Phương pháp giải:

Xác định khoảng mà đạo hàm mang dấu âm.

Lời giải chi tiết:

\(g\left( x \right) = f\left( {{x^2} - 5} \right) \Rightarrow g'\left( x \right) = 2x.f'\left( {{x^2} - 5} \right)\)

\(g'\left( x \right) \le 0 \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge 0\\f'\left( {{x^2} - 5} \right) \le 0\end{array} \right.\\\left\{ \begin{array}{l}x \le 0\\f'\left( {{x^2} - 5} \right) \ge 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge 0\\\left[ \begin{array}{l} - 1 \le {x^2} - 5 \le 2\\{x^2} - 5 \le - 4\end{array} \right.\end{array} \right.\\\left\{ \begin{array}{l}x \le 0\\\left[ \begin{array}{l} - 4 \le {x^2} - 5 \le - 1\\{x^2} - 5 \ge 2\end{array} \right.\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge 0\\\left[ \begin{array}{l}2 \le \left| x \right| \le \sqrt 7 \\\left| x \right| \le 1\end{array} \right.\end{array} \right.\\\left\{ \begin{array}{l}x \le 0\\\left[ \begin{array}{l}1 \le \left| x \right| \le 2\\\left| x \right| \ge \sqrt 7 \end{array} \right.\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2 \le x \le \sqrt 7 \\0 \le x \le 1\\ - 2 \le x \le - 1\\x \le - \sqrt 7 \end{array} \right.\)

Ta thấy: \(\left( {2;\dfrac{5}{2}} \right) \subset \left( {2;\sqrt 7 } \right) \Rightarrow \) Hàm số \(g\left( x \right) = f\left( {{x^2} - 5} \right)\) nghịch biến trên khoảng \(\left( {2;\dfrac{5}{2}} \right)\).

Chọn B.