\({\left( {x + 2} \right)^4} + 5{\left( {x + 2} \right)^2}-36 = 0.\)
Пусть \({\left( {x + 2} \right)^2} = t\), где \(t \ge 0\). Тогда уравнение примет вид:
\({t^2} + 5t-36 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 4,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t = -9 < 0.}\end{array}} \right.\)
Возвращаясь к прежней неизвестной, получим:
\({\left( {x + 2} \right)^2} = 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x + 2 = 2,}\\{x + 2 = -2}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,\,\,\,}\\{x = -4.}\end{array}} \right.} \right.\)
Ответ: –4; 0.