Решите ситему уравнений:

Решите задачу:

$\left \{ {{x^2y-4y^3=0} \atop {x+2y^2=12}} \right. \\\\ \left \{ {{y(x^2-4y^2)=0} \atop {x+2y^2=12}} \right. \\\\ \left \{ {{x^2-4y^2=0} \atop {x=12-2y^2}} \right. \\ \\ \left \{ {{(12-2y^2)^2-4y^2=0} \atop {x=12-2y^2}} \right. \\ \\ 144-52y^2+4y^4=0\\ y^4=t\\ 4t^2-52t+144=0\\ t^2-13t+36=0\\ D=169-4*36=5^2\\ t_{1}=\frac{13-5}{2}=4\\ t_{2}=\frac{13+5}{2}=9\\ x=2 ; \ x=3 \$
$x=12\\ x=+-6\\ y=0\\ y=+-2\\ y=+-3$