Verified answer

√(1-√(4x^3-5x+1))=√(1+√(4x^3-5x+1))

в квадрат

1-√(4x^3-5x+1)=1+√(4x^3-5x+1)

2√(4x^3-5x+1) = 0

4x^3-5x+1 = 0

4x^3 - 4x^2 + 4x^2 - 4x - x + 1 = 0

4x^2(x - 1) + 4x(x - 1) - 1(x - 1) = 0

(x - 1)(4x² + 4x - 1) = 0

x = 1

4x² + 4x - 1 = 0

D = 16 + 16 = 32

x23=(-4 +- √32)/8 = (-4 +- 4√2)/8 = -1/2 +- √2/2

x = {1,  -1/2 + √2/2, -1/2 - √2/2}