Cộng vế với vế:
\(x^2+y^2+2xy+x+y=6\)
\(\Leftrightarrow\left(x+y\right)^2+x+y-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=-3\\x+y=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-3-x\\y=2-x\end{matrix}\right.\)
Thay vào pt dưới:
\(\left[{}\begin{matrix}-3+x\left(-3-x\right)=2\\2+x\left(2-x\right)=2\end{matrix}\right.\)
\(\Leftrightarrow...\)
Đặt \(\begin{cases}x+y=S\\xy=P\\\end{cases} (S^2≥ 4P)\)
\(⇔ \begin{cases}(x+y)^2-xy=4\\(x+y)+xy=2\\\end{cases}\\ ⇔\begin{cases}S^2-P=4\\S+P=2\\\end{cases}\\⇔\begin{cases}S^2-2+S=4\\P=2-S\\\end{cases}\\⇔\begin{cases}\left[ \begin{array}{l}S=2\\S=-3\end{array} \right.\\\left[ \begin{array}{l}P=0\\P=5(L)\end{array} \right.\\\end{cases}\\ \Rightarrow \text{PT:} x^2-Sx+P=0 \\ \Leftrightarrow x^2-2x=0 \Leftrightarrow \left[ \begin{array}{l}x=2\\x=0\end{array} \right. \\ \Rightarrow \left[ \begin{array}{l}y=0\\y=2\end{array} \right. \)
Vậy (x;y)=(2;0);(0;2)
