Bài tập 2.60 trang 132 SBT Toán 12
Bài tập 2.60 trang 132 SBT Toán 12
Giải các bất phương trình logarit sau :
a) \({\log _{\frac{1}{3}}}(x - 1) \ge - 2\)
b) \({\log _3}(x - 3) + {\log _3}(x - 5) < 1\)
c) \({\log _{\frac{1}{2}}}\frac{{2{x^2} + 3}}{{x - 7}} < 0\)
d) \({\log _{\frac{1}{3}}}{\log _2}{x^2} > 0\)
e) \(\frac{1}{{5 - \log x}} + \frac{2}{{1 + \log x}} < 1\)
g) \(4{\log _4}x - 33{\log _x}4 \le 1\)
a) ĐK :
\(\begin{array}{l}
{\log _{\frac{1}{3}}}\left( {x - 1} \right) \ge - 2\\
\Leftrightarrow - {\log _3}\left( {x - 1} \right) \ge - 2\\
\Leftrightarrow {\log _3}\left( {x - 1} \right) \le 2\\
\Leftrightarrow x - 1 \le {3^2}\\
\Leftrightarrow x \le 10
\end{array}\)
Vậy \(x \in (1;10]\)
b) ĐK:
\(\begin{array}{l}
{\log _3}\left( {x - 3} \right) + {\log _3}\left( {x - 5} \right) < 1\\
\Leftrightarrow {\log _3}\left[ {\left( {x - 3} \right)\left( {x - 5} \right)} \right] < 1\\
\Leftrightarrow {x^2} - 8x + 15 < 3\\
\Leftrightarrow {x^2} - 8x + 12 < 0\\
\Leftrightarrow 2 < x < 6
\end{array}\)
Kết hợp điều kiện, suy ra \(5 < x < 6\)
c) ĐK:
\(\begin{array}{l}
{\log _{\frac{1}{2}}}\frac{{2{x^2} + 3}}{{x - 7}} < 0\\
\Leftrightarrow \frac{{2{x^2} + 3}}{{x - 7}} > 1\\
\Leftrightarrow \frac{{2{x^2} + 3}}{{x - 7}} - 1 > 0\\
\Leftrightarrow \frac{{2{x^2} - x + 10}}{{x - 7}} > 0\\
\Leftrightarrow \frac{{{x^2} + {x^2} - x + 1 + 9}}{{x - 7}} > 0\\
\Leftrightarrow x - 7 > 0\\
\Leftrightarrow x > 7
\end{array}\)
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:
\(\begin{array}{l}
4{\log _4}x - 33{\log _x}4 \le 1\\
\Leftrightarrow 4{\log _4}x - \frac{{33}}{{{{\log }_4}x}} - 1 \le 0\\
\Leftrightarrow \frac{{4\log _4^2x - {{\log }_4}x - 33}}{{{{\log }_4}x}} \le 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\log _4}x \le - \frac{{11}}{{40}}\\
0 < {\log _4}x \le 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \le {4^{\frac{{ - 11}}{4}}}\\
1 < x < {4^3}
\end{array} \right.
\end{array}\)
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
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