Giải hệ phương trình \(\left\{ \begin{array}{l}{x^2} + {y^2} + 4x + 2y = 0\\{x^2} + 7{y^2} - 4xy + 6y = 13\end{array} \right.\)

Lời giải:

\(\left\{ \begin{array}{l}{x^2} + {y^2} + 4x + 2y = 0\\{x^2} + 7{y^2} - 4xy + 6y = 13\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}{\left( {x + 2} \right)^2} + {\left( {y + 1} \right)^2} = 8\\{\left( {x - 2y} \right)^2} + 3{\left( {y + 1} \right)^2} = 16\end{array} \right.\)

⇔ (x – 2y)² + 3(y + 1)² = 2(x + 2)² + 2(y + 1)²

⇔ (x – 2y)² – (x + 2)² = (x + 2)² – (y + 1)²

⇔ (2x – 2y + 2)(-2y – 2) = (x – y + 1)(x + y + 3)

⇔ (x – y + 1)(-4y – 4) = (x – y + 1)(x + y + 3)

⇔ (x – y + 1)(-5y – x – 7) = 0

⇔ \(\left[ \begin{array}{l}y = x + 1\\x = - 5y - 7\end{array} \right.\)

Với y = x + 1 ta có:

(x + 1)² + x² + 4x + 2(x + 1) – 3 = 0

⇔ 2x² + 2x + 1 + 4x + 2x + 2 – 3 = 0

⇔ 2x² + 8x = 0

⇔ 2x(x + 4) = 0

⇔ \(\left[ \begin{array}{l}x = 0\\x = - 4\end{array} \right.\)

Suy ra: \(\left[ \begin{array}{l}y = 1\\y = - 3\end{array} \right.\)

+ Với x = –5y – 7 ta có:

(-5y – 7)² + y² + 4(-5y – 7) + 2y – 3 = 0

⇔ 25y² + 70y + 49 + y² – 20y – 28 + 2y – 3 = 0

⇔ 26y² + 52y + 18 = 0

⇔ \[y = \frac{{ - 13 \pm 2\sqrt {13} }}{{13}}\]

Suy ra: \[x = \frac{{ - 26 \mp 10\sqrt {13} }}{{13}}\]

Vậy (x,y)

= {(0,1);(−4,−3); \[\left( {\frac{{ - 26 - 10\sqrt {13} }}{{13}};\frac{{ - 13 + 2\sqrt {13} }}{{13}}} \right),\left( {\frac{{ - 26 + 10\sqrt {13} }}{{13}};\frac{{ - 13 - 2\sqrt {13} }}{{13}}} \right)\]