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Câu hỏi: Tập nghiệm S của bất phương trình ${{\log }_{\dfrac{1}{2}}}\left( x+1 \right)<{{\log }_{\dfrac{1}{2}}}\left( 2x-1 \right)$ là
A. $S=\left( 2;+\infty \right)$
B. $S=\left( -\infty ;2 \right)$
C. $S=\left( \dfrac{1}{2};2 \right)$
D. $S=\left( -1;2 \right)$
A. $S=\left( 2;+\infty \right)$
B. $S=\left( -\infty ;2 \right)$
C. $S=\left( \dfrac{1}{2};2 \right)$
D. $S=\left( -1;2 \right)$
Điều kiện: $\left\{ \begin{aligned}
& x+1>0 \\
& 2x-1>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>-1 \\
& x>\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow x>\dfrac{1}{2}\left( * \right)$
${{\log }_{\dfrac{1}{2}}}\left( x+1 \right)<{{\log }_{\dfrac{1}{2}}}\left( 2x-1 \right)\Leftrightarrow x+1>2x-1\Leftrightarrow x-2<0\Leftrightarrow x<2$
Kết hợp $\left( * \right)$ $\Rightarrow S=\left( \dfrac{1}{2};2 \right)$
& x+1>0 \\
& 2x-1>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>-1 \\
& x>\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow x>\dfrac{1}{2}\left( * \right)$
${{\log }_{\dfrac{1}{2}}}\left( x+1 \right)<{{\log }_{\dfrac{1}{2}}}\left( 2x-1 \right)\Leftrightarrow x+1>2x-1\Leftrightarrow x-2<0\Leftrightarrow x<2$
Kết hợp $\left( * \right)$ $\Rightarrow S=\left( \dfrac{1}{2};2 \right)$
Đáp án C.