Đáp án:
Chọn C.
Giải thích các bước giải:
\(\begin{array}{l}1 + {\log _5}\left( {{x^2} + 1} \right) \ge {\log _5}\left( {m{x^2} + 4x + m} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}m{x^2} + 4x + m > 0\,\,\,\,\forall x \in \mathbb{R}\\{\log _5}5 + {\log _5}\left( {{x^2} + 1} \right) \ge {\log _5}\left( {m{x^2} + 4x + m} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m{x^2} + 4x + m > 0\,\,\,\,\forall x \in \mathbb{R}\\{\log _5}5\left( {{x^2} + 1} \right) \ge {\log _5}\left( {m{x^2} + 4x + m} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m{x^2} + 4x + m > 0\,\,\,\,\forall x \in \mathbb{R}\\5\left( {{x^2} + 1} \right) \ge m{x^2} + 4x + m\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m > 0\\\Delta ' < 0\\\left( {5 - m} \right){x^2} - 4x + 5 - m \ge 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m > 0\\4 - {m^2} < 0\\5 - m > 0\\\Delta ' \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m > 0\\\left[ \begin{array}{l}m > 2\\m < - 2\end{array} \right.\\m < 5\\4 - {\left( {5 - m} \right)^2} \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2 < m < 5\\4 - 25 + 10m - {m^2} \le 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}2 < m < 5\\{m^2} - 10m + 21 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2 < m < 5\\\left[ \begin{array}{l}m < 3\\m > 7\end{array} \right.\end{array} \right. \Leftrightarrow 2 < m < 3.\\m \in \mathbb{Z} \Rightarrow m \in \emptyset .\end{array}\)
