x² + y² = xy + 13
x + y = √xy + 3
x, y одного знака > 0
(x + y)² = x² + y² + 2xy
x² + y² = (x + y)² - 2xy
замена
√xy = t > 0
x + y = u
Система
u² = 3t² + 13
u = t + 3
t² + 6t + 9 = 3t² + 13
2t² - 6t + 4 = 0
t² - 3t + 2 = 0
D = 9 - 8 = 1
t12 = (3 +- 1)/2 = 1 2
1. t = 1
u = 4
√xy = 1
x + y = 4
x = 4 - y
y(4 - y) = 1
y² - 4y + 1 = 0
D = 16 - 4 = 12
y12 = (4 +- √12)/2 = 2 +- √3
y1 = 2 + √3
x1 = 4 - y1 = 2 - √3
y2 = 2 - √3
x2 = 2 + √3
2. t = 2
u = 5
√xy = 2
x + y = 5
x = 5 - y
y(5 - y) = 4
y² - 5y + 4 = 0
D = (25 - 16) = 9
y34 = (5 +- 3)/2 = 1 4
y3 = 1
x3 = 5 - y3 = 4
y4 = 4
x4 = 1
ответ (1, 4) (4, 1) (2 - √3, 2 + √3) (2 + √3, 2 - √3)
ответ: x=1 y=2.
Объяснение:
{3x+7y=17 |×9 27x+63y=153
{5x-9y=-13 |×7 35x-63y=-91
Суммируем эти уравнения:
62x=62 |÷62
x=1 ⇒
3*1+7y=17
7y=17-3
7y=14 |÷7
y=2.