c. x²(x - 3) + 12 - 4x = 0
x²(x - 3) - (4x - 12) = 0
x²(x - 3) - 4(x - 3) = 0
(x² - 4)(x - 3) = 0
TH1: x² - 4 = 0
x² = 0 + 4
x² = 4
⇒ x² = 2² hay x² = (-2)²
⇒ x = 2 hay x = -2
TH2: x - 3 = 0
x = 0 + 3 = 3
d. 6x² - 15x - (2x - 5)(2x + 5) = 0
3x(2x - 5) - (2x - 5)(2x + 5) = 0
(2x - 5)[3x - (2x + 5)] = 0
(2x - 5)(3x - 2x - 5) = 0
(2x - 5)(x - 5) = 0
TH1: 2x - 5 = 0
2x = 0 + 5
2x = 5
x = 5 : 2 = $\frac{5}{2}$
e. (x - 3)(x + 3) - x(x + 5) = 15
(x² - 9) - (x² + 5x) = 15
x² - 9 - x² - 5x = 15
-9 - 5x = 15
5x = -9 - 15
5x = -24
x = -24 : 5 = $\frac{-24}{5}$
f. x² - 9 + (x + 3)(x - 9) = 0
(x² - 9) + (x + 3)(x - 9) = 0
(x + 3)(x - 3) + (x + 3)(x - 9) = 0
(x + 3)[(x - 3) + (x - 9)] = 0
(x + 3)(x - 3 + x - 9) = 0
(x + 3)(2x - 12) = 0
TH1: x + 3 = 0
x = 0 - 3 = -3
TH2: 2x - 12 = 0
2x = 0 + 12
2x = 12
x = 12 : 2 = 6
