$\frac{2}{x+1}$=$4-m$
⇒ $x+1$= $\frac{2}{4-m}$
⇔ x= $\frac{2}{4-m}-1$
⇔ x= $\frac{m-2}{4-m}$
Để x> 0 thì $\frac{m-2}{4-m}$> 0
⇔ \(\left[ \begin{array}{l}\left \{ {{m-2>0} \atop {4-m>0}} \right.\\\left \{ {{m-2<0} \atop {4-m<0}} \right.\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\left \{ {{m>2} \atop {4<m}} \right.\\\left \{ {{m<2} \atop {m>4}} \right.\end{array} \right.\)
⇒ m∈ ( 2; 4)