Giải hệ phương trình: x^2 2y 3=y^2 4x và x^ y^2=5

\(\left\{ \begin{array}{l}{x^2} - {y^2} = 4x - 2y - 3\\{x^2} + {y^2} = 5\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}{x^2} - {y^2} - 4x + 2y + 3 = 0\\{x^2} + {y^2} = 5\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}{\left( {x - 2} \right)^2} - {\left( {y - 1} \right)^2} = 0\\{x^2} + {y^2} = 5\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}\left( {x + y - 3} \right)\left( {x - y - 1} \right) = 0\\{x^2} + {y^2} = 5\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}\left[ \begin{array}{l}x = 3 - y\\x = y + 1\end{array} \right.\\{x^2} + {y^2} = 5\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}{\left( {3 - y} \right)^2} + {y^2} = 5\\{\left( {y + 1} \right)^2} + {y^2} = 5\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}2{y^2} - 6y + 4 = 0\\2{y^2} + 2y - 4 = 0\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}y = 2\\y = 1\\y = 1\\y = - 2\end{array} \right. \Rightarrow \left[ \begin{array}{l}x = 1\\x = 2\\x = 2\\x = - 1\end{array} \right.\)

Vậy (x; y) = {(1;2), (2;1) , (-1; -2)