\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=11\\2\left(x+y\right)+2xy=6+8\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)^2+2\left(x+y\right)-17-8\sqrt{2}=0\)
\(\Leftrightarrow\left(x+y-3-\sqrt{2}\right)\left(x+y+5+\sqrt{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=3+\sqrt{2}\\x+y=-5-\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}xy=3\sqrt{2}\\xy=8+5\sqrt{2}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=3+\sqrt{2}\\xy=3\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow x;y\) là nghiệm của: \(t^2-\left(3+\sqrt{2}\right)t+3\sqrt{2}=0\Rightarrow\left[{}\begin{matrix}t=3\\t=\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(3;\sqrt{2}\right);\left(\sqrt{2};3\right)\)
TH2: \(\left\{{}\begin{matrix}x+y=-5-\sqrt{2}\\xy=8+5\sqrt{2}\end{matrix}\right.\)
Do \(\left(x+y\right)^2-4xy< 0\) nên pt vô nghiệm
