Задача 25. Решите систему уравнений: \(\left\{ \begin{array}{l}3{x^2} + 2{y^2} = 50,\\12{x^2} + 8{y^2} = 50x.\end{array} \right.\)
Ответ
ОТВЕТ: \(\left( {4;\,-1} \right),\,\,\,\;\left( {4;\,1} \right).\)
Решение
Разделим обе части второго уравнения на 4. Тогда система уравнений примет вид:
\(\left\{ \begin{array}{l}3{x^2} + 2{y^2} = 50,\\3{x^2} + 2{y^2} = \dfrac{{25x}}{2}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\dfrac{{25x}}{2} = 50,\\3{x^2} + 2{y^2} = 50\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 4,\\3{x^2} + 2{y^2} = 50\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 4,\\3 \cdot {4^2} + 2{y^2} = 50\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 4,\\{y^2} = 1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 4,\\\left[ \begin{array}{l}y = -1,\\y = 1\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ \begin{array}{l}x = 4,\\y = -1,\end{array} \right.\\\left\{ \begin{array}{l}x = 4,\\y = 1.\end{array} \right.\end{array} \right.\)
Ответ: \(\left( {4;\,-1} \right),\,\,\,\;\left( {4;\,1} \right).\)