\(\left\{{}\begin{matrix}x^2+y^2=60\\x+y=4\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=60\\x+y=4\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+y=4\\4^2-2xy=60\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+y=4\\-2xy=60-16=44\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+y=4\\xy=\dfrac{44}{-2}=-22\end{matrix}\right.\)
\(\Rightarrow\) \(x;y\) là nghiệm của phương trình : \(X^2-4X-22=0\)
giải phương trình ta có : \(\left\{{}\begin{matrix}X=2+\sqrt{26}\\X=2-\sqrt{26}\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=2+\sqrt{26}\\y=2-\sqrt{26}\end{matrix}\right.\\\left\{{}\begin{matrix}x=2-\sqrt{26}\\y=2+\sqrt{26}\end{matrix}\right.\end{matrix}\right.\)
nếu \(\left\{{}\begin{matrix}x=2+\sqrt{26}\\y=2-\sqrt{26}\end{matrix}\right.\) thì \(x-y=\left(2+\sqrt{26}\right)-\left(2-\sqrt{26}\right)=2+\sqrt{26}-2+\sqrt{26}=2\sqrt{26}\)
nếu \(\left\{{}\begin{matrix}x=2-\sqrt{26}\\y=2+\sqrt{26}\end{matrix}\right.\) thì \(x-y=\left(2-\sqrt{26}\right)-\left(2+\sqrt{26}\right)=2-\sqrt{26}-2-\sqrt{26}=-2\sqrt{26}\)
vậy \(x^2+y^2=60;x+y=4\) thì \(x-y=\pm2\sqrt{26}\)
