Помогите решить систему уравнений!!! [tex]\left \{ {x^{2}}+2xy-3y^2=0 \atop {2x^2+y^2=3} \right. [/tex]. Created by Шkolnitsa. algebra-ru

1.2.Ответ (1;1) (-1;-1) (3√3/19; -√3/19) (-3√3/19; √3/19)

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Помогите решить систему уравнений!!! [tex]\left \{ {x^{2}}+2xy-3y^2=0 \atop {2x^2+y^2=3} \right. [/tex]

Answers & Comments

bretkocё $\displaystyle \left \{ {{x^2+2xy-3y^2=0} \atop {2x^2+y^2=3}} \right.\\\\ \left \{ {{(x^2+2xy+y^2)-4y^2=0} \atop {2x^2+y^2=3}} \right.\\\\ \left \{ {{(x+y)^2-4y^2=0} \atop {2x^2+y^2=3}} \right. \\\\ \left \{ {{(x+y-2y)(x+y+2y)=0} \atop {2x^2+y^2=3}} \right.\\\\ (x-y)(x+3y)=0\\\\x=y; x=-3y$

1.
$\displaystyle x=y\\\\2y^2+y^2=3\\\\3y^2=3\\\\y=\pm 1; x=\pm 1$

2.
$\displaystyle x=-3y\\\\2*9y^2+y^2=3\\\\19y^2=3\\\\y=\pm \sqrt{ \frac{3}{19}}; x=\mp 3* \sqrt{ \frac{3}{19}}$

Ответ (1;1) (-1;-1) (3√3/19; -√3/19) (-3√3/19; √3/19)

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