Question

Помогите решить систему уравнений!!! $left { {x^{2}}+2xy-3y^2=0 atop {2x^2+y^2=3} right.$

Answer 1

I hope this helps you

Answer 2

I've declared

Answer 3

Thank you very much <3

Answer 4

yw:)

Answer 5

I'm sorry, but can you help me again here: https://znanija.com/task/26563387 ? I can't understand this...

Answer 6

i missed sorry:/

Answer 7

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

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)