\({\left( {x-5} \right)^4}-3{\left( {x-5} \right)^2}-4 = 0.\)
Пусть \({\left( {x-5} \right)^2} = t\), где \(t \ge 0\). Тогда уравнение примет вид:
\({t^2}-3t-4 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 4,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t = -1 < 0.}\end{array}} \right.\)
Возвращаясь к прежней неизвестной, получим:
\({\left( {x-5} \right)^2} = 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-5 = 2,}\\{x-5 = -2}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 7,}\\{x = 3.}\end{array}} \right.} \right.\)
Ответ: 3; 7.