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Câu hỏi: Tìm tập nghiệm S của phương trình ${{\log }_{3}}\left( 2x+1 \right)-{{\log }_{3}}\left( x-1 \right)=1$.
A. $S=\left\{ 1 \right\}$
B. $S=\left\{ -2 \right\}$
C. $S=\left\{ 3 \right\}$
D. $S=\left\{ 4 \right\}$
A. $S=\left\{ 1 \right\}$
B. $S=\left\{ -2 \right\}$
C. $S=\left\{ 3 \right\}$
D. $S=\left\{ 4 \right\}$
ĐK: $\left\{ \begin{aligned}
& 2x+1>0 \\
& x-1>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>-\dfrac{1}{2} \\
& x>1 \\
\end{aligned} \right.\Leftrightarrow x>1$.
Ta có ${{\log }_{3}}\left( 2x+1 \right)-{{\log }_{3}}\left( x-1 \right)=1\Leftrightarrow {{\log }_{3}}\dfrac{2x+1}{x-1}=1\Leftrightarrow \dfrac{2x+1}{x-1}=3\Leftrightarrow x=4$.
& 2x+1>0 \\
& x-1>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>-\dfrac{1}{2} \\
& x>1 \\
\end{aligned} \right.\Leftrightarrow x>1$.
Ta có ${{\log }_{3}}\left( 2x+1 \right)-{{\log }_{3}}\left( x-1 \right)=1\Leftrightarrow {{\log }_{3}}\dfrac{2x+1}{x-1}=1\Leftrightarrow \dfrac{2x+1}{x-1}=3\Leftrightarrow x=4$.
Đáp án D.