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Bài 97. Giải các bát phương trình sau:
\(\eqalign{
& a)\,{{1 – {{\log }_4}x} \over {1 + {{\log }_2}x}} < {1 \over 2}\,; \cr
& c)\,{\log _{{1 \over 5}}}\left( {{x^2} – 6x + 18} \right) + 2{\log _5}\left( {x – 4} \right) < 0. \cr} \)
\(b)\,{\log _{{1 \over {\sqrt 5 }}}}\left( {{6^{x + 1}} – {{36}^x}} \right) \ge – 2;\)
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 \(S = \left( {4; + \infty } \right)\)