Giải hệ phương trình: $\left\{\begin{array}{l}x+y+xy=11\\ x^2+y^2+3\left(x+y\right)=28\end{array}\right.$

$\left\{\begin{array}{l}x+y+xy=11\\ x^2+y^2+3\left(x+y\right)=28\end{array}\right.$

$\iff \left\{\begin{array}{l}\left(x+y\right)+xy=11\\ {\left(x+y\right)}^2-2xy+3\left(x+y\right)=28\end{array}\right.$ (*)

Ta đặt: a = x + y và b = xy (Với a² ≥ − 4b)

Hệ phương trình (*) trở thành

+ TH1: $\left\{\begin{array}{l}a=5\\ b=6\end{array}\right.$

$$\begin{gathered} \Rightarrow \left\{\begin{array}{l}x+y=5\\ xy=6\end{array}\right.\iff \left\{\begin{array}{l}y=5-x\\ x\left(5-x\right)=6\end{array}\right. \\ \iff \left\{\begin{array}{l}y=5-x\\ x^2-5x+6=0\end{array}\right.\iff \left\{\begin{array}{l}y=5-x\\ \left(x-2\right)\left(x-3\right)=0\end{array}\right. \\ \iff \left\{\begin{array}{l}y=5-x\\ \left[\begin{array}{l}x=2\\ x=3\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x=2\\ y=3\end{array}\right.\\ \left\{\begin{array}{l}x=3\\ y=2\end{array}\right.\end{array}\right. \end{gathered}$$

+ TH2: $\left\{\begin{array}{l}a=-10\\ b=21\end{array}\right.$

$$\begin{gathered} \Rightarrow \left\{\begin{array}{l}x+y=-10\\ xy=21\end{array}\right.\iff \left\{\begin{array}{l}y=-10-x\\ x\left(-10-x\right)=21\end{array}\right. \\ \iff \left\{\begin{array}{l}y=-10-x\\ x^2+10x+21=0\end{array}\right.\iff \left\{\begin{array}{l}y=-10-x\\ \left(x+3\right)\left(x+7\right)=0\end{array}\right. \\ \iff \left\{\begin{array}{l}y=-10-x\\ \left[\begin{array}{l}x=-3\\ x=-7\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x=-3\\ y=-7\end{array}\right.\\ \left\{\begin{array}{l}x=-7\\ y=-3\end{array}\right.\end{array}\right. \end{gathered}$$

Vậy cặp nghiệm (x; y) của hệ phương trình là: $\left(x;y\right)=\left\{\left(2;3\right),\left(3;2\right),\left(-3;-7\right),\left(-7;-3\right)\right\}$