(x² - 5x + 6)(x² - 1) > 0
первая скобка
D=25-24
x12=(5+-1)/2 = 2 3
x² - 5x + 6 = (x - 2)(x - 3)
a² - b² = (a - b)(a + b)
x² - 1= (x - 1)(x + 1)
(x - 2)(x - 3)(x - 1)(x + 1) > 0
(-1) (1) (2) (3)
x∈(-∞, -1) U (1, 2) U (3, +∞)
(x² - 5x + 6)(x² - 1) > 0
первая скобка
D=25-24
x12=(5+-1)/2 = 2 3
x² - 5x + 6 = (x - 2)(x - 3)
a² - b² = (a - b)(a + b)
x² - 1= (x - 1)(x + 1)
(x - 2)(x - 3)(x - 1)(x + 1) > 0
(-1) (1) (2) (3)
x∈(-∞, -1) U (1, 2) U (3, +∞)