Giải thích các bước giải:
Ta có:
1,
\(\begin{array}{l}
4{x^2} + {y^2} < 2xy + 2x + y + 1\\
\Leftrightarrow 8{x^2} + 2{y^2} - 4xy - 4x - 2y - 2 < 0\\
\Leftrightarrow \left( {4{x^2} - 4xy + {y^2}} \right) + \left( {4{x^2} - 4x + 1} \right) + \left( {{y^2} - 2y + 1} \right) < 4\\
\Leftrightarrow {\left( {2x - y} \right)^2} + {\left( {2x - 1} \right)^2} + {\left( {y - 1} \right)^2} < 4\\
x,y \in {Z^ + } \Rightarrow 2x - 1 \ne 0\\
\Rightarrow 0 < {\left( {2x - 1} \right)^2} < 4 \Rightarrow {\left( {2x - 1} \right)^2} = 1 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = - 1\\
2x - 1 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\left( L \right)\\
x = 1
\end{array} \right. \Rightarrow x = 1\\
0 \le {\left( {y - 1} \right)^2} < 4 \Rightarrow \left[ \begin{array}{l}
y - 1 = 0\\
y - 1 = - 1\\
y - 1 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = 1 \Rightarrow {\left( {2x - y} \right)^2} = 1\left( {t/m} \right)\\
y = 2 \Rightarrow {\left( {2x - y} \right)^2} = 0\,\,\left( {t/m} \right)
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( {1;1} \right);\left( {1;0} \right)} \right\}
\end{array}\)
