Задача 27. Решите систему уравнений: \(\left\{ \begin{array}{l}2{x^2} + 3{y^2} = 11,\\4{x^2} + 6{y^2} = 11x.\end{array} \right.\)
Ответ
ОТВЕТ: \(\left( {2;\,-1} \right),\,\,\,\;\left( {2;\,1} \right).\)
Решение
Разделим обе части второго уравнения на 2. Тогда система уравнений примет вид:
\(\left\{ \begin{array}{l}2{x^2} + 3{y^2} = 11,\\2{x^2} + 3{y^2} = \dfrac{{11x}}{2}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\dfrac{{11x}}{2} = 11,\\2{x^2} + 3{y^2} = 11\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 2,\\2{x^2} + 3{y^2} = 11\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 2,\\2 \cdot {2^2} + 3{y^2} = 11\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 2,\\{y^2} = 1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 2,\\\left[ \begin{array}{l}y = -1,\\y = 1\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ \begin{array}{l}x = 2,\\y = -1,\end{array} \right.\\\left\{ \begin{array}{l}x = 2,\\y = 1.\end{array} \right.\end{array} \right.\)
Ответ: \(\left( {2;\,-1} \right),\,\,\,\;\left( {2;\,1} \right).\)