\(\frac{-3}{x+2}< \frac{2}{3-x}\)
⇔\(\frac{-3\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}< \frac{2\left(x+2\right)}{\left(3-x\right)\left(x+2\right)}\)
➞\(-9+3x< 2x+4\)
⇔\(3x-2< 4+9\)
⇔ \(x< 13\)
\(\frac{-3}{x+2}\)<\(\frac{2}{3-x}\)
⇔ \(\frac{-3\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}< \frac{2\left(x+2\right)}{\left(3-x\right)\left(x+2\right)}\)
➞ \(-9+3x< 2x+4\)
⇔ \(3x-2x< 4+9\)
⇔ \(x< 13\)