[tex]|x+2|\cdot(|x+3|-2)\ \textless \ 0 \\ \\ |x+2|\ \textgreater \ 0, \forall x\in \mathbb_{R} \quad $ \ deci, trebuie sa ne legam doar de $ |x+3|-2 \\ \\ |x+3|-2 \ \textless \ 0 \Rightarrow |x+3| \ \textless \ 2\Rightarrow -2\ \textless \ x+3\ \textless \ 2\Big{|}-3 \\ \Rightarrow -5\ \textless \ x\ \textless \ -1 \Rightarrow x\in(-5,-1)[/tex]
