Đề bài

Giải mỗi bất phương trình sau: a) \({2^{5x + 1}} > 0,25;\) b) \({\left( {\frac{4}{9}} \right)^{x - 1}} < {\left( {\frac{3}{2}} \right)^{x + 2}};\) c) \({\log _{16}}\left( {3x + 4} \right) <  - \frac{1}{4};\) d) \({\log _{0,2}}\left( {{x^2} - 6x + 9} \right) \ge {\log _{0,2}}\left( {x - 3} \right).\)

Lời giải chi tiết

a) \({2^{5x + 1}} > 0,25 \Leftrightarrow {2^{5x + 1}} > {2^{ - 2}} \Leftrightarrow 5x + 1 >  - 2 \Leftrightarrow x >  - \frac{3}{5}.\)b) \({\left( {\frac{4}{9}} \right)^{x - 1}} < {\left( {\frac{3}{2}} \right)^{x + 2}} \Leftrightarrow {\left( {\frac{2}{3}} \right)^{2\left( {x - 1} \right)}} < {\left( {\frac{3}{2}} \right)^{x + 2}} \Leftrightarrow {\left( {\frac{3}{2}} \right)^{2\left( {1 - x} \right)}} < {\left( {\frac{3}{2}} \right)^{x + 2}}\)\( \Leftrightarrow 2\left( {1 - x} \right) < x + 2 \Leftrightarrow 3x > 0 \Leftrightarrow x > 0.\)c) Điều kiện: \(3x + 4 > 0 \Leftrightarrow x >  - \frac{4}{3}.\)\({\log _{16}}\left( {3x + 4} \right) <  - \frac{1}{4} \Leftrightarrow {\rm{l}}o{g_{{2^4}}}\left( {3x + 4} \right) <  - \frac{1}{4} \Leftrightarrow \frac{1}{4}{\rm{l}}o{g_2}\left( {3x + 4} \right) <  - \frac{1}{4}\)\( \Leftrightarrow {\rm{l}}o{g_2}\left( {3x + 4} \right) <  - 1 \Leftrightarrow 3x + 4 < \frac{1}{2} \Leftrightarrow x <  - \frac{7}{6}.\)Suy ra nghiệm của bất phương trình là: \( - \frac{4}{3} < x <  - \frac{7}{6}.\) d) \({\log _{0,2}}\left( {{x^2} - 6x + 9} \right) \ge {\log _{0,2}}\left( {x - 3} \right) \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 6x + 9 \le x - 3\\{x^2} - 6x + 9 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 7x + 12 \le 0\\{\left( {x - 3} \right)^2} > 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}\left( {x - 3} \right)\left( {x - 4} \right) \le 0\\x \ne 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3 \le x \le 4\\x \ne 3\end{array} \right. \Leftrightarrow 3 < x \le 4\).