Задача 30. Решите систему уравнений: \(\left\{ \begin{array}{l}{x^2} + 4{y^2} = 25,\\3{x^2} + 12{y^2} = 25x.\end{array} \right.\)
Ответ
ОТВЕТ: \(\left( {3;\,-2} \right),\,\,\,\;\left( {3;\,2} \right).\)
Решение
Разделим обе части второго уравнения на 3. Тогда система уравнений примет вид:
\(\left\{ \begin{array}{l}{x^2} + 4{y^2} = 25,\\{x^2} + 4{y^2} = \dfrac{{25x}}{3}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\dfrac{{25x}}{3} = 25,\\{x^2} + 4{y^2} = 25\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 3,\\{x^2} + 4{y^2} = 25\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 3,\\{3^2} + 4{y^2} = 25\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 3,\\{y^2} = 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 3,\\\left[ \begin{array}{l}y = -2,\\y = 2\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ \begin{array}{l}x = 3,\\y = -2,\end{array} \right.\\\left\{ \begin{array}{l}x = 3,\\y = 2.\end{array} \right.\end{array} \right.\)
Ответ: \(\left( {3;\,-2} \right),\,\,\,\;\left( {3;\,2} \right).\)