Đáp án:
\[A = 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2{x^2} + {y^2} + 9 = 6x + 2xy\\
\Leftrightarrow 2{x^2} + {y^2} + 9 - 6x - 2xy = 0\\
\Leftrightarrow \left( {{x^2} - 6x + 9} \right) + \left( {{x^2} - 2xy + {y^2}} \right) = 0\\
\Leftrightarrow {\left( {x - 3} \right)^2} + {\left( {x - y} \right)^2} = 0\\
\left. \begin{array}{l}
{\left( {x - 3} \right)^2} \ge 0\\
{\left( {x - y} \right)^2} \ge 0
\end{array} \right\} \Rightarrow {\left( {x - 3} \right)^2} + {\left( {x - y} \right)^2} = 0\\
\Rightarrow \left\{ \begin{array}{l}
x - 3 = 0\\
x - y = 0
\end{array} \right. \Rightarrow x = y = 3\\
A = {x^{2017}}.{y^{2018}} - {x^{2018}}.{y^{2017}} + \frac{1}{9}xy\\
= {3^{2017}}{.3^{2018}} - {3^{2018}}{.3^{2017}} + \frac{1}{9}.3.3\\
= 1
\end{array}\)
