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Câu hỏi: Tìm tập nghiệm $S$ của bất phương trình ${{\log }_{2}}\left( 3x-2 \right)+{{\log }_{\dfrac{1}{2}}}\left( 6-5x \right)>0$ là
A. $S=\left( 1;+\infty \right).$
B. $S=\left( \dfrac{2}{3};1 \right).$
C. $S=\left( 1;\dfrac{6}{5} \right).$
D. $S=\left( 1;\dfrac{6}{5} \right].$
A. $S=\left( 1;+\infty \right).$
B. $S=\left( \dfrac{2}{3};1 \right).$
C. $S=\left( 1;\dfrac{6}{5} \right).$
D. $S=\left( 1;\dfrac{6}{5} \right].$
$\begin{aligned}
& \centerdot BPT\Leftrightarrow {{\log }_{2}}\left( 3x-2 \right)>{{\log }_{2}}\left( 6-5x \right) \\
& \Leftrightarrow \left\{ \begin{aligned}
& 3x-2>6-5x \\
& 6-5x>0 \\
\end{aligned} \right. \\
& \Leftrightarrow 1<x<\dfrac{6}{5}\cdot \\
\end{aligned}$
Vậy $S=\left( 1;\dfrac{6}{5} \right)$.
& \centerdot BPT\Leftrightarrow {{\log }_{2}}\left( 3x-2 \right)>{{\log }_{2}}\left( 6-5x \right) \\
& \Leftrightarrow \left\{ \begin{aligned}
& 3x-2>6-5x \\
& 6-5x>0 \\
\end{aligned} \right. \\
& \Leftrightarrow 1<x<\dfrac{6}{5}\cdot \\
\end{aligned}$
Vậy $S=\left( 1;\dfrac{6}{5} \right)$.
Đáp án C.
