a) Điều kiện: \(\displaystyle x - 1 > 0 \Leftrightarrow x > 1\).
\(\displaystyle {\log _{\frac{1}{3}}}(x - 1) \ge - 2\)\(\displaystyle \Leftrightarrow x - 1 \le {\left( {\frac{1}{3}} \right)^{ - 2}}\)\(\displaystyle \Leftrightarrow x - 1 \le 9\)\(\displaystyle \Leftrightarrow x \le 10\)
Kết hợp điều kiện ta được \(\displaystyle 1 < x \le 10\).
b) Điều kiện: \(\displaystyle \left\{ \begin{array}{l}x - 3 > 0\\x - 5 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 3\\x > 5\end{array} \right. \Leftrightarrow x > 5\).
Khi đó bpt\(\displaystyle \Leftrightarrow {\log _3}{\rm{[}}(x - 3)(x - 5){\rm{]}} < {\log _3}3\) \(\displaystyle \Leftrightarrow \left( {x - 3} \right)\left( {x - 5} \right) < 3\) \(\displaystyle \Leftrightarrow {x^2} - 8x + 15 < 3\)
\(\displaystyle \Leftrightarrow {x^2} - 8x + 12 < 0\) \(\displaystyle \Leftrightarrow 2 < x < 6\).
Kết hợp điều kiện ta được \(\displaystyle 5 < x < 6\).
c) Điều kiện: \(\displaystyle \frac{{2{x^2} + 3}}{{x - 7}} > 0\) \(\displaystyle \Leftrightarrow x - 7 > 0 \Leftrightarrow x > 7\).
Khi đó bpt\(\displaystyle \Leftrightarrow \frac{{2{x^2} + 3}}{{x - 7}} > {\left( {\frac{1}{2}} \right)^0} = 1\) \(\displaystyle \Leftrightarrow 2{x^2} + 3 > x - 7\) \(\displaystyle \Leftrightarrow 2{x^2} - x + 10 > 0\) (luôn đúng).
Vậy bất phương trình có nghiệm \(\displaystyle x > 7\).
d) Điều kiện: \(\displaystyle \left\{ \begin{array}{l}{x^2} > 0\\{\log _2}{x^2} > 0\end{array} \right.\) \(\displaystyle \Leftrightarrow \left\{ \begin{array}{l}x \ne 0\\{x^2} > {2^0} = 1\end{array} \right.\) \(\displaystyle \Leftrightarrow \left\{ \begin{array}{l}x \ne 0\\\left[ \begin{array}{l}x > 1\\x < - 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x > 1\\x < - 1\end{array} \right.\)
Khi đó bpt\(\displaystyle \Leftrightarrow {\log _{\frac{1}{3}}}{\log _2}{x^2} > {\log _{\frac{1}{3}}}1\) \(\displaystyle \Leftrightarrow {\log _2}{x^2} < 1 \Leftrightarrow {x^2} < 2\) \(\displaystyle \Leftrightarrow - \sqrt 2 < x < \sqrt 2 \)
Kết hợp điều kiện ta được \(\displaystyle \left[ \begin{array}{l}1 < x < \sqrt 2 \\ - \sqrt 2 < x < - 1\end{array} \right.\).
e) Điều kiện: \(\displaystyle \left\{ \begin{array}{l}x > 0\\\log x \ne 5\\\log x \ne - 1\end{array} \right.\).
Đặt \(\displaystyle t = \log x\) với điều kiện \(\displaystyle t \ne 5,t \ne - 1\) ta có:
\(\displaystyle \frac{1}{{5 - t}} + \frac{2}{{1 + t}} < 1\)\(\displaystyle \Leftrightarrow \frac{{t + 1 + 10 - 2t}}{{5 + 4t - {t^2}}} - 1 < 0\) \(\displaystyle \Leftrightarrow \frac{{{t^2} - 5t + 6}}{{{t^2} - 4t - 5}} > 0\) \(\displaystyle \Leftrightarrow \frac{{(t - 2)(t - 3)}}{{(t + 1)(t - 5)}} > 0\)\(\displaystyle \Leftrightarrow \left[ \begin{array}{l}t < - 1\\2 < t < 3\\t > 5\end{array} \right.\)
TH1: \(\displaystyle t < - 1\) suy ra \(\displaystyle \log x < - 1 \Leftrightarrow x < \frac{1}{{10}}\).
TH2: \(\displaystyle 2 < t < 3\) suy ra \(\displaystyle 2 < \log x < 3 \Leftrightarrow 100 < x < 1000\).
TH3: \(\displaystyle t > 5\) suy ra \(\displaystyle \log x > 5 \Leftrightarrow x > {10^5}\).
Vậy \(\displaystyle x < \frac{1}{{10}}\) hoặc \(\displaystyle 100 < x < 1000\) hoặc \(\displaystyle x > 100000\).
g) Điều kiện \(\displaystyle x > 0,x \ne 1\).
Đặt \(\displaystyle t = {\log _4}x\), ta có: \(\displaystyle 4t - \frac{{33}}{t} \le 1\)
\(\displaystyle \Leftrightarrow \frac{{4{t^2} - t - 33}}{t} \le 0\)\(\displaystyle \Leftrightarrow \frac{{(4t + 11)(t - 3)}}{t} \le 0\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}t \le - \frac{{11}}{4}\\0 < t \le 3\end{array} \right.\)
\(\displaystyle \Rightarrow \left[ \begin{array}{l}{\log _4}x \le - \frac{{11}}{4}\\0 < {\log _4}x \le 3\end{array} \right.\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}0 < x \le {4^{ - \frac{{11}}{4}}}\\1 < x \le 64\end{array} \right.\)
