ПОМОГИТЕ ПОЖ!!! Решите неравенство: log^2 по основанию 1/6 x>4

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ПОМОГИТЕ ПОЖ!!! Решите неравенство: log^2 по основанию 1/6 x>4


Алгебра (15 баллов) | 236 просмотров
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image4, \\ x>0, \\ \log^2_{\frac{1}{6}}x-4>0, \\ (\log_{\frac{1}{6}}x-2)(\log_{\frac{1}{6}}x+2)>0, " alt="\log^2_{\frac{1}{6}}x>4, \\ x>0, \\ \log^2_{\frac{1}{6}}x-4>0, \\ (\log_{\frac{1}{6}}x-2)(\log_{\frac{1}{6}}x+2)>0, " align="absmiddle" class="latex-formula">
image0,} \atop {\log_{\frac{1}{6}}x+2>0;}} \right.}} \right. \left [ {{ \left \{ {{\log_{\frac{1}{6}}x<2,} \atop {\log_{\frac{1}{6}}x<-2;}} \right. } \atop {\left \{ {{\log_{\frac{1}{6}}x>2,} \atop {\log_{\frac{1}{6}}x>-2;}} \right.}} \right. \left [ {{\log_{\frac{1}{6}}x<-2,} \atop {\log_{\frac{1}{6}}x>2;}} \right. " alt=" \left [ {{ \left \{ {{\log_{\frac{1}{6}}x-2<0,} \atop {\log_{\frac{1}{6}}x+2<0;}} \right. } \atop {\left \{ {{\log_{\frac{1}{6}}x-2>0,} \atop {\log_{\frac{1}{6}}x+2>0;}} \right.}} \right. \left [ {{ \left \{ {{\log_{\frac{1}{6}}x<2,} \atop {\log_{\frac{1}{6}}x<-2;}} \right. } \atop {\left \{ {{\log_{\frac{1}{6}}x>2,} \atop {\log_{\frac{1}{6}}x>-2;}} \right.}} \right. \left [ {{\log_{\frac{1}{6}}x<-2,} \atop {\log_{\frac{1}{6}}x>2;}} \right. " align="absmiddle" class="latex-formula"> 
image(\frac{1}{6})^{-2},} \atop {x<(\frac{1}{6})^{2};}} \right. \left [ {{x>36,} \atop {x<\frac{1}{36};}} \right. \\ x\in(0;\frac{1}{36})\cup(36;+\infty)" alt="\left [ {{x>(\frac{1}{6})^{-2},} \atop {x<(\frac{1}{6})^{2};}} \right. \left [ {{x>36,} \atop {x<\frac{1}{36};}} \right. \\ x\in(0;\frac{1}{36})\cup(36;+\infty)" align="absmiddle" class="latex-formula">
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