(x-3/4)² + (x-3/4)(x-1/2) = 0
⇔ [(x-3)/4].[(x-3)/4 + (x-1)/2] = 0
⇔ [(x-3)/4].[(x-3)/4 + 2.(x-1)/(2.2)] = 0
⇔ [(x-3)/4].[(x-3)/4 + (2x -2).4] = 0
⇔ [(x-3)/4].[(x-3 +2x -2)/4] = 0
⇔ [(x-3)/4].[(3x -5)/4] = 0
TH1: (x-3)/4 = 0
⇔ x-3 = 0
⇔ x = 3
TH2: (3x -5)/4 = 0
⇔ 3x -5 = 0
⇔ 3x = 5
⇔ x = 5/3
Vậy S= {5/3; 3}