\(3x-\dfrac{1}{x-2}=\dfrac{x-1}{2-x}\)
\(ĐK:x\ne2\)
\(\Rightarrow\dfrac{3x\left(2-x\right)+1}{2-x}=\dfrac{x-1}{2-x}\)
\(\Leftrightarrow3x\left(2-x\right)+1=x-1\)
\(\Leftrightarrow6x-3x^2+1-x+1=0\)
\(\Leftrightarrow-3x^2+5x+2=0\)
\(\Leftrightarrow3x^2-5x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{3}\right\}\)
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