\({\left( {x-1} \right)^5} + {\left( {x + 3} \right)^5} = 242\left( {x + 1} \right).\)
Пусть \(x + 1 = t\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x = t-1\). Тогда уравнение примет вид:
\({\left( {t-2} \right)^5} + {\left( {t + 2} \right)^5} = 242t\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,{t^5}-10{t^4} + 40{t^3}-80{t^2} + 80t-32 + {t^5} + 10{t^4} + 40{t^3} + 80{t^2} + 80t + 32 = 242t\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,2{t^5} + 80{t^3}-82t = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2t\left( {{t^4} + 40{t^2}-41} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{t^4} + 40{t^2}-41 = 0.}\end{array}} \right.\)
Уравнение \({t^4} + 40{t^2}-41 = 0\) является биквадратным. Пусть \({t^2} = y,\) где \(y \ge 0.\) Тогда:
\({y^2} + 40y-41 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{y = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{y = -41 < 0.}\end{array}} \right.\)
Возвращаясь к прежней переменной t, получим: \({t^2} = 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,\,\,}\\{t = -1.}\end{array}} \right.\)
Возвращаясь к прежней переменной x, получим:
\(\left[ {\begin{array}{*{20}{c}}{x + 1 = 0,}\\{x + 1 = 1,}\\{x + 1 = -1}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = 0,\,\,}\\{x = -2.}\end{array}} \right.\)
Ответ: \(-2;\,-1;\,\,\,0\).