Đáp án:
\(\left[ \begin{array}{l}
y = 1\\
y = - \dfrac{2}{7}\\
y = \sqrt {\dfrac{3}{7}} \\
y = - \sqrt {\dfrac{3}{7}}
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = - \dfrac{{13}}{7}\\
x = - \dfrac{{2\sqrt {21} }}{7}\\
x = \dfrac{{2\sqrt {21} }}{7}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - xy + x - 6{y^2} + 2y = 0\\
{x^2} - xy + {y^2} = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x - 6{y^2} + 2y - {y^2} = - 3\\
{x^2} - xy + {y^2} = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - 3 - 2y + 7{y^2}\\
{\left( { - 3 - 2y + 7{y^2}} \right)^2} - \left( { - 3 - 2y + 7{y^2}} \right).y + {y^2} = 3\left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to 9 + 4{y^2} + 49{y^4} + 12y - 28{y^3} - 42{y^2} + 3y + 2{y^2} - 7{y^3} + {y^2} = 3\\
\to 49{y^4} - 35{y^3} - 35{y^2} + 15y + 6 = 0\\
\to 49{y^4} - 49{y^3} + 14{y^3} - 14{y^2} - 21{y^2} + 21y - 6y + 6 = 0\\
\to 49{y^3}\left( {y - 1} \right) + 14{y^2}\left( {y - 1} \right) - 21y\left( {y - 1} \right) - 6\left( {y - 1} \right) = 0\\
\to \left[ \begin{array}{l}
y - 1 = 0\\
49{y^3} + 14{y^2} - 21y - 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = 1\\
7{y^2}\left( {7y + 2} \right) - 3\left( {7y + 2} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = 1\\
7y + 2 = 0\\
7{y^2} - 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = 1\\
y = - \dfrac{2}{7}\\
y = \sqrt {\dfrac{3}{7}} \\
y = - \sqrt {\dfrac{3}{7}}
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = - \dfrac{{13}}{7}\\
x = - \dfrac{{2\sqrt {21} }}{7}\\
x = \dfrac{{2\sqrt {21} }}{7}
\end{array} \right.
\end{array}\)
