Задача 4. Решите систему уравнений \(\left\{ {\begin{array}{*{20}{c}} {{2^{\frac{{x-y}}{2}}} + {2^{\frac{{y-x}}{2}}} = 2,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;} \\ {\lg \left( {2x-y} \right) + 1 = \lg \left( {y + 2x} \right) + \lg 6} \end{array}} \right.\)
Ответ
ОТВЕТ: \(\left( {4;\,2} \right).\)
Решение
\(\left\{ {\begin{array}{*{20}{c}}{{2^{\frac{{x-y}}{2}}} + {2^{\frac{{y-x}}{2}}} = 2,5,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\lg \left( {2x-y} \right) + 1 = \lg \left( {y + 2x} \right) + \lg 6.}\end{array}} \right.\)
Рассмотрим первое уравнение системы: \({2^{\frac{{x-y}}{2}}} + {2^{-\,\frac{{x-y}}{2}}} = 2,5.\)
Пусть \({2^{\frac{{x-y}}{2}}} = t,\,\,\,\,\,\,\,t > 0.\) Тогда:
\(t + \dfrac{1}{t}-\frac{5}{2} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}2{t^2}-5t + 2 = 0,\\t \ne 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}t = \frac{1}{2},\\t = 2.\end{array} \right.\)
Вернёмся к прежним переменным:
\(\left[ \begin{array}{l}{2^{\frac{{x-y}}{2}}} = 2,\\{2^{\frac{{x-y}}{2}}} = {2^{-1}}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\dfrac{{x-y}}{2} = 1,\\\dfrac{{x-y}}{2} = -1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}y = x-2,\\y = x + 2.\end{array} \right.\)
Преобразуем второе уравнение исходной системы:
\(\lg \left( {2x-y} \right) + \lg 10 = \lg \left( {y + 2x} \right) + \lg 6\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\lg \left( {20x-10y} \right) = \lg \left( {6y + 12x} \right)\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}20x-10y = 6y + 12x,\\2x-y > 0,\\y + 2x > 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 2y,\\2x-y > 0,\\y + 2x > 0.\end{array} \right.\)
Таким образом, исходная система равносильна совокупности двух систем:
\(\left[ \begin{array}{l}\left\{ \begin{array}{l}y = x-2,\\x = 2y,\\2x-y > 0,\\y + 2x > 0\end{array} \right.\\\left\{ \begin{array}{l}y = x + 2,\\x = 2y,\\2x-y > 0,\\y + 2x > 0\end{array} \right.\end{array} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ \begin{array}{l}\left\{ \begin{array}{l}y = 2y-2,\\x = 2y,\\2x-y > 0,\\y + 2x > 0\end{array} \right.\\\left\{ \begin{array}{l}y = 2y + 2,\\x = 2y,\\2x-y > 0,\\y + 2x > 0\end{array} \right.\end{array} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ \begin{array}{l}\left\{ \begin{array}{l}x = 4,\\y = 2,\\2x-y > 0,\\y + 2x > 0\end{array} \right.\\\left\{ \begin{array}{l}x = -4,\\y = -2,\\2x-y > 0,\\y + 2x > 0\end{array} \right.\end{array} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ \begin{array}{l}x = 4,\\y = 2.\end{array} \right.\)
Ответ: \(\left( {4;2} \right).\)