Решите пожалуйста неравенства:

Question 1

Question 2

А) ОТВЕТ: $\mathtt{x\in(1;\frac{11}{3})}$

$\mathtt{\log_3^2(4-x)-1\ \textless \ 0;~(\log_3(4-x)-1)(\log_3(4-x)+1)\ \textless \ 0;~}\\\mathtt{-1\ \textless \ \log_3(4-x)\ \textless \ 1;~\frac{1}{3}\ \textless \ 4-x\ \textless \ 3;~-3\ \textless \ x-4\ \textless \ -\frac{1}{3};~1\ \textless \ x\ \textless \ \frac{11}{3}}$

б) ОТВЕТ: $\mathtt{x\in(\frac{1}{125};\frac{1}{25})}$

$\mathtt{\log_{0,2}^2x-5\log_{0,2}x\ \textless \ -6;~\log_5^2x+5\log_5x+6\ \textless \ 0;~}\\\mathtt{(\log_5x+3)(\log_5x+2)\ \textless \ 0;~-3\ \textless \ \log_5x\ \textless \ -2;~\frac{1}{5^3}\ \textless \ x\ \textless \ \frac{1}{5^2}}$

Question 3

(1)
$log_3^2(4-x)\ \textless \ 1\\\\ log_3^2(4-x)-1^2\ \textless \ 0\\\\ (log_3(4-x)+1)*(log_3(4-x)-1)\ \textless \ 0\\\\ -1\ \textless \ log_3(4-x)\ \textless \ 1\\\\ \begin{equation*} \begin{cases} log_3(4-x)\ \textgreater \ -1\\ log_3(4-x)\ \textless \ 1 \end{cases} \end{equation*}\\\\ \begin{equation*} \begin{cases} log_3(4-x)\ \textgreater \ log_3(3^{-1})\\ log_3(4-x)\ \textless \ log_3(3) \end{cases} \end{equation*} \begin{equation*} \begin{cases} 4-x\ \textgreater \ \frac{1}{3}\\ 4-x\ \textless \ 3\\ 4-x\ \textgreater \ 0 \end{cases} \end{equation*}$

$\begin{equation*} \begin{cases} x\ \textless \ \frac{11}{3}\\ x\ \textgreater \ 1\\ x\ \textless \ 4 \end{cases} \end{equation*}$

$x\in(1;\ \frac{11}{3})$
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(2)
$log_{0.2}^2(x)-5*log_{0.2}(x)\ \textless \ -6\\\\ log_{0.2}^2(x)-5*log_{0.2}(x)+6\ \textless \ 0\\\\ (log_{0.2}(x)-2)*(log_{0.2}(x)-3)\ \textless \ 0\\\\ 2\ \textless \ log_{0.2}(x)\ \textless \ 3\\\\ 2\ \textless \ log_{5^{-1}}(x)\ \textless \ 3\\\\ 2\ \textless \ -log_{5}}(x)\ \textless \ 3\\\\ \left \{ {{log_{5}}(x)\ \textless \ -2} \atop {log_{5}}(x)\ \textgreater \ -3}} \right. ;\\\\ \left \{ {{log_{5}}(x)\ \textless \ log_5(5^{-2})} \atop {log_{5}}(x)\ \textgreater \ log_5(5^{-3})}} \right. \\\\ \left \{ {{x\ \textless \ \frac{1}{25}} \atop {x\ \textgreater \ \frac{1}{125}}}\\ \atop {x\ \textgreater \ 0} \right. \\\\ x\in(\frac{1}{125}};\ \frac{1}{25}})$