\(\left( {{{\log }_3}\dfrac{3}{x}} \right){\log _2}x-{\log _3}\dfrac{{{x^3}}}{{\sqrt 3 }} = \dfrac{1}{2} + {\log _2}\sqrt x .\)
Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{\dfrac{3}{x} > 0,}\\{x > 0}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,x > 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {0;\infty } \right).} \right.\)
\(\left( {{{\log }_3}3-{{\log }_3}x} \right){\log _2}x-\left( {{{\log }_3}{x^3}-{{\log }_3}\sqrt 3 } \right) = \dfrac{1}{2} + \dfrac{1}{2}{\log _2}x\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\log _2}x-{\log _3}x \cdot {\log _2}x-3{\log _3}x + \dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}{\log _2}x = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\log _2}x-2{\log _3}x \cdot {\log _2}x-6{\log _3}x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{{\log }_3}x}}{{{{\log }_3}2}}-2{\log _3}x \cdot {\log _2}x-6{\log _3}x = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\log _3}x\left( {\dfrac{1}{{{{\log }_3}2}}-2{{\log }_2}x-6} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\log }_3}x = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\log }_2}3-2{{\log }_2}x-6 = 0}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\log }_2}x = \dfrac{1}{2}{{\log }_2}3-{{\log }_2}8}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.\,\left[ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,\,\,}\\{x = \dfrac{{\sqrt 3 }}{8}.}\end{array}} \right.\)
Ответ: \(1;\,\,\,\dfrac{{\sqrt 3 }}{8}\).