Question

решите систему уравнений :
{y^2+x=2
{x^2+2y^2=3

Answer 1

$left { {{y^2+x=2} atop {x^2+2y^2=3}} right. \y^2=2-x \x^2+2(2-x)=3 \x^2+4-2x-3=0 \x^2-2x+1=0 \(x-1)^2=0 \x-1=0 \x=1 \y^2=2-1=1 \y_1=1 \y_2=-1$
Ответ: (1;1), (1;-1)

Answer 2

$y^2+x=2 \ x^2 +2y^2=3 \ y^2=frac{3-x^2}{2} \ frac{3-x^2}{2}+x=2 \ 3-x^2+2x=4 \ x^2-2x+1=0 \ D=4-4=0 \ x_{1,2}=frac{2}{2}=1 \ y^2=frac{3-1}{2}=1 \ y_1=1 \ y_2=-1$