Đáp án:
$\left \{ {{(xy+7)(x^2y^2+6)=4y^3} \atop {y^3+15=xy^4}} \right.$
⇔$\left \{ {{(xy+7)(x^2y^2+6)=4y^3} \atop {y^3(xy-1)=15}} \right.$
⇔$\left \{ {{(xy+7)(x^2y^2+6)=4.15:(xy-1)} \atop {y^3=15:(xy-1)}} \right.$
⇔$\left \{ {{(xy+7)(xy-1)(x^2y^2+6)=60} \atop {y^3=15:(xy-1)}} \right.$
⇔$\left \{ {{x^4y^4+6x^3y^3-x^2y^2+36xy-42=60} \atop {y^3=15:(xy-1)}} \right.$
⇔$\left \{ {{\left[ \begin{array}{l}xy=1,754824798\\xy=-7,129853629\end{array} \right.} \atop {y^3=15:(xy-1)}} \right.$
Từ đó suy ra x,y
