ĐK: \(\displaystyle 7 - x \ge 0 \Leftrightarrow x \le 7\).
Khi đó \(\displaystyle \frac{{{2^x}}}{2} < {2^{\sqrt {7 - x} }}\)\(\displaystyle \Leftrightarrow {2^{x - 1}} < {2^{\sqrt {7 - x} }}\) \(\displaystyle \Leftrightarrow x - 1 < \sqrt {7 - x} \) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x - 1 < 0\\\left\{ \begin{array}{l}x - 1 \ge 0\\{\left( {x - 1} \right)^2} < 7 - x\end{array} \right.\end{array} \right.\)
\(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x < 1\\\left\{ \begin{array}{l}x \ge 1\\{x^2} - x - 6 < 0\end{array} \right.\end{array} \right.\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x < 1\\\left\{ \begin{array}{l}x \ge 1\\ - 2 < x < 3\end{array} \right.\end{array} \right.\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x < 1\\1 \le x < 3\end{array} \right. \Leftrightarrow x < 3\).
Chọn A.
