Question

(х-2) (х-3) (х-4) > 0​

Answer 1

Ответ:

1(x-2)*1(x-3)*1(x-4)>0

x-2*x-3*x-4>0

x+x+x=2+3+4

3x=9

x=9:3

x=3

Объяснение:

Я не понял зачем тут >0 поэтому вот решение это примера

Answer 2

To solve the inequality $(x-2)(x-3)(x-4) > 0$, we can use the concept of interval notation and the properties of polynomial functions.

First, let's find the critical points of the polynomial, where it changes sign. These critical points occur when any of the factors $(x-2)$, $(x-3)$, or $(x-4)$ equals zero. So, we have three critical points:

1. $x = 2$
2. $x = 3$
3. $x = 4$

Now, we can create a sign chart to determine the intervals where the inequality is satisfied. We'll consider four intervals based on these critical points: $(-\infty, 2)$, $(2, 3)$, $(3, 4)$, and $(4, \infty)$.

1. For $x < 2$: All three factors are negative, so the product $(x-2)(x-3)(x-4)$ is negative.

2. For $2 < x < 3$: The factor $(x-2)$ is positive, while $(x-3)$ and $(x-4)$ are negative, so the product is positive.

3. For $3 < x < 4$: All three factors are positive, so the product is positive.

4. For $x > 4$: All three factors are positive, so the product is positive.

So, the solution to the inequality $(x-2)(x-3)(x-4) > 0$ is the union of the intervals where the product is greater than zero:

$(-\infty, 2) \cup (2, 3) \cup (3, 4) \cup (4, \infty)$

In interval notation, this solution can be written as:

$(- \infty, 2) \cup (2, 3) \cup (3, 4) \cup (4, \infty)$