Question

Решите ситему уравнений:

 

$left { {{x^2y-4y^3=0} atop {x+2y^2=12} left$

Answer 1

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