\(\left\{ {\begin{array}{*{20}{c}}{y \cdot {x^{{{\log }_y}x}} = {x^{\frac{5}{2}}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\log }_4}y \cdot {{\log }_y}\left( {y-3x} \right) = 1.}\end{array}} \right.\)
Запишем ОДЗ: \(\left\{ \begin{array}{l}x > 0,\\y > 0,\\y \ne 1,\\y-3x > 0.\end{array} \right.\)
Прологарифмируем обе части первого уравнения по основанию «x»:
\({\log _x}\left( {y \cdot {x^{{{\log }_y}x}}} \right) = {\log _x}{x^{\frac{5}{2}}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\log _x}y + {\log _x}{x^{{{\log }_y}x}} = \dfrac{5}{2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\log _x}y + {\log _y}x = \dfrac{5}{2}\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,{\log _x}y + \dfrac{1}{{{{\log }_x}y}}-\dfrac{5}{2} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2\log _x^2y-5{\log _x}y + 2 = 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}{\log _x}y = 2,\\{\log _x}y = \dfrac{1}{2}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}y = {x^2},\\y = \sqrt x .\end{array} \right.\)
Преобразуем второе уравнение:
\({\log _4}y \cdot {\log _y}\left( {y-3x} \right) = 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{{{\log }_y}\left( {y-3x} \right)}}{{{{\log }_y}4}} = 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\log _4}\left( {y-3x} \right) = 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = 3x + 4.\)
Таким образом, исходная система равносильна совокупности двух систем:
\(\left[ \begin{array}{l}\left\{ \begin{array}{l}y = {x^2},\\y = 3x + 4,\end{array} \right.\\\left\{ \begin{array}{l}y = \sqrt x ,\\y = 3x + 4.\end{array} \right.\end{array} \right.\)
Рассмотрим первую систему совокупности:
\(\left\{ \begin{array}{l}y = {x^2},\\y = 3x + 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2} = 3x + 4,\\y = 3x + 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-3x-4 = 0,\\y = 3x + 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ \begin{array}{l}x = -1,\\y = 1,\end{array} \right.\\\left\{ \begin{array}{l}x = 4,\\y = 16.\end{array} \right.\end{array} \right.\)
Решение \(\left( {-1;\,1} \right)\) не удовлетворяет ОДЗ.
Рассмотрим вторую систему совокупности:
\(\left\{ \begin{array}{l}y = \sqrt x ,\\y = 3x + 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\sqrt x = 3x + 4,\\y = 3x + 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 9{x^2} + 24x + 16,\\3x + 4 \ge 0,\\y = 3x + 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}9{x^2} + 23x + 16 = 0,\\3x + 4 \ge 0,\\y = 3x + 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\emptyset .\)
Ответ: \(\left( {4;16} \right).\)