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Câu hỏi: Tập nghiệm của bất phương trình ${{\log }_{\dfrac{1}{2}}}\left( x-1 \right)>{{\log }_{2}}\dfrac{1}{{{x}^{2}}-1}$ là:
A. $\left[ 2;+\infty \right)$
B. $\varnothing $
C. $\left( 0;1 \right)$
D. $\left( 1;+\infty \right)$
A. $\left[ 2;+\infty \right)$
B. $\varnothing $
C. $\left( 0;1 \right)$
D. $\left( 1;+\infty \right)$
Ta có: ${{\log }_{\dfrac{1}{2}}}\left( x-1 \right)>{{\log }_{2}}\dfrac{1}{{{x}^{2}}-1}\Leftrightarrow {{\log }_{2}}\dfrac{1}{x-1}>{{\log }_{2}}\dfrac{1}{{{x}^{2}}-1}$
$\Leftrightarrow \dfrac{1}{x-1}>\dfrac{1}{{{x}^{2}}-1}>0\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{x+1-1}{{{x}^{2}}-1}>0 \\
& {{x}^{2}}-1>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>0 \\
& \left[ \begin{aligned}
& x>1 \\
& x<-1 \\
\end{aligned} \right.\Leftrightarrow x>1 \\
\end{aligned} \right.$.
$\Leftrightarrow \dfrac{1}{x-1}>\dfrac{1}{{{x}^{2}}-1}>0\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{x+1-1}{{{x}^{2}}-1}>0 \\
& {{x}^{2}}-1>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>0 \\
& \left[ \begin{aligned}
& x>1 \\
& x<-1 \\
\end{aligned} \right.\Leftrightarrow x>1 \\
\end{aligned} \right.$.
| ${{\log }_{a}}f\left( x \right)>{{\log }_{a}}g\left( x \right)\Leftrightarrow f\left( x \right)>g\left( x \right)>0$ (với $a>1$ ). |
Đáp án D.