\({\left( {\sqrt {2 + \sqrt 3 } } \right)^x} + {\left( {\sqrt {2-\sqrt 3 } } \right)^x} = 4.\)
Заметим, что:
\(2-\sqrt 3 = \dfrac{{\left( {2-\sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}}{{2 + \sqrt 3 }} = \dfrac{{4-3}}{{2 + \sqrt 3 }} = \dfrac{1}{{2 + \sqrt 3 }}.\)
Тогда исходное уравнение примет вид:
\({\left( {\sqrt {2 + \sqrt 3 } } \right)^x} + \dfrac{1}{{{{\left( {\sqrt {2 + \sqrt 3 } } \right)}^x}}} = 4.\)
Пусть \({\left( {\sqrt {2 + \sqrt 3 } } \right)^x} = t\), где \(t > 0\). Тогда:
\(t + \dfrac{1}{t} = 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{t^2}-4t + 1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 2-\sqrt 3 ,}\\{t = 2 + \sqrt 3 .}\end{array}} \right.\)
Возвращаясь к прежней переменной, получим:
\(\left[ {\begin{array}{*{20}{c}}{{{\left( {\sqrt {2 + \sqrt 3 } } \right)}^x} = 2-\sqrt 3 ,}\\{{{\left( {\sqrt {2 + \sqrt 3 } } \right)}^x} = 2 + \sqrt 3 \,}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\,} \right.\left[ {\begin{array}{*{20}{c}}{{{\left( {2 + \sqrt 3 } \right)}^{\frac{x}{2}}} = {{\left( {2 + \sqrt 3 } \right)}^{-1}}}\\{{{\left( {2 + \sqrt 3 } \right)}^{\frac{x}{2}}} = {{\left( {2 + \sqrt 3 } \right)}^1}}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\dfrac{x}{2} = -1,}\\{\dfrac{x}{2} = 1\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -2,}\\{x = 2.\,\,\,}\end{array}} \right.} \right.\)
Ответ: \( \pm 2\).