$$\left\lbrace\begin{array}{l} x^2+y^2-3x+4y=1 \\ 3x^2-2y^2-9x-8y=3 \end{array}\right.$$

Đề bài:

Giải hệ phương trình \begin{equation} \label{eq:42.I} \left\lbrace\begin{array}{l} x^2+y^2-3x+4y=1 \\ 3x^2-2y^2-9x-8y=3 \end{array}\right. \end{equation}

Lời giải:

\begin{align} \eqref{eq:42.I} & \Leftrightarrow \left\lbrace\begin{array}{l} 3x^2+3y^2-9x+12y=3 \\ 3x^2-2y^2-9x-8y=3 \end{array}\right. \nonumber \\ & \Leftrightarrow \left\lbrace\begin{array}{l} x^2+y^2-3x+4y=1 \\ 3x^2+3y^2-9x+12y-3x^2+2y^2+9x+8y=0 \end{array}\right.\nonumber \\ & \Leftrightarrow \left\lbrace\begin{array}{l} x^2+y^2-3x+4y=1 \\ 5y^2+20y=0 \end{array}\right.\nonumber \\ & \Leftrightarrow \left\lbrace\begin{array}{l} x^2+y^2-3x+4y=1 \\ 5y(y+4)=0 \end{array}\right.\nonumber \\ & \Leftrightarrow \left\lbrace\begin{array}{l} x^2+y^2-3x+4y=1 \\ \left[ \begin{array}{l} y=0 \\ y=-4 \end{array} \right. \end{array}\right.\nonumber \\ & \Leftrightarrow \left[\begin{array}{l} \left\lbrace \begin{array}{l} x^2+y^2-3x+4y=1 \\ y=0 \end{array} \right. \\ \left\lbrace \begin{array}{l} x^2+y^2-3x+4y=1 \\ y=-4 \end{array} \right. \end{array}\right.\nonumber \\ & \Leftrightarrow \left[\begin{array}{l} \left\lbrace \begin{array}{l} x^2-3x-1=0 \quad (\Delta=9+4=13>0) \\ y=0 \end{array} \right. \\ \left\lbrace \begin{array}{l} x^2-3x-1=0 \quad (\Delta=9+4=13>0) \\ y=-4 \end{array} \right. \end{array}\right.\nonumber \\ & \Leftrightarrow \left[\begin{array}{l} \left\lbrace \begin{array}{l} \left[ \begin{array}{l} x= \dfrac{3+\sqrt{13}}{2} \\ x= \dfrac{3-\sqrt{13}}{2} \end{array} \right. \\ y=0 \end{array} \right. \\ \left\lbrace \begin{array}{l} \left[ \begin{array}{l} x= \dfrac{3+\sqrt{13}}{2} \\ x= \dfrac{3-\sqrt{13}}{2} \end{array} \right. \\ y=-4 \end{array} \right. \end{array}\right. \nonumber \\ & \Leftrightarrow \left[\begin{array}{l} \left\lbrace \begin{array}{l} x= \dfrac{3+\sqrt{13}}{2} \\ y=0 \end{array} \right. \\ \left\lbrace \begin{array}{l} x= \dfrac{3-\sqrt{13}}{2} \\ y=0 \end{array} \right. \\ \left\lbrace \begin{array}{l} x= \dfrac{3+\sqrt{13}}{2} \\ y=-4 \end{array} \right. \\ \left\lbrace \begin{array}{l} x= \dfrac{3-\sqrt{13}}{2} \\ y=-4 \end{array} \right. \end{array}\right.\nonumber \end{align}
$$S=\left\lbrace \left( \dfrac{3+\sqrt{13}}{2}; 0\right); \left( \dfrac{3-\sqrt{13}}{2}; 0\right); \left( \dfrac{3+\sqrt{13}}{2}; -4\right); \left( \dfrac{3-\sqrt{13}}{2}; -4\right) \right\rbrace$$