\({\left( {\sqrt {5\sqrt 2 -7} } \right)^x} + 6{\left( {\sqrt {5\sqrt 2 + 7} } \right)^x} = 7.\)
Заметим, что:
\(5\sqrt 2 + 7 = \dfrac{{\left( {5\sqrt 2 + 7} \right)\left( {5\sqrt 2 -7} \right)}}{{5\sqrt 2 -7}} = \dfrac{{50-49}}{{5\sqrt 2 -7}} = \dfrac{1}{{5\sqrt 2 -7}}.\)
Тогда исходное уравнение примет вид:
\({\left( {\sqrt {5\sqrt 2 -7} } \right)^x} + \dfrac{6}{{{{\left( {\sqrt {5\sqrt 2 -7} } \right)}^x}}} = 7.\)
Пусть \({\left( {\sqrt {5\sqrt 2 -7} } \right)^x} = t\), где \(t > 0\). Тогда:
\(t + \frac{6}{t} = 7\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{t^2}-7t + 6 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,}\\{t = 6.}\end{array}} \right.\)
Возвращаясь к прежней переменной получим:
\(\left[ {\begin{array}{*{20}{c}}{{{\left( {\sqrt {5\sqrt 2 -7} } \right)}^x} = 1,}\\{{{\left( {\sqrt {5\sqrt 2 -7} } \right)}^x} = 6}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\left( {\sqrt {5\sqrt 2 -7} } \right)}^x} = {{\left( {\sqrt {5\sqrt 2 -7} } \right)}^0},\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {\sqrt {5\sqrt 2 -7} } \right)}^x} = {{\left( {\sqrt {5\sqrt 2 -7} } \right)}^{{{\log }_{\sqrt {5\sqrt 2 -7} }}6}}}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.} \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = {{\log }_{\sqrt {5\sqrt 2 -7} }}6}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = \dfrac{2}{{{{\log }_6}\left( {5\sqrt 2 -7} \right)}}.}\end{array}} \right.} \right.\)
Ответ: \(0;\,\,\,\,\,\dfrac{2}{{{{\log }_6}\left( {5\sqrt 2 -7} \right)}}\).