3x^2 - 7x + 6 = 2(2x - 3) sqrt(3x - 2) ĐK : x >= 2/3
<=>3x^2 - 7x + 6 - 4(2x - 3)= 2(2x - 3) sqrt(3x - 2) - 4(2x - 3) 
<=> 3x^2 - 15x + 18 = 2(2x - 3) . (sqrt(3x - 2) - 2)
<=> 3x^2 -6x - 9x + 18 = 2(2x - 3) . (sqrt(3x - 2) - 2)
<=> 3x(x - 2) - 9(x - 2) = 2(2x - 3) . (sqrt(3x - 2) - 2)
<=> (x - 2).(3x - 9) = 2 (2x - 3) . (3x - 2 - 4)/(sqrt(3x - 2) + 2)
<=> (x - 2)(3x - 9) = 2 (2x - 3) . (3(x - 2))/(sqrt(3x - 2) + 2)
<=> (x - 2)(3x - 9) -2(2x - 3) . (3(x - 2))/(sqrt(3x - 2) + 2) = 0
<=> (x - 2).[3x - 9 - 2(2x - 3) . 3/(sqrt(3x - 2) + 2)] = 0
<=> (x - 2) . [3x - 9 - (6(2x - 3))/(sqrt(3x - 2) + 2)] = 0
<=> (x - 2) . (3x - 9 - (12x - 18)/(sqrt(3x - 2) + 2)) = 0
TH1 : x - 2 = 0 <=> x = 2(tm)
TH2 : 3x - 9 - (12x - 18)/(sqrt(3x - 2) + 2) = 0
<=> ((3x - 9)(sqrt(3x - 2) + 2) - 12x + 18)/(sqrt(3x - 2) + 2) = 0
<=> (3x - 9)(sqrt(3x - 2) + 2) - 12x + 18 = 0
<=> 3xsqrt(3x - 2) + 6x - 9sqrt(3x - 2) - 18 - 12x + 18 = 0
<=> 3xsqrt(3x - 2)  - 9sqrt(3x - 2) - 6x = 0
<=> xsqrt(3x - 2) - 3sqrt(3x - 2) - 2x = 0
<=> sqrt(3x - 2)(x - 3) = 2x
<=> sqrt(3x - 2)(x - 3) - 4(x - 3) = -2x + 12
<=> (x - 3) . (sqrt(3x - 2) - 4) = 2(6 - x)
<=> (x - 3) . (3x - 2 - 16)/(sqrt(3x - 2) + 4) + 2(x - 6) = 0
<=> 3(x - 3) . (x - 6)/(sqrt(3x - 2) + 4) + 2(x - 6) = 0
<=> (x - 6).[(3(x - 3))/(sqrt(3x - 2) + 4) + 2] = 0
TH3 : x - 6 = 0 <=> x = 6=(tm)
TH4 : (3(x - 3))/(sqrt(3x - 2) + 4) + 2 = 0
<=> 3x - 9 + 2sqrt(3x - 2) + 8 = 0
<=> 3x - 1 + 2sqrt(3x - 2)   = 0
Ta có : x >= 2/3 => 3x - 1 > 0
=> 3x - 1 + 2sqrt(3x - 2) > 0 (Vì : 2sqrt(3x - 2) >= 0)
-> Loại
Vậy S = {2 ; 6}