\(\dfrac{4-3x}{x-1}< 2\)
\(\Leftrightarrow4-3x< 2\left(x-1\right)\)
\(\Leftrightarrow4-3x< 2x-2\)
\(\Leftrightarrow2x+3x>4+2\)
\(\Leftrightarrow5x>6\)
\(\Leftrightarrow x>\dfrac{6}{5}\)
Vậy: \(x>\dfrac{6}{5}\)
\(\dfrac{4-3x}{x-1}< 2\left(ĐK:x>1\right)\\\Rightarrow\dfrac{4-3x}{x-1}-2< 0\\ \Rightarrow\dfrac{4-3x}{x-1}-\dfrac{2x-2}{x-1}< 0 \\ \Rightarrow\dfrac{6-5x}{x-1}< 0\\ \Rightarrow \left[{}\begin{matrix}6-5x< 0\\x-1< 0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-5x< -6\\x< 1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x>\dfrac{6}{5}\\x< 1\end{matrix}\right.\)
