a) \(0,{1^{2 - x}} > 0,{1^{4 + 2x}}\)
\( \Leftrightarrow 2 - x < 4 + 2x \) (vì 0 < 0,1 < 1)
\(\Leftrightarrow 3x > - 2\)
\(\Leftrightarrow x > \frac{{ - 2}}{3}\).
b) \({2.5^{2x + 1}} \le 3\)
\( \Leftrightarrow {5^{2x + 1}} \le \frac{3}{2}\)
\(\Leftrightarrow 2x + 1 \le {\log _5}\frac{3}{2} \)
\(\Leftrightarrow 2x \le {\log _5}\frac{3}{2} - 1\)
\(\Leftrightarrow x \le \frac{1}{2}\left( {{{\log }_5}\frac{3}{2} - 1} \right) = \frac{1}{2}.{\log _5}\frac{3}{{10}} = {\log _5}\frac{{\sqrt {30} }}{{10}}\).
c) \({\log _3}\left( {x + 7} \right) \ge - 1\) (ĐK: x > -7)
\( \Leftrightarrow x + 7 \ge {3^{ - 1}}\)
\(\Leftrightarrow x + 7 \ge \frac{1}{3} \)
\(\Leftrightarrow x \ge \frac{{ - 20}}{3}\).
Kết hợp điều kiện ta có \(x \ge \frac{{ - 20}}{3}\).
d) \({\log _{0,5}}\left( {x + 7} \right) \ge {\log _{0,5}}\left( {2x - 1} \right)\) (ĐK: \(x > \frac{1}{2}\))
\(\Leftrightarrow x + 7 \le 2x - 1\) (vì 0 < 0,5 < 1)
\(\Leftrightarrow x \ge 8\).
Kết hợp điều kiện ta có \(x \ge 8\).
