ĐK: \(x\ne\dfrac{1}{2};x\ne-\dfrac{1}{3}\)
\(\dfrac{x+2}{3x+1}\ge\dfrac{x-2}{2x-1}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(2x-1\right)-\left(x-2\right)\left(3x+1\right)}{\left(3x+1\right)\left(2x-1\right)}\ge0\)
\(\Leftrightarrow\dfrac{2x^2+3x-2-3x^2+5x+2}{6x^2-x-1}\ge0\)
\(\Leftrightarrow\dfrac{-x^2+8x}{6x^2-x-1}\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x^2+8x\ge0\\6x^2-x-1>0\end{matrix}\right.\left(1\right)\) hoặc \(\left\{{}\begin{matrix}-x^2+8x\le0\\6x^2-x-1< 0\end{matrix}\right.\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}0\le x\le8\\\left[{}\begin{matrix}x>\dfrac{1}{2}\\x< -\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x\le8\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le0\\x\ge8\end{matrix}\right.\\-\dfrac{1}{3}< x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow-\dfrac{1}{3}< x\le0\)
Vậy ...
