посчитайте уравнение номер 3

(√(3 + 2√2))^x + (√(3 - 2√2))^x = 6

3 ± 2√2 = 2 ± 2√2 + 1 = √2² ± 2√2 + 1² = (√2 ± 1)²

(√(√2 + 1)²)^x + (√(√2 - 1)²)^x = 6

|√2 + 1|^x + |√2 - 1|^x = 6

√2 > 1

√2 + 1 = 1/(√2 - 1)

(√2 + 1)^x + (√2 - 1)^x = 6

(√2 + 1)^x = y > 0

y + 1/y = 6

y² - 6y + 1 = 0

D = 36 - 4 = 32

y12 = (6 +- √32)/2 = (6 +- 4√2) = 3 +- 2√2 = (√2 ± 1)²

1. y1 = (√2 + 1)²

(√2 + 1)^x = (√2 + 1)²

x = 2

2. y2 = (√2 - 1)²

(√2 + 1)^x = 1/(√2 + 1)²

x = -2

x = {-2, 2}