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