$\left\{\begin{matrix}$\left\{\begin{array}{l}1+x^3y^3=19x^3\\y+xy^2=-6x^2\end{array}\right.$
\frac{1}{x^{3}}+y^{3}=19\\ \frac{y}{x^{2}}+\frac{y^{2}}{x}=-6
\end{matrix}\right.$
Đặt $a= \frac{1}{x}$
$\left\{\begin{matrix}
a^{3}+y^{3}=19 (1)\\ a^{2}b+ab^{2}=-6 (2)
\end{matrix}\right.$
Lấy (1)+3.(2)=> $(a+y)^{3}=1$.....
- tri2308 yêu thích