$\begin{cases} 6x^2 y+2y^3+35=0 \ (1) \\ 5x^2+5y^2+2xy+5x+13y=0\end{cases} \ \ (2)$
$(1) +3. (2)$ ta được
$6x^2y+2y^3+35+15x^2+15y^2+6x(2y+5)+39y=0$
$\Leftrightarrow 6x^2y+15x^2+2y^3+5y^2+10y^2+25y+14y+35+6x(2y+5)=0$
$\Leftrightarrow (2y+5)(6x^2+2y^2+10y+14+6x)$
$\Leftrightarrow (2y+5)[6(x+\frac{1}{2}^2)+2(y+\frac{5}{2})^2]=0$ dễ rồi nhé