10 баллов. решить неравенство. log1/2(2x+4) больше log1/2(x+1);

[latex]log_{_\frac{1}{2}}(2x+4)\ \textgreater \ log_{_\frac{1}{2}}(x+1) \ \ \ \ \ \ \ \ \ ODZ: \left \{ {{2x+4\ \textgreater \ 0} \atop {x+1\ \textgreater \ 0}} \right. \left \{ {{x\ \ \textgreater \ \ -2} \atop {x\ \textgreater \ -1}} \right. x\ \textgreater \ -1 \\f(t)=log_{_\frac{1}{2}}t\ -\ spadayushaya\ funkziya,\ poetomu: \\2x+4\ \textless \ x+1 \\x\ \textless \ -3\ -\ ne\ vhodit\ v\ ODZ \\OTBET: neravenstvo\ ne \ imeet\ resheniy[/latex]