Question

решить систему уравнений
$\left \{ {{x+y^{2} =2} \atop {2y^{2}+ x^{2}=3}} \right.$

Answer 1

x = 2 - y^2 
x^2 = 4 - 4y^2 + y^4 
2y^2 + 4 - 4y^2 + y^4 = 3 
y^4 - 2y^2 + 1 = 0 
( y^2 - 1 )^2 = 0 
y^2 - 1 = 0
y^2 = 1 
y1 = 1 
y2 = - 1 
x1 = 2 - 1 = 1
x2 = 2 - 1 = 1 
Ответ ( 1 ; - 1 ) ; ( 1 ; 1 )