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Đặt $a=x+y,b=xy$ thì HPT
$\Leftrightarrow \begin{cases}a^3-3ab=19 \\ a(b+8)=2 \end{cases}\Leftrightarrow \begin{cases}a^3-3ab=19 \\ b=\frac{2}{a}-8 \end{cases}\Leftrightarrow \begin{cases}a^3-3a\left (\frac{2}{a}-8 \right )-19=0 \\ b=\frac{2}{a}-8 \end{cases}$
$\Leftrightarrow \begin{cases}a^3+24a-25= 0\\ b=\frac{2}{a}-8 \end{cases}\Leftrightarrow \begin{cases}a=1 \\ b=-6 \end{cases}\Leftrightarrow (x,y)\in \{(3,-2),(-2,3) \}$
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