Question

Помогите решить эту задачу

Answer 1

$2 {x}^{2} + {y}^{2} + 3xy = 12 \ 2 {x}^{2} + 2 {y}^{2} + 4xy - {y}^{2} = 14 \ \ 2 {x}^{2} + {y}^{2} + 3xy = 12 \ 2 {x}^{2} + {y}^{2} + 4xy = 14 \ \ xy = 2 \ 2 {x}^{2} + {y}^{2} = 6 \ \ x = frac{2}{y} \ 2 times frac{4}{ {y}^{2} } + {y}^{2} = 6 \ \ x = frac{2}{y} \ frac{8 + {y}^{4} - 6 {y}^{2} }{ {y}^{2} } = 0 \ \ {y}^{2} = t \ {t}^{2} - 6t + 8 = 0 \ t1 = 4 \ t2 = 2 \ 1) {y}^{2} = 4 \ y1 = 2 \ x1 = 1 \ y2 = - 2 \ x2 = - 1\ \ 2) {y}^{2} = 2 \ y3 = sqrt{2} \ x3 = sqrt{2} \ y4 = - sqrt{2} \ x4 = - sqrt{2}$
Получили решения:
(2; 1), (-2; -1),
(sqrt(2), sqrt(2)), (-sqrt2; -sqrt2)

x1*y2 + x2*y1 = 2*(-1) + (-2)*1 = -2 - 2 = -4