+ Biến đổi hệ phương trình đã cho, ta được: $\left\{ \begin{array}{l}xy\left( x+y \right)=30\\\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)=35\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l}xy\left( x+y \right)=30\\\left( x+y \right)\left( {{\left( x+y \right)}^{2}}-3xy \right)=35\end{array} \right..$ + Đặt$\left\{ \begin{array}{l}x+y=S\\xy=P\end{array} \right.$ ta thu được hệ phương trình sau $\begin{array}{l}\left\{ \begin{array}{l}S.P=30\\S\left( {{S}^{2}}-3P \right)=35\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l}S.P=30\\{{S}^{3}}-3S.P=35\end{array} \right.\\\Leftrightarrow \left\{ \begin{array}{l}S.P=30\\{{S}^{3}}-3.30=35\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l}S.P=30\\{{S}^{3}}=125\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l}S=5\\P=6\end{array} \right..\end{array}$ Quay trở lại phép đặt, ta có$\left\{ \begin{array}{l}x+y=5\\xy=6\end{array} \right.\Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x=2\\y=3\end{array} \right.\\\left\{ \begin{array}{l}x=3\\y=2\end{array} \right.\end{array} \right..$
