Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} \frac{x^2+y^2}{xy}=\frac{13}{6}\\ x+y=6\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{(x+y)^2-2xy}{xy}=\frac{13}{6}\\ x+y=6\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \frac{6^2}{xy}-2=\frac{13}{6}\\ x+y=6\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} xy=\frac{216}{25}\\ x+y=6\end{matrix}\right.\)
\(\Rightarrow x(6-x)=\frac{216}{25}\)
\(\Leftrightarrow x^2-6x+\frac{216}{25}=0\)
\(\Leftrightarrow (x-3)^2=\frac{9}{25}\Rightarrow \left[\begin{matrix} x=\frac{-3}{5}+3=\frac{12}{5}\rightarrow y=6-x=\frac{18}{5}\\ x=\frac{3}{5}+3=\frac{18}{5}\rightarrow y=6-x=\frac{12}{5}\end{matrix}\right.\)
Vậy $(x,y)=(\frac{12}{5}, \frac{18}{5})$ và hoán vị.
