\({\left( {x + 3} \right)^4} + {\left( {x + 5} \right)^4} = 16.\)
Пусть \(x + 4 = t\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = t-4\). Тогда уравнение примет вид:
\({\left( {t-1} \right)^4} + {\left( {t + 1} \right)^4} = 16\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\)
\( \Leftrightarrow \,\,\,\,\,\,\,{t^4}-4{t^3} + 6{t^2}-4t + 1 + {t^4} + 4{t^3} + 6{t^2} + 4t + 1 = 16\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,2{t^4} + 12{t^2}-14 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{t^4} + 6{t^2}-7 = 0.\)
Получили биквадратное уравнение. Пусть \({t^2} = y\), где \(y \ge 0\). Тогда:
\({y^2} + 6y-7 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{y = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{y = -7 < 0.}\end{array}} \right.\)
Вернёмся к переменной \(t\):
\({t^2} = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,\,\,}\\{t = -1.}\end{array}} \right.\)
Вернёмся к переменной
\(\left[ {\begin{array}{*{20}{c}}{x + 4 = 1,\,\,}\\{x + 4 = -1}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -3,}\\{x = -5.}\end{array}} \right.\)
Ответ: – 5; – 3.