Điều kiện xác định của \(\sqrt {8{\rm{x}} - {x^2} - 15} \) là 8x – x² – 15 ≥ 0

⇔ x² – 8x + 15 ≤ 0

⇔ x² – 8x + 16 – 1 ≤ 0

⇔ (x – 4)² – 1 ≤ 0

⇔ (x – 4 – 1)(x – 4 + 1) ≤ 0

⇔ (x – 5)(x – 3) ≤ 0

\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x - 3 \le 0\\x - 5 \ge 0\end{array} \right.\\\left\{ \begin{array}{l}x - 3 \ge 0\\x - 5 \le 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \le 3\\x \ge 5\end{array} \right.\\\left\{ \begin{array}{l}x \ge 3\\x \le 5\end{array} \right.\end{array} \right. \Leftrightarrow 3 \le x \le 5\)

Vậy \(\sqrt {8{\rm{x}} - {x^2} - 15} \) xác định khi 3 ≤ x ≤ 5.