Bài tập 2.60 trang 132 SBT Toán 12

a) ĐK : x > 1

\(\begin{array}{l}
{\log _{\frac{1}{3}}}\left( {x - 1} \right) \ge  - 2\\
 \Leftrightarrow  - {\log _3}\left( {x - 1} \right) \ge  - 2\\
 \Leftrightarrow {\log _3}\left( {x - 1} \right) \le 2\\
 \Leftrightarrow x - 1 \le {3^2}\\
 \Leftrightarrow x \le 10
\end{array}\)

Vậy \(x \in (1;10]\)

b) ĐK: x > 5

\(\begin{array}{l}
{\log _3}\left( {x - 3} \right) + {\log _3}\left( {x - 5} \right) < 1\\
 \Leftrightarrow {\log _3}\left[ {\left( {x - 3} \right)\left( {x - 5} \right)} \right] < 1\\
 \Leftrightarrow {x^2} - 8x + 15 < 3\\
 \Leftrightarrow {x^2} - 8x + 12 < 0\\
 \Leftrightarrow 2 < x < 6
\end{array}\)

Kết hợp điều kiện, suy ra \(5 < x < 6\)

c) ĐK: x > 7

\(\begin{array}{l}
{\log _{\frac{1}{2}}}\frac{{2{x^2} + 3}}{{x - 7}} < 0\\
 \Leftrightarrow \frac{{2{x^2} + 3}}{{x - 7}} > 1\\
 \Leftrightarrow \frac{{2{x^2} + 3}}{{x - 7}} - 1 > 0\\
 \Leftrightarrow \frac{{2{x^2} - x + 10}}{{x - 7}} > 0\\
 \Leftrightarrow \frac{{{x^2} + {x^2} - x + 1 + 9}}{{x - 7}} > 0\\
 \Leftrightarrow x - 7 > 0\\
 \Leftrightarrow x > 7
\end{array}\)

Vậy x > 7

d) ĐK: \(x \in R\)

\(\begin{array}{l}
{\log _{\frac{1}{3}}}{\log _2}{x^2} > 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\log _2}{x^2} > 0\\
{\log _2}{x^2} < 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} > 1\\
{x^2} < 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x <  - 1\\
x > 1
\end{array} \right.\\
 - \sqrt 2  < x < \sqrt 2 
\end{array} \right.\\
 \Leftrightarrow x \in \left( { - \sqrt 2 ; - 1} \right) \cup \left( {1;\sqrt 2 } \right)
\end{array}\)

e) ĐK : \(x > 0;x \ne {10^5};x \ne \frac{1}{{10}}\)

\(\begin{array}{l}
\frac{1}{{5 - \log x}} + \frac{2}{{1 + \log x}} < 1\\
 \Leftrightarrow \frac{{1 + \log x + 2(5 - \log x) - (5 - \log x)(1 + \log x)}}{{(5 - \log x)(1 + \log x)}} < 0\\
 \Leftrightarrow \frac{{{{\log }^2}x - 5\log x + 6}}{{(5 - \log x)(1 + \log x)}} < 0\\
 \Leftrightarrow \frac{{(\log x - 2)(\log x - 3)}}{{(5 - \log x)(1 + \log x)}} < 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\log x <  - 1\\
2 < \log x < 3\\
\log x > 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x < \frac{1}{{10}}\\
100 < x < 1000\\
x > {10^5}
\end{array} \right.
\end{array}\)

Kết hợp điều kiện, ta có:
\(\left[ \begin{array}{l}
0 < x < \frac{1}{{10}}\\
100 < x < 1000\\
x > 100000
\end{array} \right.\)

g) ĐK: x > 0

\(\begin{array}{l}
4{\log _4}x - 33{\log _x}4 \le 1\\
 \Leftrightarrow 4{\log _4}x - \frac{{33}}{{{{\log }_4}x}} - 1 \le 0\\
 \Leftrightarrow \frac{{4\log _4^2x - {{\log }_4}x - 33}}{{{{\log }_4}x}} \le 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{\log _4}x \le  - \frac{{11}}{{40}}\\
0 < {\log _4}x \le 3
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x \le {4^{\frac{{ - 11}}{4}}}\\
1 < x < {4^3}
\end{array} \right.
\end{array}\)

Kết hợp điều kiện, ta có: \(\left[ \begin{array}{l}

0 < x \le {4^{ - \frac{{11}}{4}}}\\
1 < x \le 64
\end{array} \right.\)

 

-- Mod Toán 12 HỌC247