Question

(2-х)^3(х+2)^2(х-3)≥0​

Answer 1

$(2 - x)^{3} (x + 2)^{2} (x - 3) \geqslant 0 \\ - (x - 2)^{3} (x + 2)^{2} (x - 3) \geqslant 0 \\ (x - 2)^{3} (x + 2)^{2} (x - 3) \leqslant 0$

Ответ:

$x \in[2;3]\cup(-2)$

Answer 2

To solve the inequality $(2-x)^3 (x+2)^2 (x-3) \geq 0$, we need to find the intervals on the number line where the expression is non-negative (greater than or equal to zero).

First, let's find the critical points where the expression changes sign. These are the values of $x$ where any of the factors change from negative to positive or vice versa.

1. $(2 - x)^3 = 0$ when $x = 2$.
2. $(x + 2)^2 = 0$ when $x = -2$.
3. $(x - 3) = 0$ when $x = 3$.

Now we have four intervals to consider based on these critical points: $(-\infty, -2)$, $(-2, 2)$, $(2, 3)$, and $(3, \infty)$.

Next, we need to determine the sign of each factor in each interval.

1. In the interval $(-\infty, -2)$:

   • $(2 - x)^3$ is positive (since $2 - x$ is negative and odd power doesn't change the sign).
   • $(x + 2)^2$ is positive (since $x + 2$ is negative and even power preserves the sign).
   • $(x - 3)$ is negative (since $x - 3$ is negative).

   So, $(2 - x)^3 (x + 2)^2 (x - 3) < 0$ in this interval.

2. In the interval $(-2, 2)$:

   • $(2 - x)^3$ is negative.
   • $(x + 2)^2$ is positive.
   • $(x - 3)$ is negative.

   So, $(2 - x)^3 (x + 2)^2 (x - 3) > 0$ in this interval.

3. In the interval $(2, 3)$:

   • $(2 - x)^3$ is positive.
   • $(x + 2)^2$ is positive.
   • $(x - 3)$ is negative.

   So, $(2 - x)^3 (x + 2)^2 (x - 3) < 0$ in this interval.

4. In the interval $(3, \infty)$:

   • $(2 - x)^3$ is positive.
   • $(x + 2)^2$ is positive.
   • $(x - 3)$ is positive.

   So, $(2 - x)^3 (x + 2)^2 (x - 3) > 0$ in this interval.

To summarize:

• The expression $(2 - x)^3 (x + 2)^2 (x - 3)$ is less than zero ($< 0$) in the intervals $(-\infty, -2)$ and $(2, 3)$.
• The expression $(2 - x)^3 (x + 2)^2 (x - 3)$ is greater than zero ($> 0$) in the interval $(-2, 2)$ and $(3, \infty)$.

So, the solution to the inequality $(2 - x)^3 (x + 2)^2 (x - 3) \geq 0$ is $x \in (-2, 2] \cup [3, \infty)$.