(x+2) (x² - 1) = 0
TH1: x+2 = 0 ⇔ x = -2
TH2: x² -1 = 0
⇔ x² = 1
⇔ x = ±√1
Vậy S= {-2; ±√1}
(2-3x) (x+11) = (3x-2) (2-5x)
⇔ (2-3x).(x+11) -(3x-2).(2-5x) = 0
⇔ (2-3x).(x+11) +(2-3x).(2-5x) = 0
⇔ (2-3x).(x+11 +2-5x) = 0
⇔ (2-3x).(-4x +13) = 0
TH1: 2-3x = 0 ⇔ x = 2/3
TH2: -4x +13 = 0 ⇔ 13/4
Vậy S= {2/3; 13/4}
(5x-3)² - (4x-7)² = 0
⇔ 25x² -30x +9 -16x² +56x -49 = 0
⇔ 9x² +26x -40 = 0
⇔ 9x² +36x -10x -40 = 0
⇔ 9x.(x +4) -10.(x+4) = 0
⇔ (x+4).(9x -10) = 0
TH1: x+4 = 0 ⇔ x = -4
TH2: 9x -10 = 0 ⇔ x = 10/9
Vậy S= {10/9; -4}
4x² - 12x + 5 = 0
⇔ 4x² -2x -10x +5 = 0
⇔ 2x(2x -1) -5.(2x -1) = 0
⇔ (2x-1).(2x-5) = 0
TH1: 2x -1 = 0 ⇔ x = 1/2
TH2: 2x- 5 = 0 ⇔ x = 5/2
Vậy S= {1/2; 5/2}
