Решим первое неравенство:
\({\log _2}\dfrac{{3x-2}}{{x-1}} + {\log _2}\dfrac{{{{\left( {x-1} \right)}^3}}}{{3x-2}} < 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{{3x-2}}{{x-1}} > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}\dfrac{{{{\left( {x-1} \right)}^3}}}{{3x-2}} > 0,\\{\log _2}\left( {\dfrac{{3x-2}}{{x-1}} \cdot \dfrac{{{{\left( {x-1} \right)}^3}}}{{3x-2}}} \right) < {\log _2}2\end{array}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\dfrac{{3x-2}}{{x-1}} > 0,\\\dfrac{{{{\left( {x-1} \right)}^3}}}{{3x-2}} > 0,\\{\left( {x-1} \right)^2} < 2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x \in \left( {-\infty ;\dfrac{2}{3}} \right) \cup \left( {1;\,\infty } \right),\\x \in \left( {1-\sqrt 2 ;\,1 + \sqrt 2 } \right)\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,\left( {1-\sqrt 2 ;\dfrac{2}{3}} \right) \cup \left( {1;1 + \sqrt 2 } \right).\)
Решим второе неравенство:
\(\dfrac{{\sqrt {8-2x-{x^2}} }}{{2x + 9}}-\dfrac{{\sqrt {8-2x-{x^2}} }}{{x + 10}} \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{\sqrt {8-2x-{x^2}} \left( {x + 10-2x-9} \right)}}{{\left( {2x + 9} \right)\left( {x + 10} \right)}} \ge 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\dfrac{{\sqrt {8-2x-{x^2}} \left( {1-x} \right)}}{{\left( {2x + 9} \right)\left( {x + 10} \right)}} \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}8-2x-{x^2} = 0,\\\left\{ \begin{array}{l}8-2x-{x^2} \ge 0,\\\dfrac{{1-x}}{{\left( {2x + 9} \right)\left( {x + 10} \right)}} \ge 0\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\left[ \begin{array}{l}x = -4,\\x = 2,\\\left\{ \begin{array}{l}x \in \left[ {-4;2} \right],\\x \in \left( {-\infty ;-10} \right) \cup \,\left( {-4,5;1} \right]\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,\left[ {-4;1} \right] \cup \left\{ 2 \right\}.\)
Найдём общее решение:
Таким образом, решение исходной системы неравенств: \(x \in \left( {1-\sqrt 2 ;\dfrac{2}{3}} \right) \cup \left\{ {-2} \right\}.\)
Ответ: \(\left( {1-\sqrt 2 ;\dfrac{2}{3}} \right) \cup \left\{ {-2} \right\}.\)