# Bài 81 trang 129 SGK giải tích 12 nâng cao

##### Hướng dẫn giải

\eqalign{ & a)\,{\log _5}\left( {3x - 1} \right) < 1 \Leftrightarrow {\log _5}\left( {3x - 1} \right) < {\log _5}5 \cr & \Leftrightarrow 0 < 3x - 1 < 5 \Leftrightarrow {1 \over 3} < x < 2 \cr}

Vậy $$S = \left( {{1 \over 3};2} \right)$$

\eqalign{ & b)\,{\log _{{1 \over 3}}}\left( {5x - 1} \right) > 0 \Leftrightarrow {\log _{{1 \over 3}}}\left( {5x - 1} \right) > {\log _{{1 \over 3}}}1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow 0 < 5x - 1 < 1 \Leftrightarrow {1 \over 5} < x < {2 \over 5} \cr}

Vậy $$S = \left( {{1 \over 5};{2 \over 5}} \right)$$

\eqalign{ & c)\,{\log _{0,5}}\left( {{x^2} - 5x + 6} \right) \ge - 1 \Leftrightarrow \,{\log _{0,5}}\left( {{x^2} - 5x + 6} \right) \ge {\log _{0,5}}2 \cr & \Leftrightarrow 0 < {x^2} - 5x + 6 \le 2 \Leftrightarrow \left\{ \matrix{ {x^2} - 5x + 6 > 0 \hfill \cr {x^2} - 5x + 4 \le 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x < 2\,\text { hoặc }\,x > 3 \hfill \cr 1 \le x \le 4 \hfill \cr} \right. \Leftrightarrow 1 \le x < 2\,\,\text { hoặc }\,\,3 < x \le 4 \cr}

Vậy tập nghiệm của bất phương trình là: $$S = \left[ {1;2} \right) \cup \left( {3;4} \right]$$

\eqalign{ & d)\,{\log _3}{{1 - 2x} \over x} \le 0 \Leftrightarrow {\log _3}{{1 - 2x} \over x} \le {\log _3}1 \cr & \Leftrightarrow 0 < {{1 - 2x} \over x} \le 1 \Leftrightarrow \left\{ \matrix{ {{1 - 2x} \over x} > 0 \hfill \cr {{1 - 2x} \over x} - 1 \le 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ 0 < x < {1 \over 2} \hfill \cr {{1 - 3x} \over x} \le 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 0 < x < {1 \over 2} \hfill \cr x \le 0\,\text { hoặc }\,x \ge {1 \over 3} \hfill \cr} \right. \cr & \Leftrightarrow {1 \over 3} \le x < {1 \over 2} \cr}

Vậy $$S = \left[ {{1 \over 3};{1 \over 2}} \right)$$