Giải hệ phương trình: $\left\{\begin{array}{l}xy+x+y=11\\ x^{2}y+x{y}^2=30\end{array}\right.$

$\left\{\begin{array}{l}xy+x+y=11\\ x^{2}y+x{y}^2=30\end{array}\right.\iff \left\{\begin{array}{l}xy+\left(x+y\right)=11\\ xy\left(x+y\right)=30\end{array}\right.$ (*)

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

Hệ phương trình (*) trở thành

$$\begin{gathered} \left(*\right)\iff \left\{\begin{array}{l}a+b=11\\ ab=30\end{array}\right.\iff \left\{\begin{array}{l}b=11-a\\ a\left(11-a\right)=30\end{array}\right. \\ \iff \left\{\begin{array}{l}b=11-a\\ a^2-11a+30=0\end{array}\right.\iff \left\{\begin{array}{l}b=11-a\\ \left(a-5\right)\left(a-6\right)=0\end{array}\right. \\ \iff \left\{\begin{array}{l}b=11-a\\ \left[\begin{array}{l}a=5\\ a=6\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}a=5\\ b=6\end{array}\right.\\ \left\{\begin{array}{l}a=6\\ b=5\end{array}\right.\end{array}\right. \end{gathered}$$

+ TH1:

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

+ TH2: $\left\{\begin{array}{l}a=6\\ b=5\end{array}\right.$

$$\begin{gathered} \Rightarrow \left\{\begin{array}{l}x+y=6\\ xy=5\end{array}\right.\iff \left\{\begin{array}{l}y=6-x\\ x\left(6-x\right)=5\end{array}\right. \\ \iff \left\{\begin{array}{l}y=6-x\\ x^2-6x+5=0\end{array}\right.\iff \left\{\begin{array}{l}y=6-x\\ \left(x-1\right)\left(x-5\right)=0\end{array}\right. \\ \iff \left\{\begin{array}{l}y=6-x\\ \left[\begin{array}{l}x=1\\ x=5\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x=1\\ y=5\end{array}\right.\\ \left\{\begin{array}{l}x=5\\ y=1\end{array}\right.\end{array}\right. \end{gathered}$$

Vậy cặp nghiệm (x; y) của hệ phương trình là: $\left(x;y\right)=\left\{\left(2;3\right),\left(3;2\right),\left(1;5\right),\left(5;1\right)\right\}$