ОДЗ :
0\\\\(2+x)(2-x)<0" alt="\frac{2+x}{2-x} >0\\\\(2+x)(2-x)<0" align="absmiddle" class="latex-formula">
+ - +
__________₀__________₀_________
- 2 2
///////////////////////
x ∈ (- 2 , 2)
, значит
0\\\\(x-\frac{2}{3})(x-2)>0" alt="\frac{2+x}{2-x}<2\\\\\frac{2+x}{2-x}-2<0\\\\\frac{2+x-4+2x}{2-x}<0\\\\\frac{3x-2}{2-x}<0\\\\3(x-\frac{2}{3})(x-2)>0\\\\(x-\frac{2}{3})(x-2)>0" align="absmiddle" class="latex-formula">
+ - +
_____________₀______________₀_____________
2/3 2
x ∈ (- ∞ ; 2/3) ∪ (2 ; + ∞)
С учётом ОДЗ, ответ : x ∈ (- 2 ; 2/3)