Question

Решите уравнение:
$\sqrt[3]{(3x+\sqrt{1+x^2})^2 } +3\sqrt[3]{8x^2-1}+2\sqrt[3]{(3x-\sqrt{1+x^2})^2 }=0$

Answer 1

a = ∛(3x + √(1 + x²))

b = ∛(3x - √(1 + x²))

ab = ∛(3x + √(1 + x²))∛(3x - √(1 + x²)) = ∛(9x² - 1 - x²) = ∛(8x² - 1)  

∛(3x + √(1 + x²))² + 3∛(8x² - 1 ) + 2∛(3x - √(1 + x²))²= 0

a² + 3ab + 2b² = 0

D = 9b² - 8b² = b²

a12 = (-3b +- b)/2 = -2b   -b

1. a = -b

∛(3x + √(1 + x²)) = -∛(3x - √(1 + x²))

3x + √(1 + x²) = - (3x - √(1 + x²))

6x = 0

x = 0

2. a = -2b

∛(3x + √(1 + x²)) = -2∛(3x - √(1 + x²))

3x + √(1 + x²) = -8 (3x - √(1 + x²))

3x + √(1 + x²) = -24x + 8√(1 + x²)

27x = 7√(1 + x²)

x > 0

729x² = 49 + 49x²

680x² = 49

x² = 49/680

x = - √49/680 < 0 нет

x = √49/680 = 7/2√170 = 7√170/340

ответ 7√170/340, 0