Log2(3x-1)- log2(5x+1) меньше log2(x-1)-2

[latex] log_{2} (3x-1)- log_{2}(5x+1)\ \textless \ log_{2} (x-1)-2 [/latex] ОДЗ: [latex] \left \{ {{3x-1\ \textgreater \ 0} \atop {5x+1}\ \textgreater \ 0} \atop {x-1}\ \textgreater \ 0\right. [/latex] [latex] \left \{ {{x\ \textgreater \ \frac{1}{3} } \atop {x}\ \textgreater \ - \frac{1}{5} } \atop {x}\ \textgreater \ 1\right. [/latex] [latex]x[/latex] ∈ [latex](1;+[/latex] ∞ [latex])[/latex] [latex] log_{2} (3x-1)- log_{2}(5x+1)\ \textless \ log_{2} (x-1)- log_{2} 4[/latex] [latex] log_{2} (3x-1)+ log_{2} 4\ \textless \ log_{2} (x-1)+ log_{2}(5x+1)[/latex] [latex] log_{2}(4 (3x-1))\ \textless \ log_{2} ((x-1)(5x+1))[/latex] [latex] log_{2}(12x-4)\ \textless \ log_{2} (5 x^{2} +x-5x-1)[/latex] [latex] log_{2}(12x-4)\ \textless \ log_{2} (5 x^{2} -4x-1)[/latex] [latex]12x-4\ \textless \ 5 x^{2} -4x-1[/latex] [latex]-5x^2+16x-3\ \textless \ 0[/latex] [latex]5x^2-16x+3\ \textgreater \ 0[/latex] [latex]D=(-16)^2-4*5*3=256-60=196=14^2[/latex] [latex]x_1= \frac{16+14}{10}=3 [/latex] [latex]x_2= \frac{16-14}{10}=0.2 [/latex]        +                     -               + ---------------(0.2)----------(3)------------ ////////////////                       /////////////// С учётом ОДЗ получаем Ответ: [latex](3;+[/latex] ∞ [latex])[/latex]