Решить систему уравнений 2 х^2-3 ху+2 у^2=4 2 х^2+3 у^2=14

1. Введем замену переменной:

y = px;

{2 х^2 - 3 ху + 2 у^2 = 4;

{2 х^2 + 3 у^2 = 14; {2 х^2 - 3pх^2 + 2p^2x^2 = 4;

{2 х^2 + 3p^2x^2 = 14; {2 х^2 - 3pх^2 + 2p^2x^2 = 4;

{2 х^2 + 3p^2x^2 = 14; {x^2 (2 - 3p + 2p^2) = 4;

{х^2 (2 + 3p^2) = 14.

2. Разделим уравнения:

(2 - 3p + 2p^2) / (2 + 3p^2) = 4/14 = 2/7; 7 (2 - 3p + 2p^2) = 2 (2 + 3p^2); 14 - 21p + 14p^2 = 4 + 6p^2; 8p^2 - 21p + 10 = 0; D = 21^2 - 4 * 8 * 10 = 441 - 320 = 121; p = (21 ± √121) / 16 = (21 ± 11) / 16; p1 = (21 - 11) / 16 = 10/16 = 5/8; p2 = (21 + 11) / 16 = 32/16 = 2.

3. Вернемся к системе:

х^2 (2 + 3p^2) = 14; х^2 = 14 / (3p^2 + 2);

1) p = 5/8;

3p^2 + 2 = 3 * (5/8) ^2 + 2 = (3 * 25 + 128) / 64 = 203/64; x^2 = 14 / (203/64) = (14 * 64) / (7 * 29) = 2 * 64/29; x = ±8√2/√29; y = px = 5/8 * x = 5/8 * (±8√2/√29) = ±5√2/√29.

2) p = 2;

3p^2 + 2 = 3 * 2^2 + 2 = 14; x^2 = 14/14 = 1; x = ±1; y = px = ±2.

Ответ: (±8√2/√29; ±5√2/√29), (±1; ±2).