Найти x*y,если (x,y)-решение системы уравнений {2x^2+3xy-y^2=4 3x^2+2xy-2y^2=3

Найти x*y,если (x,y)-решение системы уравнений {2x^2+3xy-y^2=4
3x^2+2xy-2y^2=3
- Предмет:
  Математика
- Автор:
  greysondaniel
- 6 лет назад

Ответы 1

Выравним коэффициенты $\left \{ {{6 x^{2}+9xy - 3y^{2} =12} \atop {-12 x^{2} -8xy+8 y^{2} =12}} ight.$ Сложим первое и второе уравнение, получим $-6 x^{2} +xy +5 y^{2} =0$ Разделим всё на xy $6 \frac{x}{y} -1 -5 \frac{y}{x} =0$ Пусть $\frac{x}{y} =t$ $6t -1 -5 \frac{1}{t}=0$ $6 t^{2} -t -5 =0$ t= 1 или t = - 5/6Если t = 1, то x = yподставив это в любое уравнение, получимx = 1 y = 1 или x = -1 y = -1xy = 1Другое решение является посторонним, так как квадрат становится отрицательным
- Автор:
  tanner57
- 6 лет назад
- 0
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Найти x*y,если (x,y)-решение системы уравнений {2x^2+3xy-y^2=4 3x^2+2xy-2y^2=3

Ответы 1

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Найти x*y,если (x,y)-решение системы уравнений {2x^2+3xy-y^2=4
3x^2+2xy-2y^2=3