Иррациональное уравнение

3√(3x² + 2x - 4) - 2x = 3x² - 2

3√(3x² + 2x - 4) = 3x² + 2x - 2

ограничения 3x² + 2x - 4 >= 0

D = 4 + 4*4*3 = 4 + 48 = 52

x12 = (-2 +- √52)/6 = (-1 +- √13)/3

+++++++(-1 +- √13)/3 ------------- (-1 + √13)/3 +++++++++

x ∈ (-∞, (-1 +- √13)/3] U [(-1 +- √13)/3, + ∞) ≈ (-∞, -4.6/3] U [2.6/3, +∞)

3√(3x² + 2x - 4) = (3x² + 2x - 4) + 2

√(3x² + 2x - 4) = t (>=0)

3t = t² + 2

t² - 3t + 1 = 0

t1 = 1

t2 = 2

1. t1 = 1

√(3x² + 2x - 4) = 1

3x² + 2x - 4 = 1

3x² + 2x - 5 = 0

D = 4 + 4*5*3 = 64

x12 = (-2 +- 8)/6 = 1 - 5/3

2. t2 = 2

√(3x² + 2x - 4) = 2

3x² + 2x - 4 = 4

3x² + 2x - 8 = 0

D = 4 + 4*8*3 = 100

x12 = (-2 +- 10)/6 = 4/3 - 2

x = {-2, -5/3, 1, 4/3}