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Hệ tương đương với:
$\left\{ \begin{array}{l} xy+1=7y-x\\ (xy+1)^2-xy=13y^2 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} xy+1=7y-x\\ (7y-x)^2-xy-13y^2=0 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} xy+1=7y-x\\36y^2-15xy+x^2=0 \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x=3y\\ xy+1+x-7y=0 \end{array} \right.\\\left\{ \begin{array}{l} x=12y\\ xy+1+x-7y=0 \end{array} \right. \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x=3y\\ 3y^2-4y+1=0 \end{array} \right.\\ \left\{ \begin{array}{l} x=12y\\ 12y^2+5y+1=0 \end{array} \right. \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x=3\\ y=1 \end{array} \right.\\\left\{ \begin{array}{l} x=3\\ y=\frac{1}{3} \end{array} \right. \end{array} \right.$
Vậy: $(x,y)\in\{(3;1),(1;\frac{1}{3})\}$
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