Giải bài 2.60 trang 132 - SBT Giải tích lớp 12
Giải các bất phương trình lôgarit sau :
\(\begin{align} & a)\,{{\log }_{\frac{1}{3}}}\left( x-1 \right)\ge -2; \\ & b)\,{{\log }_{3}}\left( x-3 \right)+{{\log }_{3}}\left( x-5 \right)<1; \\ & c)\,{{\log }_{\frac{1}{2}}}\dfrac{2{{x}^{2}}+3}{x-7}<0; \\ & d)\,{{\log }_{\frac{1}{3}}}{{\log }_{2}}{{x}^{2}}>0; \\ & e)\dfrac{1}{5-\log x}+\dfrac{2}{1+\log x}<1; \\ & g)4{{\log }_{4}}x-33{{\log }_{x}}4\le 1. \\ \end{align} \)
a) ĐK : \( x > 1\)
\(\begin{aligned} & {{\log }_{\frac{1}{3}}}\left( x-1 \right)\ge -2 \\ & \Leftrightarrow -{{\log }_{3}}\left( x-1 \right)\ge -2 \\ & \Leftrightarrow {{\log }_{3}}\left( x-1 \right)\le 2 \\ & \Leftrightarrow x-1\le {{3}^{2}} \\ & \Leftrightarrow x\le 10 \\ \end{aligned}\)
Vậy \(x\in (1;10] \)
b) ĐK: \(x>5 \)
\(\begin{aligned} & {{\log }_{3}}\left( x-3 \right)+{{\log }_{3}}\left( x-5 \right)<1 \\ & \Leftrightarrow {{\log }_{3}}\left[ \left( x-3 \right)\left( x-5 \right) \right]<1 \\ & \Leftrightarrow {{x}^{2}}-8x+15<3 \\ & \Leftrightarrow {{x}^{2}}-8x+12<0 \\ & \Leftrightarrow 2< x<6 \\ \end{aligned} \)
Kết hợp điều kiện, suy ra \(5< x<6 \)
c) ĐK: \(x>7 \)
\(\begin{aligned} & {{\log }_{\frac{1}{2}}}\dfrac{2{{x}^{2}}+3}{x-7}<0 \\ & \Leftrightarrow \dfrac{2{{x}^{2}}+3}{x-7}>1 \\ & \Leftrightarrow \dfrac{2{{x}^{2}}+3}{x-7}-1>0 \\ & \Leftrightarrow \dfrac{2{{x}^{2}}-x+10}{x-7}>0 \\ & \Leftrightarrow \dfrac{{{x}^{2}}+{{x}^{2}}-x+1+9}{x-7}>0 \\ & \Leftrightarrow x-7>0 \\ & \Leftrightarrow x>7 \\ \end{aligned} \)
Vậy x> 7
d) ĐK: \(x\in \mathbb{R} \)
\(\begin{aligned} & {{\log }_{\frac{1}{3}}}{{\log }_{2}}{{x}^{2}}>0 \\ & \Leftrightarrow \left\{ \begin{aligned} & {{\log }_{2}}{{x}^{2}}>0 \\ & {{\log }_{2}}{{x}^{2}}<1 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & {{x}^{2}}>1 \\ & {{x}^{2}}<2 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & \left[ \begin{aligned} & x<-1 \\ & x>1 \\ \end{aligned} \right. \\ & -\sqrt{2}< x<\sqrt{2} \\ \end{aligned} \right.\Leftrightarrow x\in \left( -\sqrt{2};-1 \right)\cup \left( 1;\sqrt{2} \right) \\ \end{aligned} \)
e) ĐK \(x> 0; x\ne {{10}^{5}};x\ne \dfrac{1}{10} \)
\(\begin{aligned} & \dfrac{1}{5-\log x}+\dfrac{2}{1+\log x}<1 \\ & \Leftrightarrow \dfrac{1+\log x+2\left( 5-\log x \right)-\left( 5-\log x \right)\left( 1+\log x \right)}{\left( 5-\log x \right)\left( 1+\log x \right)}<0 \\ & \Leftrightarrow \dfrac{{{\log }^{2}}x-5\log x+6}{\left( 5-\log x \right)\left( 1+\log x \right)}<0 \\ & \Leftrightarrow \dfrac{\left( \log x-2 \right)\left( \log x-3 \right)}{\left( 5-\log x \right)\left( 1+\log x \right)}<0 \\ & \Leftrightarrow \left[ \begin{aligned} & \log x<-1 \\ & 2<\log x<3 \\ & \log x>5 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x<\dfrac{1}{10} \\ & 100< x<1000 \\ & x>{{10}^{5}} \\ \end{aligned} \right. \\ \end{aligned}\)
Kết hợp điều kiện, ta có:
\(\left[ \begin{aligned} & 0< x<\dfrac{1}{10} \\ & 100< x<1000 \\ & x>100000 \\ \end{aligned} \right. \)
g) ĐK: \(x> 0\)
\(\begin{aligned} & 4{{\log }_{4}}x-33{{\log }_{x}}4\le 1 \\ & \Leftrightarrow 4{{\log }_{4}}x-\dfrac{33}{{{\log }_{4}}x}-1\le 0 \\ & \Leftrightarrow \dfrac{4\log _{4}^{2}x-{{\log }_{4}}x-33}{{{\log }_{4}}x}\le 0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{\log }_{4}}x\le -\dfrac{11}{4} \\ & 0<{{\log }_{4}}x\le 3 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x\le {{4}^{-\frac{11}{4}}} \\ & 1< x<{{4}^{3}} \\ \end{aligned} \right. \\ \end{aligned} \)
Kết hợp điều kiện, ta có: \(\left[ \begin{aligned} & 0 < x\le {{4}^{-\frac{11}{4}}} \\ & 1< x\le 64 \\ \end{aligned} \right. \)