Question

как решить? можно поподробней

Answer 1

$\frac{3}{2+x}\ \textgreater \ 2\\\\\frac{3}{2+x}-2\ \textgreater \ 0\\\\\frac{3-4-2x}{2+x}\ \textgreater \ 0\\\\\frac{-1-2x}{2+x}\ \textgreater \ 0\\\\ \frac{-(2x+1)}{2+x}\ \textgreater \ 0\\\\\frac{2x+1}{x+2}\ \textless \ 0\\\\+++(-2)---(-\frac{1}{2})+++\\\\x\in (-2,-\frac{1}{2})$