a) \(\displaystyle {(8,4)^{\frac{{x - 3}}{{{x^2} + 1}}}} < 1\)\(\displaystyle \Leftrightarrow 8,{4^{\frac{{x - 3}}{{{x^2} + 1}}}} < 8,{4^0}\)\(\displaystyle \Leftrightarrow \frac{{x - 3}}{{{x^2} + 1}} < 0 \Leftrightarrow x < 3\)
b) \(\displaystyle {2^{|x - 2|}} > {4^{|x + 1|}}\)\(\displaystyle \Leftrightarrow {2^{|x - 2|}} > {2^{2|x + 1|}}\)\(\displaystyle \Leftrightarrow |x - 2| > 2|x + 1|\) \(\displaystyle \Leftrightarrow {x^2} - 4x + 4 > 4({x^2} + 2x + 1)\)
\(\displaystyle \Leftrightarrow 3{x^2} + 12x < 0\)\(\displaystyle \Leftrightarrow - 4 < x < 0\).
c) \(\displaystyle \frac{{{4^x} - {2^{x + 1}} + 8}}{{{2^{1 - x}}}} < {8^x}\)\(\displaystyle \Leftrightarrow {2^{2x}} - {2.2^x} + 8 < {2^{3x}}{.2^{1 - x}}\)\(\displaystyle \Leftrightarrow {2^{2x}} + {2.2^x} - 8 > 0\)
Đặt \(\displaystyle t = {2^x} > 0\) ta được: \(\displaystyle {t^2} + 2t - 8 > 0\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}t < - 4\\t > 2\end{array} \right.\).
Kết hợp với \(\displaystyle t > 0\) ta được \(\displaystyle t > 2\).
Suy ra \(\displaystyle {2^x} > 2 \Leftrightarrow x > 1\).
d) Đặt \(\displaystyle t = {3^x}\left( {t > 0} \right)\), ta có bất phương trình \(\displaystyle \frac{1}{{t + 5}} \le \frac{1}{{3t - 1}}\)
\(\displaystyle \Leftrightarrow \frac{1}{{t + 5}} - \frac{1}{{3t - 1}} \le 0\) \(\displaystyle \Leftrightarrow \frac{{2t - 6}}{{\left( {t + 5} \right)\left( {3t - 1} \right)}} \le 0\) \(\displaystyle \Leftrightarrow \frac{{2t - 6}}{{3t - 1}} \le 0\) (do \(\displaystyle t + 5 > 0\))
\(\displaystyle \Leftrightarrow \frac{1}{3} < t \le 3\)
Do đó \(\displaystyle \frac{1}{3} < {3^x} \le 3 \Leftrightarrow - 1 < x \le 1\) .
Vậy \(\displaystyle - 1 < x \le 1\).
