`a, (2x - 1)^2 - 25 = 0`
`(2x - 1)^2 - 5^2 = 0`
`(2x - 1 - 5)(2x - 1 + 5) = 0`
`=> 2x - 1 - 5 = 0 => x = 3`
`=> 2x - 1 + 5 = 0 => x = -2`
Vậy `x ∈ {3;-2}`
`b, 8x^3 - 50x = 0`
` x(8x^2 - 50) = 0`
`=> x = 0`
`=> 8x^2 - 50 = 0 => 8x^2 = 50 => x^2 = 25/4 => x = 5/2`
Vậy `x ∈ {0;5/2}`