Đáp án:
$\begin{array}{l}
+ Khi\,y = 0 \Rightarrow {x^2} - 5x + 6 = 0 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\\
+ Khi\,y = 1 \Rightarrow {x^2} - 5x + 4 = 0 \Rightarrow \left[ \begin{array}{l}
x = 1\\
x = 4
\end{array} \right.\\
+ Khi\,y \ge 2\\
\Rightarrow {x^2} - 2x + 1 - 3x + 6 = {3^y}\\
\Rightarrow {x^2} - 2x + 1 - 3\left( {x - 2} \right) = {3^y}\\
\Rightarrow x = 3k + 1\,\,hay\,\,\,9{k^2} - 9k + 3 = {3^y}\left( {ktm} \right)\\
Vậy\,\left( {x;y} \right) = {\rm{\{ }}\left( {2;0} \right);\left( {3;0;\left( {1;1} \right);\left( {4;1} \right)} \right)
\end{array}$
