\(5x^2+y^2=17+2xy\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+4x^2=17\)
\(\Leftrightarrow\left(x-y\right)^2+4x^2=17=1+4\times2^2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
\(5x^2+y^2=17+2xy\Leftrightarrow\left(x-y\right)^2+4x^2=17\)
\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\x^2\ge0\end{matrix}\right.\) \(\Rightarrow4x^2\le17;x^2\le\dfrac{17}{4};\dfrac{-\sqrt{17}}{2}\le x\in\dfrac{\sqrt{17}}{2}\)
\(x\in Z\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
\(\left[{}\begin{matrix}x=0\Rightarrow\left(x-y\right)^2=17\left(y\notin Z\right)\left(l\right)\\x=\pm1\Rightarrow\left(x-y\right)^2=13\left(y\notin Z\right)\left(l\right)\\x=2\Rightarrow\left(2-y\right)^2=1\Rightarrow y=\left\{1;3\right\}\\x=-2\Rightarrow-2-y=\pm1\Rightarrow y=\left\{-1;-3;\right\}\end{matrix}\right.\)
\(\left(x;y\right)=\left(2;1\right);\left(2;3\right);\left(-2;-1\right);\left(-2;-3\right)\)
