\(x^2+5y^2-4xy-4y+3=0\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2-4y+4\right)=1\)
\(\Leftrightarrow\left(x-2y\right)^2+\left(y-2\right)^2=1\)
Vì \(x;y\in Z\)\(\Rightarrow\left(x-2y\right)^2\ge0;\left(y-2\right)^2\ge0\) và \(\left(x-2y\right)^2;\left(y-2\right)^2\in N\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\\left(y-2\right)^2=1\end{matrix}\right.\\\left\{{}\begin{matrix}\left(x-2y\right)^2=1\\\left(y-2\right)^2=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x=6\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=5\\y=2\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2-4y+4\right)-1=0\)
\(\Leftrightarrow\left(x-2y\right)^2+\left(y-2\right)^2=1=0^2+1^2\)
Vì \(x,y\in Z\) nên ta có các trường hợp sau:
+ TH1 : \(\left\{{}\begin{matrix}x-2y=0\\y-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=3\end{matrix}\right.\left(TM\right)\)
+ TH2 : \(\left\{{}\begin{matrix}x-2y=0\\y-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(TM\right)\)
+ TH3 : \(\left\{{}\begin{matrix}x-2y=1\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=2\end{matrix}\right.\) (TM )
+ TH4 : \(\left\{{}\begin{matrix}x-2y=-1\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\left(TM\right)\)
Vậy có 4 cặp số (x,y) thỏa mãn yêu cầu bài toán là
( 6 ; 3 ) ; ( 2 ; 1 ) ; ( 5 ; 2 ) ; ( 3 ; 2 ).
