# Bài 97 trang 132 SGK giải tích 12 nâng cao

##### Hướng dẫn giải

a) Ta có $${\log _4}x = {1 \over 2}{\log _2}x$$. Đặt $$t = {\log _2}x$$

Ta có

\eqalign{ & {{1 - {1 \over 2}t} \over {1 + t}} - {1 \over 2} \le 0 \Leftrightarrow {{2 - t - 1 - t} \over {2\left( {1 + t} \right)}} \le 0 \Leftrightarrow {{1 - 2t} \over {1 + t}} \le 0 \cr & \Leftrightarrow t < - 1\,\,\text{ hoặc }\,\,t \ge {1 \over 2} \Leftrightarrow {\log _2}x < - 1\,\,\text{ hoặc }\,\,{\log _2}x \ge {1 \over 2} \cr & \Leftrightarrow 0 \le x \le {1 \over 2}\,\,\text{ hoặc }\,\,x \ge \sqrt 2 \cr}

Vậy $$S = \left( {0;{1 \over 2}} \right) \cup \left[ {\sqrt 2 ; + \infty } \right)$$
b) Ta có $${\log _{{1 \over {\sqrt 5 }}}}\left( {{6^{x + 1}} - {{36}^x}} \right) \ge - 2$$

$$\Leftrightarrow 0 < {6^{x + 1}} - {36^x} \le {\left( {{1 \over {\sqrt 5 }}} \right)^{ - 2}} = 5 \Leftrightarrow \left\{ \matrix{ {6.6^x} - {36^x} > 0 \hfill \cr {6.6^x} - {36^x} \le 5 \hfill \cr} \right.$$

Đặt $$t = {6^x}\,\,\left( {t > 0} \right)$$. Ta có hệ:

$$\left\{ \matrix{ 6t - {t^2} > 0 \hfill \cr {t^2} - 6t + 5 \ge 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 0 < t < 6 \hfill \cr t \le 1\,\,\text{ hoặc }\,\,t \ge 5 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ 0 < t \le 1 \hfill \cr 5 \le t < 6 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ {6^x} \le 1 \hfill \cr 5 \le {6^x} < 6 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x \le 0 \hfill \cr {\log _6}5 \le x < 1 \hfill \cr} \right.$$

Vậy $$S = \left( { - \infty ;0} \right] \cup \left[ {{{\log }_6}5;1} \right)$$
c) Điều kiện:

$$\left\{ \matrix{ {x^2} - 6x + 18 > 0 \hfill \cr x - 4 > 0 \hfill \cr} \right. \Leftrightarrow x > 4$$

\eqalign{ & {\log _{{1 \over 5}}}\left( {{x^2} - 6x + 18} \right) + 2{\log _5}\left( {x - 4} \right) < 0 \Leftrightarrow {\log _5}{\left( {x - 4} \right)^2} < {\log _5}\left( {{x^2} - 6x + 18} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {\left( {x - 4} \right)^2} < {x^2} - 6x + 18 \Leftrightarrow x > 1 \cr}

Kết hợp điều kiện ta có $$x > 4$$
Vậy $$S = \left( {4; + \infty } \right)$$