• Предмет: Математика
  • Автор: SkorLine2016
  • Вопрос задан 8 лет назад

12.12 Решите систему уравнений 2)x^3-y^3=218; x^2+xy+y^2=109;

4)x+xy=3; xy^2+xy^3=12; 5)x^2+y^2=29; xy=10; 6)√x+√y=5; x+y=13; 7)x-y+xy=5;

x-y-xy= -7

Ответы

Ответ дал: AntonTasakov
0

Ответ: (x; y)

2) (7; 5), (5; 7)

4) (1; 2), (-3; -2)

5) ( 5; 2), ( -2; -5), ( 2; 5), ( -5; -2)

6) (4; 9), (9; 4)

7) (2; 3), (-3; -2)

Объяснение:

2)

x^3 - y^3 = 218

x^2 + xy + y^2 = 109

(x - y)(x^2 + xy + y^2) = 218

x^2 + xy + y^2 = 109

(x-y) * 109 = 218

x^2 - 2xy + 3xy + y^2 = 109

x - y = 2

(x - y)^2 + 3xy = 109

x - y = 2

4 + 3xy = 109 -> 3xy = 105 -> xy = 35

x = 2 + y

xy = 35 -> (2 + y)y = 35 -> y^2 +2y - 35 = 0

y = -1 ± 6

x = 1 ± 6

Ответ: (7; 5), (5; 7)

4)

x + xy = 3

xy^2 + xy^3 = 12 -> y^2(x + xy) = 12 -> 3y^2 = 12

x + xy = 3

y^2 = 4 -> y = ±2

x1 = 1, y1 = 2

x2 = -3, y2 = -2

Ответ: (1; 2), (-3; -2)

5)

x^2 + y^2 = 29

xy = 10

x^2 + y^2 - 2xy = 29 - 2*10

xy = 10

(x - y)^2 = 9

xy = 10

x - y = ±3

xy = 10 -> (y ± 3)y = 10 -> y^2 ± 3y - 10 = 0

при +3: у = ( -3 ± 7)/2, x = (3 ± 7)/2

при -3: y = (3 ± 7)/2, x = ( -3 ± 7)/2

Ответ: ( 5; 2), ( -2; -5), ( 2; 5), ( -5; -2)

6)

√x + √y = 5 | ^2

x + y = 13

x + 2√(xy) + y = 25 -> 2√(xy) + 13 = 25 -> 2√(xy) = 12

x + y = 13

√(xy) = 6

x + y = 13 -> x = 13 - y

√((13 - y)y) = 6 -> 13y - y^2 = 36 -> y^2 -13y + 36 = 0

y = (13 ± 5)/2

y1 = 9, x1 = 4

y2 = 4, x2 = 9

Ответ: (4; 9), (9; 4)

7)

x - y + xy = 5 (1)

x - y - xy = - 7 (2)

(1) - (2) : 2xy = 12 -> xy = 6

(1) + (2) : 2x - 2y = -2 -> x - y = -1

x = y - 1

xy = 6 -> (y - 1)y = 6 -> y^2 - y - 6 = 0

y = (1 ± 5)/2, x = (-1 ± 5)/2

Ответ: (2; 3), (-3; -2)

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