tìm các cặp số nguyên (x y) thỏa mãn x^2 +2y^2 -2xy+2x-6y+1=0

Đáp án: $\left( {x;y} \right) \in \left\{ {\left( {3;4} \right);\left( { - 1;0} \right);\left( {3;2} \right);\left( { - 1;2} \right)} \right\}$

Giải thích các bước giải:

$\begin{array}{l}
{x^2} + 2{y^2} - 2xy + 2x - 6y + 1 = 0\\
 \Leftrightarrow {x^2} + {y^2} + 1 - 2xy + 2x - 2y\\
 + {y^2} - 4y + 4 = 4\\
 \Leftrightarrow {\left( {x - y + 1} \right)^2} + {\left( {y - 2} \right)^2} = 4 = 0 + 4\\
Do:x;y \in Z\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x - y + 1} \right)^2} = 0\\
{\left( {y - 2} \right)^2} = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
{\left( {x - y + 1} \right)^2} = 4\\
{\left( {y - 2} \right)^2} = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - y + 1 = 0\\
\left[ \begin{array}{l}
y - 2 = 2\\
y - 2 =  - 2
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x - y + 1 = 2\\
x - y + 1 =  - 2
\end{array} \right.\\
y - 2 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = y - 1\\
\left[ \begin{array}{l}
y = 4\\
y = 0
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = y + 1\\
x = y - 3
\end{array} \right.\\
y = 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3;y = 4\\
x =  - 1;y = 0\\
x = 3;y = 2\\
x =  - 1;y = 2
\end{array} \right.\\
Vay\,\left( {x;y} \right) \in \left\{ {\left( {3;4} \right);\left( { - 1;0} \right);\left( {3;2} \right);\left( { - 1;2} \right)} \right\}
\end{array}$