Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x\, \in \,\left( {0;1} \right) \cup \left( {1; + \infty } \right),}\\{y\, \in \,\left( {0;1} \right) \cup \left( {1; + \infty } \right).}\end{array}} \right.\)
Рассмотрим первое уравнение:
\({\log _y}x + {\log _x}y = 2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _y}x + \dfrac{1}{{{{\log }_y}x}} = 2.\)
Пусть \({\log _y}x = t.\) Тогда:
\(t + \dfrac{1}{t} = 2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{t^2}-2t + 1 = 0,}\\{t \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,t = 1.\)
Вернёмся к прежним переменным:
\({\log _y}x = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,y = x.\)
\(\left\{ {\begin{array}{*{20}{c}}{y = x,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-y = 20}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-x-20 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = x,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{x = 5,\,\,\,}\\{x = -4}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = 5,}\\{y = 5,}\end{array}\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = -4,}\\{y = -4.}\end{array}} \right.}\end{array}} \right.\)
Решение \(\left( {-4;-4} \right)\) не удовлетворяет ОДЗ.
Ответ: \(\left( {5;5} \right).\)