Học lớp hướng dẫn giải
\(\begin{array}{l}1 + {\log _5}\left( {{x^2} + 1} \right) \ge {\log _5}\left( {m{{\rm{x}}^2} + 4{\rm{x}} + m} \right) \Leftrightarrow \left\{ \begin{array}{l}m{{\rm{x}}^2} + 4{\rm{x}} + m > 0\\5\left( {{x^2} + 1} \right) \ge m{{\rm{x}}^2} + 4{\rm{x}} + m\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left\{ \begin{array}{l}m > 0\\4 - {m^2} < 0\end{array} \right.\\\left( {m - 5} \right){x^2} + 4{\rm{x}} + m - 5 \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m > 2\\\left( {m - 5} \right){x^2} + 4{\rm{x}} + m - 5 \le 0\,\,\left( * \right)\end{array} \right.\end{array}\)
TH1: \(m - 5 = 0 \Leftrightarrow m = 5 \Rightarrow \left( * \right) \Leftrightarrow x \le 0.\)
TH2: \(m - 5 \ne 0 \Leftrightarrow m \ne 5 \Rightarrow \left( * \right) \Leftrightarrow \left\{ \begin{array}{l}m - 5 < 0\\4 - {\left( {m - 5} \right)^2} \le 0\end{array} \right. \Rightarrow m \le 3.\)
Suy ra: \(2 < m \le 3 \Rightarrow X = \left( {2;3} \right].\)