Question

2){2x2 – 3y2 = -19, {xy = -6; 3){x2+y2=65 {xy=28 Смотреть ответ daraprelj daraprelj 2) | * y² 72-3y⁴=-19y² -3y⁴+19y²+72=0 | :(-1) 3y⁴-19y²-72=0 Пусть y²=t, t>0 3t²-19t-72=0 D = (-19)²-4*3*(-72) = 361+864 = 1225 = 35² <0 Вернёмся к замене y² = 9 y = ±3 x₁ = - (6 : 3) = -2 x₂ = - (6 : (-3)) = 2 Ответ: (-2;3),(2;-3) 3) | * y² 784 + y⁴ = 65y² y⁴ - 65y² + 784 = 0 Пусть y²=t, t>0 t²-65t+784=0 D = (-65)²-4*1*784 = 4225 - 3136 = 1089 = 33² Вернёмся к замене x₁ = 28:7 = 4 x₂ = 28:(-7) = -4 x₃ = 28:4 = 7 x₄ = 28:(-4) = -7 Ответ: (4;7),(-4;-7),(7;4),(-7;-4)

Answer 1

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