ĐKXĐ:\(x\ne2\)
\(3x-\dfrac{1}{x-2}=\dfrac{x-1}{2-x}\\ \Leftrightarrow\dfrac{3x\left(x-2\right)}{x-2}-\dfrac{1}{x-2}=\dfrac{-\left(x-1\right)}{x-2}\\ \Leftrightarrow\dfrac{3x^2-6x}{x-2}-\dfrac{1}{x-2}+\dfrac{x-1}{x-2}=0\\ \Leftrightarrow\dfrac{3x^2-6x-1+x-1}{x-2}=0\\ \Rightarrow3x^2-5x-2=0\\ \Leftrightarrow\left(3x^2-6x\right)+\left(x-2\right)=0\\ \Leftrightarrow3x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=\dfrac{-1}{3}\end{matrix}\right.\)
\(ĐK:x\ne2\)
\(\Rightarrow3x=\dfrac{1}{x-2}-\dfrac{x-1}{x-2}\)
\(\Leftrightarrow\dfrac{3x\left(x-2\right)}{x-2}=\dfrac{1-x+1}{x-2}\)
\(\Leftrightarrow3x\left(x-2\right)=1-x+1\)
\(\Leftrightarrow3x^2-6x=2-x\)
\(\Leftrightarrow3x^2-5x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)
