ĐKXĐ: \(\left\{{}\begin{matrix}\frac{3x-2}{x+1}\ge0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-2\ge0\\x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-2< 0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\\x\ne-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\frac{2}{3}\\x< -1\end{matrix}\right.\\x\ne-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\frac{2}{3}\\x< -1\end{matrix}\right.\)
Ta có: \(\sqrt{\frac{3x-2}{x+1}}=3\)
\(\Leftrightarrow\sqrt{\frac{3x-2}{x+1}}=\sqrt{9}\)
\(\Leftrightarrow\frac{3x-2}{x+1}=9\)
\(\Leftrightarrow\frac{3x-2}{x+1}=\frac{9\left(x+1\right)}{x+1}\)
Suy ra: \(3x-2=9\left(x+1\right)\)
\(\Leftrightarrow3x-2-9x-9=0\)
\(\Leftrightarrow-6x-11=0\)
\(\Leftrightarrow-6x=11\)
hay \(x=\frac{-11}{6}\)(nhận)
Vậy: \(S=\left\{-\frac{11}{6}\right\}\)
Ỏooo :<
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