Question

ПОМОГИТЕ ПОЖАЛУЙСТА

Answer 1

1)
$x + y = 7 \ xy = 12 \ \ y = 7 - x \ x(7 - x) = 12 \ \ y = 7 - x \ 7x - {x}^{2} = 12 \ \ y = 7 - x \ {x}^{2} - 7x + 12 = 0 \ x1 = 4 \ x2 = 3 \ y1 = 3 \ y2 = 4$
(x + y) = 7
Ответ: 7.

2)
${y}^{2} + {x}^{2} = 17 \ y - 3x - 1 = 0 \ \ y = 3x + 1 \ (3x + 1) ^{2} + {x}^{2} = 17 \ \ y = 3x + 1 \ 9 {x}^{2} + 6x + 1 + {x}^{2} = 17 \ \ y = 3x + 1 \ 10 {x}^{2} + 6x - 16 = 0 \ \ y = 3x + 1 \ 5 {x}^{2} + 3x - 8 = 0 \ d = {b}^{2} - 4ac = 9 - 4 times 5 times ( - 8) = 169 \ x1 = frac{ - 3 + 13}{10} = 1 \ x2 = frac{ - 3 - 13}{10} = - frac{8}{5} \ y1 = 4 \ y2 = 3 times ( - frac{8}{5} ) + 1 = - frac{24}{5} + frac{5}{5} = - frac{19}{5} \ x1 + y1 = 5 \ x2 + y2 = - frac{27}{5}$
Ответ: 5.

3)
$x - y = 4 \ xy = 5 \ \ x = 4 + y \ y(4 + y) = 5 \ \ x = 4 + y \ {y}^{2} + 4y - 5 = 0 \ \ x = 4 + y \ y1 = - 5 \ y2 = 1 \ x1 = - 1 \ x2 = 5 \ x1 + y1 = - 6 \ x2 + y2 = 6$
Ответ: 6.

4)
${x}^{2} + {y}^{2} = 10 \ x - y = 2 \ \ x = 2 + y \ {(2 + y)}^{2} + {y}^{2} = 10 \ \ x = 2 + y \ 4 + 4y + {y}^{2} + {y}^{2} = 10 \ \ x = 2 + y \ 2 {y}^{2} + 4y - 6 = 0 \ \ x = 2 + y \ {y}^{2} + 2y - 3 = 0 \ \ x = 2 + y \ y1 = - 3 \ y2 = 1 \ x1 = - 1 \ x2 = 3 \ x1 + y1 = - 4 \ x2 + y2 = 4$
Ответ: 4.