\(x^2-4xy+5y^2=100\\ \left(x^2-4xy+4y^2\right)+y^2=100\\ \left(x-2y\right)^2+y^2=100=0+10^2=6^2+8^2=10^2+0=8^2+6^2\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2y=0\\y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x-2y=6\\y=8\end{matrix}\right.\\\left\{{}\begin{matrix}x-2y=10\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2y=8\\y=6\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=20;y=10\\x=22;y=8\\x=10;y=0\\x=20;y=6\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(20;10\right);\left(22;8\right);\left(10;0\right);\left(20;6\right)\right\}\)
x2-4xy+5y2=100
⇔ x2-4xy+4y2+y2=100
⇔ (x-2y)2+y2=102+0=(-10)2+0
⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2y=10\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2y=0\\y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x-2y=-10\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2y=0\\y=-10\end{matrix}\right.\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=10\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=20\\y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x=-10\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=-20\\y=-10\end{matrix}\right.\end{matrix}\right.\)
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