Đáp án:

$\left( {x;y} \right) \in \left\{ {\left( {0;0} \right),\left( {1;2} \right),\left( { - 1;3} \right)} \right\}$

Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
5\left( {{x^2} + xy + {y^2}} \right) = 7\left( {x + 2y} \right)\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + xy + {y^2} = 7k\\
x + 2y = 5k
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + xy + {y^2} = 7k\\
x = 5k - 2y
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {5k - 2y} \right)^2} + \left( {5k - 2y} \right)y + {y^2} = 7{k^2}\\
x = 5k - 2y
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
25{k^2} - 20ky + 4{y^2} + 5ky - 2{y^2} + {y^2} = 7{k^2}\\
x = 5k - 2y
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
3{y^2} - 15ky + 25{k^2} - 7k = 0\left( 1 \right)\\
x = 5k - 2y
\end{array} \right.\left( I \right)
\end{array}$

Coi $(1)$ là phương trình ẩn $y$ tham số $k$

Để $\left( 1 \right)$ có nghiệm $y$ 

$\begin{array}{l}
 \Leftrightarrow \Delta  = {\left( { - 15k} \right)^2} - 4.3.\left( {25{k^2} - 7k} \right) \ge 0\\
 \Leftrightarrow  - 75{k^2} + 84k \ge 0\\
 \Leftrightarrow 0 \le k \le \frac{{84}}{{75}}
\end{array}$

Mà $k \in Z \Rightarrow \left[ \begin{array}{l}
k = 0\\
k = 1
\end{array} \right.$

+) TH1: Nếu $k=0$

$\begin{array}{l}
\left( I \right)tt:\left\{ \begin{array}{l}
3{y^2} = 0\\
x =  - 2y
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y = 0\\
x =  - 2y
\end{array} \right. \Leftrightarrow x = y = 0\\
 \Rightarrow \left( {x;y} \right) = \left( {0;0} \right)
\end{array}$

+) TH2: Nếu $k=1$

$\begin{array}{l}
\left( I \right)tt:\left\{ \begin{array}{l}
3{y^2} - 15y + 18 = 0\\
x = 5 - 2y
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {y - 2} \right)\left( {y - 3} \right) = 0\\
x = 5 - 2y
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
y = 2\\
y = 3
\end{array} \right.\\
x = 5 - 2y
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = 2;x = 1\\
y = 3;x =  - 1
\end{array} \right.\\
 \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {1;2} \right),\left( { - 1;3} \right)} \right\}
\end{array}$

Như vậy có: $\left( {x;y} \right) \in \left\{ {\left( {0;0} \right),\left( {1;2} \right),\left( { - 1;3} \right)} \right\}$ thỏa mãn đề.

Vậy $\left( {x;y} \right) \in \left\{ {\left( {0;0} \right),\left( {1;2} \right),\left( { - 1;3} \right)} \right\}$