ĐK: \(\hept{\begin{cases}2-6x\ge0\\x+2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}6x\le2\\x\ge-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{1}{3}\\x\ge-2\end{cases}}\Leftrightarrow-2\le x\le\frac{1}{3}.\)
Vậy TXĐ: \(D=\left[-2;\frac{1}{3}\right]\)