Đáp án:
$\begin{array}{l}
{a^2} + {b^2} + 1 = ab + a + b\\
\Leftrightarrow 2{a^2} + 2{b^2} + 2 = 2ab + 2a + 2b\\
\Leftrightarrow {a^2} - 2ab + {b^2} + {a^2} - 2a + 1\\
+ {b^2} - 2b + 1 = 0\\
\Leftrightarrow {\left( {a - b} \right)^2} + {\left( {a - 1} \right)^2} + {\left( {b - 1} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {a - b} \right)^2} \ge 0\\
{\left( {a - 1} \right)^2} \ge 0\\
{\left( {b - 1} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a - b = 0\\
a - 1 = 0\\
b - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = b\\
a = 1\\
b = 1
\end{array} \right.\\
\Leftrightarrow a = b = 1\\
Vậy\,a = b = 1
\end{array}$
