Question

Помогите пожалуйста

Answer 1

Ответ:

(3;2) (-3;-2) (2;3) (-2;-3)

Пошаговое объяснение:

первое уравнение:

$xy - ( {x}^{2} - 2xy + {y}^{2} ) = 5 \\ xy - {(x - y)}^{2} = 5 \\ {(x - y)}^{2} = xy - 5$

во втором уравнении, подставляя из первого

$5 {(xy)}^{2} - ( {x}^{4} - 2 {x}^{2} {y}^{2} + {y}^{4} ) = 155 \\ 5 {(xy)}^{2} - {( {x}^{2} - {y}^{2} )}^{2} = 155 \\ 5 {(xy)}^{2} - {((x - y)(x + y))}^{2} = 155 \\5 {(xy)}^{2} -(xy - 5) {(x + y)}^{2} = 155 \\ 5 {(xy)}^{2} -(xy - 5) ({(x - y)}^{2} + 4xy) = 155 \\ 5 {(xy)}^{2} -(xy - 5) ((xy - 5)+ 4xy) = 155 \\ 5 {(xy)}^{2} -(xy - 5) (5xy - 5) = 155 \\ {(xy)}^{2} -(xy - 5) (xy - 1) = 31$

сделаем замену xy=t

${t}^{2} - (t - 5)(t - 1) = 31 \\ {t}^{2} - {t}^{2} + 5t + t - 5 = 31 \\ 6t = 36 \\ t = 6$

откуда

$y = \frac{6}{x}$

$3 \times 6 - {x}^{2} - {( \frac{6}{x} )}^{2} = 5 \\ {x}^{4} -13 {x}^{2} + 36 = 0 \\ d = {13}^{2} - 4 \times 36 = 25 \\ \sqrt{d} = 5 \\ {x}^{2} = \frac{13 + 5}{2} = 9 \\ {x}^{2} = \frac{13 - 5}{2} = 4 \\ x_{1} = 3 \\ x_{2} = - 3 \\ x_{3} = 2 \\ x_{4} = - 2$