√3+2x/4-x>-√3 решите

$\frac{ \sqrt{3} +2x}{4-x}\ \textgreater \ - \sqrt{3} \\ \\ \frac{ \sqrt{3} +2x}{4-x}+\sqrt{3}\ \textgreater \ 0 \\ \\ \frac{ \sqrt{3} +2x+ \sqrt{3}\cdot(4-x) }{4-x}\ \textgreater \ 0 \\ \\ \frac{ \sqrt{3} +2x+ 4\sqrt{3}-x\cdot \sqrt{3} }{4-x}\ \textgreater \ 0 \\ \\ \frac{ 5\sqrt{3} +x\cdot (2-\sqrt{3}) }{4-x}\ \textgreater \ 0$

$\left \{ {{ 5\sqrt{3}+x\cdot(2- \sqrt{3})\ \textgreater \ 0 } \atop {4-x\ \textgreater \ 0}} \right.$

$- \frac{5 \sqrt{3}}{2- \sqrt{3} }\ \textless \ x\ \textless \ 4$

или

$\left \{ {{ 5\sqrt{3}+x\cdot(2- \sqrt{3})\ \textless \ 0 } \atop {4-x\ \textless \ 0}} \right.$

система не имеет решения

Ответ. $- \frac{5 \sqrt{3}}{2- \sqrt{3} }\ \textless \ x\ \textless \ 4$