\({\left( {x-2} \right)^4} + {\left( {x-3} \right)^4} = 1.\)
Пусть \(x-\dfrac{5}{2} = t\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = t + \dfrac{5}{2}\). Тогда уравнение примет вид:
\({\left( {t + \dfrac{1}{2}} \right)^4} + {\left( {t-\dfrac{1}{2}} \right)^4} = 1\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{t^4} + 2{t^3} + \dfrac{3}{2}{t^2} + \dfrac{1}{2}t + \dfrac{1}{{16}} + {t^4}-2{t^3} + \dfrac{3}{2}{t^2}-\dfrac{1}{2}t + \dfrac{1}{{16}} = 1\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,2{t^4} + 3{t^2}-\dfrac{7}{8} = 0.\)
Получили биквадратное уравнение. Пусть \({t^2} = y\), где \(y \ge 0\). Тогда:
\(2{y^2} + 3y-\frac{7}{8} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{y = \dfrac{1}{4},\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{y = -\dfrac{7}{4} < 0.}\end{array}} \right.\)
Вернёмся к переменной \(t\):
\({t^2} = \dfrac{1}{4}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = \dfrac{1}{2},\,\,\,\,}\\{t = -\dfrac{1}{2}.}\end{array}} \right.\)
Вернёмся к переменной
\(\left[ {\begin{array}{*{20}{c}}{x-\dfrac{5}{2} = \dfrac{1}{2},\,\,}\\{x-\dfrac{5}{2} = -\dfrac{1}{2}}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,}\\{x = 2.}\end{array}} \right.\)
Ответ: 2; 3.