]{(x-\frac{1}{2})^2}-\sqrt[3]{x-\frac{1}{2}}+1)(4x^2-2x+1)\)
Đặt \(\sqrt[3]{x-\frac{1}{2}}=a\) thì vế phải là:
$4(a^2-a+1)(4x^2-2x+1)=4[(a-\frac{1}{2})^2+\frac{3}{4}][(2x-\frac{1}{2})^2+\frac{3}{4}]$
$\geq 4.\frac{3}{4}.\frac{3}{4}=\frac{9}{4}>1$ nên TH đang xét bị loại
Vậy $x=\frac{-1}{2}$
