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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) \(4{\log _4}x – 33{\log _x}4 \le 1\)
Hướng dẫn làm bài:
a) \(0 < x – 1 \le {(\frac{1}{3})^{ – 2}} \Leftrightarrow 1 < x \le 10\)
b)
\(\eqalign{& \left\{ {\matrix{{x > 5} \cr {{{\log }_3}{\rm{[}}(x – 3)(x – 5){\rm{]}} < {{\log }_3}3} \cr} } \right. \cr & \Leftrightarrow \left\{ {\matrix{{x > 5} \cr {{x^2} – 8x + 12 < 0} \cr} \Leftrightarrow \left\{ {\matrix{{x > 5} \cr {2 < x < 6} \cr} } \right.} \right. \cr & \Leftrightarrow 5 < x < 6 \cr} \)
c)
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\(\eqalign{& \left\{ {\matrix{{x – 7 > 0} \cr {{{2{x^2} + 3} \over {x – 7}} > 1} \cr} \Leftrightarrow \left\{ {\matrix{{x > 7} \cr {2{x^2} + 3 > x – 7} \cr} } \right.} \right. \cr & \Leftrightarrow \left\{ {\matrix{{x > 7} \cr {2{x^2} – x + 10 > 0} \cr} \Leftrightarrow \left\{ {\matrix{{x > 7} \cr {x \in R} \cr} \Leftrightarrow x > 7} \right.} \right. \cr} \)
d)
\(\eqalign{
& {\log _{{1 \over 3}}}{\log _2}{x^2} > {\log _{{1 \over 3}}}1 \cr
& \Leftrightarrow {\log _2}{x^2} < 1 \cr
& \Leftrightarrow {\log _2}{x^2} < {\log _2}2 \cr
& \Leftrightarrow 0 < {x^2} < 2 \cr} \)
\(\Leftrightarrow 0 < |x| < \sqrt 2 \Leftrightarrow \left[ {\matrix{{ – \sqrt 2 < x < 0} \cr {0 < x < \sqrt 2 } \cr} } \right.\)
e) Đặt \(t = \log x\) với điều kiện \(t \ne 5,t \ne – 1\) ta có:
\(\eqalign{
& {1 \over {5 – t}} + {2 \over {1 + t}} < 1 \Leftrightarrow {{t + 1 + 10 – 2t} \over {5 + 4t – {t^2}}} – 1 < 0 \cr
& \Leftrightarrow {{{t^2} – 5t + 6} \over {{t^2} – 4t – 5}} > 0 \Leftrightarrow {{(t – 2)(t – 3)} \over {(t + 1)(t – 5)}} > 0 \cr & \Leftrightarrow \left[ {\matrix{{t < – 1} \cr {2 < t < 3} \cr {t > 5} \cr} } \right. \cr} \)
Suy ra log x < -1 hoặc 2 < log x < 3 hoặc log x > 5.
Vậy \(x < \frac{1}{{10}}\) hoặc 100 < x < 1000 hoặc x > 100 000.
g) Với điều kiện \(x > 0,x \ne 1\) đặt \(t = {\log _4}x\) , ta có: \(4t – \frac{{33}}{t} \le 1\)
\(\eqalign{& \Leftrightarrow {{4{t^2} – t – 33} \over t} \le 0 \Leftrightarrow {{(4t + 11)(t – 3)} \over t} \le 0 \cr & \Leftrightarrow \left[ {\matrix{{t \le – {{11} \over 4}} \cr {0 < t \le 3} \cr} } \right. \cr & \Leftrightarrow \left[ {\matrix{{{{\log }_4}x \le – {{11} \over 4}} \cr {0 < {{\log }_4}x \le 3} \cr} } \right. \Leftrightarrow \left[ {\matrix{{0 < x \le {4^{ – {{11} \over 4}}}} \cr {1 < x \le 64} \cr} } \right. \cr} \)