$⇔\dfrac{-3}{x+2}-\dfrac{2}{3-x}<0$
$⇔\dfrac{-3(3-x)-2(x+2)}{(x+2)(3-x)}<0$
$⇔\dfrac{x-13}{(x+2)(3-x)}<0$
$⇔$\(\left[ \begin{array}{l}\begin{cases}x-13>0\\(x+2)(3-x)<0\end{cases}\\\begin{cases}x-13<0\\(x+2)(3-x)>0\end{cases}\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}\begin{cases}x>13\\x+2>0\end{cases}\\\begin{cases}x<13\\\left[ \begin{array}{l}\begin{cases}x+2>0\\3-x>0\end{cases}\\\begin{cases}x+2<0\\3-x<0\end{cases}\end{array} \right.\ \end{cases}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x>13\\\begin{cases}x<13\\\left[ \begin{array}{l}\begin{cases}x+2>0\\3-x>0\end{cases}⇒CHỌN\\\begin{cases}x+2<0\\3-x<0\end{cases}⇒LOẠI\end{array} \right.\ \end{cases}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x>13\\\begin{cases}x<13\\-2<x<3\end{cases}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x>13\\-2<x<3\end{array} \right.\)
