(х+1)(х+2)(х+3)/(х+4)(3-х)(2х+1)>0

Предыдущий
вопрос Следующий
вопрос

Let's find the critical points first by setting each denominator equal to zero:

x + 4 = 0
x = -4

3 - x = 0
x = 3

2x + 1 = 0
2x = -1
x = -1/2

Now, we create a number line with these critical points (-4, -1/2, 3) and test the intervals between them to find where the expression is greater than zero.

Test x = -5:
(0)(-1)(-9)/(0)(8)(-1) = 0 < 0, so this interval is not included in the solution.

Test x = -1:
(0)(1)(2)/(0)(4)(0) = 0 < 0, so this interval is not included in the solution.

Test x = 0:
(1)(2)(3)/(4)(3)(1) = 6/12 > 0, so this interval is included in the solution.

Test x = 2:
(3)(4)(5)/(6)(1)(5) = 60/30 > 0, so this interval is included in the solution.

Test x = 4:
(5)(6)(7)/(8)(-1)(9) = -210/ -72 > 0, so this interval is included in the solution.

Therefore, the solution to the inequality (х+1)(х+2)(х+3)/(х+4)(3-х)(2х+1) > 0 is:
x ∈ (-4, -1/2) U (3, ∞)