Lời giải:
ĐKXĐ: \(x^2-16>0\Leftrightarrow (x-4)(x+4)>0\Leftrightarrow \left[\begin{matrix}
x>4\\
x< -4\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x\in (4;+\infty)\\ x\in (-\infty; -4)\end{matrix}\right.\)
Vậy TXĐ \(D=(-\infty;-4)\cup (4;+\infty)\)