Bài tập 97 trang 132 SGK Toán 12 NC

a) Ta có: \({\log _4}x = \frac{1}{2}{\log _2}x\)

Đặt \(t = {\log _2}x\)

Ta có: 

\(\frac{{1 - \frac{1}{2}t}}{{1 + t}} - \frac{1}{2} \le 0 \Leftrightarrow \frac{{2 - t - 1 - t}}{{2(1 + t)}} \le 0 \Leftrightarrow \frac{{1 - 2t}}{{1 + t}} \le 0\)

\(\begin{array}{l}
\frac{{1 - \frac{1}{2}t}}{{1 + t}} - \frac{1}{2} \le 0\\
 \Leftrightarrow \frac{{2 - t - 1 - t}}{{2(1 + t)}} \le 0\\
 \Leftrightarrow \frac{{1 - 2t}}{{1 + t}} \le 0\\
 \Leftrightarrow t <  - 1 \vee t \ge \frac{1}{2}\\
 \Leftrightarrow {\log _2}x <  - 1 \vee {\log _2}x \ge \frac{1}{2}\\
 \Leftrightarrow 0 \le x \le \frac{1}{2} \vee x \ge \sqrt 2 
\end{array}\)

Vậy \(S = \left( {0;\frac{1}{2}} \right) \cup \left[ {\sqrt 2 ; + \infty } \right)\)

b) Ta có: 

\(\begin{array}{l}
{\log _{\frac{1}{{\sqrt 5 }}}}({6^{x + 1}} - {36^x}) \ge  - 2\\
\begin{array}{*{20}{l}}
{ \Leftrightarrow 0 < {6^{x + 1}} - {{36}^x} \le {{\left( {\frac{1}{{\sqrt 5 }}} \right)}^{ - 2}} = 5}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{{{6.6}^x} - {{36}^x} > 0}\\
{{{6.6}^x} - {{36}^x} \le 5}
\end{array}} \right.}
\end{array}
\end{array}\)

Đặt  t = 6^x (t > 0). Ta có hệ:

\(\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{6t - {t^2} > 0}\\
{{t^2} - 6t + 5 \ge 0}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{0 < t < 6}\\
{t \le 1,t \ge 5}
\end{array}} \right.}\\
\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{0 < t \le 1}\\
{5 \le t < 6}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{6^x} \le 1}\\
{5 \le {6^x} < 6}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x \le 0}\\
{{{\log }_6}5 \le x < 1}
\end{array}} \right.
\end{array}
\end{array}\)

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

c) Điều kiện 

\(\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{{x^2} - 6x + 18 > 0}\\
{x - 4 > 0}
\end{array}} \right. \Leftrightarrow x > 4}\\
{{{\log }_{\frac{1}{5}}}({x^2} - 6x + 18) + 2{{\log }_5}(x - 4) < 0}\\
{ \Leftrightarrow {{\log }_5}{{(x - 4)}^2} < {{\log }_5}({x^2} - 6x + 18)}\\
{ \Leftrightarrow {{(x - 4)}^2} < {x^2} - 6x + 18 \Leftrightarrow x > 1}
\end{array}\)

Kết hợp điều kiện ta có x > 4.

Vậy \(S = (4; + \infty )\).

-- Mod Toán 12 HỌC247