`sqrt (5x - 1) + sqrt (12x - 8) = x^2 + 3` `(ĐK: x >= 2/3)`

`<=> sqrt (5x - 1) - (x + 1) + sqrt (12x - 8) - 2x = x^2 - 3x + 2`

`<=> (5x - 1 - x^2 - 2x - 1)/(sqrt (5x - 1) + (x + 1)) + (-4x^2 + 12x - 8)/(sqrt (12x - 8) + 2x) - (x - 1) (x - 2) = 0`

`<=> (-x^2 + 3x - 2)/(sqrt (5x - 1) + (x + 1)) + (-4x^2 + 12x - 8)/(sqrt (12x - 8) + 2x) - (x - 1) (x - 2) = 0`

`<=> ((x - 2)(1 - x))/(sqrt (5x - 1) + (x + 1)) + (-4 (x - 1) (x - 2))/(sqrt (12x - 8) + 2x) - (x - 1) (x - 2) = 0`

`<=> (x - 1) (x - 2) (-1/(sqrt (5x - 1) + (x + 1)) - 4/(sqrt (12x - 8) + 2x) - 1) = 0`

Mà `x >= 2/3 => (-1/(sqrt (5x - 1) + (x + 1)) - 4/(sqrt (12x - 8) + 2x) - 1) ne 0`

`<=> (x - 1) (x - 2) = 0`

`<=>` \(\left[ \begin{array}{l}x = 1 (t/m)\\x = 2 (t/m)\end{array} \right.\) 

Vậy `S = {1; 2}`