Решите уравнение: [latex] log_{2} x^{2} (3x+1)-3 log_{1/2} \frac{4}{3x+1} = \frac{2}{7} ^{log_{2/7}1,5+log_{2/7} 4 } [/latex]

[latex]log_2(x^2*(3x+1))-3*log_{1/2}( \frac{4}{3x+1})= \frac{2}{7}^{log_{2*7}1.5+log_{2/7}4 [/latex] [latex]ODZ: \left \{ {{(3x+1)\ \textgreater \ 0} \atop {3x+1 \neq 0}} \right. \left \{ {{x\ \textgreater \ -1/3} \atop {x \neq -1/3}} \right. (-1/3: +oo) [/latex] [latex]log_2x^2+log_2(3x+1)-3Log_{1/2^{-1}} \frac{4}{3x+1}= \frac{2}{7}^{log_{2/7}1.5*4 [/latex] [latex]log_2 x^2+log_2(3x+1)+3(log_2 \frac{4}{3x+1})= \frac{2}{7}^{log_{2/7}6 [/latex] [latex]log_2x^2+log_2(3x+1)+3log_24-3log_2(3x+1)=6[/latex] [latex]log_2x^2+log_2(3x+1)+6-3log_2(3x+1)=6[/latex] [latex]log_2x^2-2log(3x+1)=0[/latex] [latex]log_2x^2-log_2(3x+1)^2=0[/latex] [latex]log_2( \frac{x^2}{(3x+1)^2})=0 [/latex] [latex] \frac{x^2}{(3x+1)^2}=2^0 [/latex] [latex] \frac{x^2}{(3x+1)^2}=1 [/latex] [latex]x^2=(3x+1)^2 x^2=9x^2+6x+1 8x^2+6x+1=0 D=36-32=4=2^2 x= (-6-2)/16=-8/16=-1/2 x= (-6+2)/16=-4/16=-1/4[/latex] [latex]x= -1/2 \neq (-1/3; +oo)[/latex]  не подходит Ответ [latex]x= - \frac{1}{4} [/latex]