Решить систему Уравнений ∫x² + 2y² = 17,∫6x²-xy-12y² = 0

1. Решим второе уравнение относительно x:

{x^2 + 2y^2 = 17;

{6x^2 - xy - 12y^2 = 0; 6x^2 - yx - 12y^2 = 0; D = y^2 + 4 * 6 * 12y^2 = y^2 + 288y^2 = 289y^2 = (17y) ^2; x = (y ± 17y) / 12; x1 = (y - 17y) / 12 = - 16y/12 = - 4y/3; x2 = (y + 17y) / 12 = 18y/12 = 3y/2.

2. Решим систему для каждого значения:

1) x = - 4y/3;

(-4y/3) ^2 + 2y^2 = 17; 16y^2/9 + 2y^2 = 17; 16y^2 + 18y^2 = 9 * 17; 34y^2 = 9 * 17; 2y^2 = 9; y^2 = 9/2 = 18/4; y = ±3√2/2; x = - 4/3 * (±3√2/2) = ∓2√2.

2) x = 3y/2;

(3y/2) ^2 + 2y^2 = 17; 9y^2/4 + 2y^2 = 17; 9y^2 + 8y^2 = 4 * 17; 17y^2 = 4 * 17; y^2 = 4; y = ±2; x = 3/2 * (±2) = ±3.

Ответ: (±2√2; ∓3√2/2), (±3; ±2).