Giúp với mọi người ơi

$\begin{array}{l}
2\log _9^2x = {\log _3}x.{\log _3}\left( {\sqrt {2x + 1}  - 1} \right)\left( 1 \right)\\
ĐK:\left\{ \begin{array}{l}
x > 0\\
\sqrt {2x + 1}  - 1 > 0\\
2x + 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
2x + 1 > 1\\
x \ge  - \frac{1}{2}
\end{array} \right. \Leftrightarrow x > 0\\
\left( 1 \right) \Leftrightarrow 2{\left( {\frac{1}{2}{{\log }_3}x} \right)^2} = {\log _3}x.{\log _3}.\left( {\sqrt {2x + 1}  - 1} \right)\\
 \Leftrightarrow \frac{1}{2}\log _3^2x = {\log _3}x.{\log _3}\left( {\sqrt {2x + 1}  - 1} \right)\\
 \Leftrightarrow {\log _3}x\left( {\frac{1}{2}{{\log }_3}x - {{\log }_3}\left( {\sqrt {2x + 1}  - 1} \right)} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{\log _3}x = 0\\
\frac{1}{2}{\log _3}x - {\log _3}\left( {\sqrt {2x + 1}  - 1} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
{\log _3}\sqrt x  = {\log _3}\left( {\sqrt {2x + 1}  - 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\sqrt x  = \sqrt {2x + 1}  - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\sqrt x  + 1 = \sqrt {2x + 1} 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x + 2\sqrt x  + 1 = 2x + 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x - 2\sqrt x  = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\sqrt x \left( {\sqrt x  - 2} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\sqrt x  = 0\\
\sqrt x  - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 0(L)\\
x = 4
\end{array} \right.\\
 \Rightarrow S = \left\{ {1;4} \right\}
\end{array}$