Предмет: Алгебра, автор: Kitota20

3х²-2у²=1
{
2х²-2у=1
(эти 2 уравнения система!)

Пж, помогите, срочно

Ответы

Автор ответа: Аноним
1

У вас есть система уравнений:

3x^2 - 2y^2 = 1

2x^2 - 2y = 1

Давайте попробуем решить эту систему методом исключения. Сначала мы можем выразить x из второго уравнения:

2x^2 - 2y = 1

2x^2 = 2y + 1

x^2 = (2y + 1)/2

x = ±√((2y + 1)/2)

Теперь, подставим это значение x в первое уравнение:

3((2y + 1)/2) - 2y^2 = 1

Упростим это уравнение:

3(y + 1) - 2y^2 = 1

3y + 3 - 2y^2 = 1

Теперь перенесем все члены на одну сторону:

2y^2 - 3y + 2 = 0

Это уравнение второй степени. Мы можем решить его с помощью дискриминанта:

D = (-3)^2 - 4 * 2 * 2 = 9 - 16 = -7

Дискриминант отрицательный, что означает, что уравнение имеет два комплексных корня. Следовательно, система уравнений имеет комплексные решения.

Итак, решения системы в комплексных числах будут:

x = ±√((2y + 1)/2)

y - любое комплексное число, которое удовлетворяет уравнению 2y^2 - 3y + 2 = 0.

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