Question

(x^2-3x+2)(x^3-2x^2)(4-x^2)<-0

Answer 1

(x²-3x+2)(x³-2x²)(4-x²)≤0

x²-3x+2=0, по теореме, обратной теореме Виета х=1; х=2

(х-2)(х-1)х²(х-2)(2-х)(2+х)≤0

(х-2)³(х-1)х²(2+х)≥0

Решим неравенство методом интервалов.

_______-2_____0___1____2________

-                 +         +       -              +

х∈[-2;1]∪[2;+∞)

Answer 2

Ответ:

Пошаговое объяснение:

Решение на фото

Answer 3

To find the values of x that satisfy the inequality $(x^2-3x+2)(x^3-2x^2)(4-x^2) < 0$, we can follow these steps:

Step 1: Determine the critical points by setting each factor equal to zero and solving for x.

For the factor $x^2 - 3x + 2$, we have: $x^2 - 3x + 2 = 0$ Factoring the quadratic, we get: $(x-1)(x-2) = 0$

So, the critical points for this factor are $x = 1$ and $x = 2$.

For the factor $x^3 - 2x^2$, we have: $x^3 - 2x^2 = 0$ Factoring out $x^2$, we get: $x^2(x - 2) = 0$

So, the critical points for this factor are $x = 0$ and $x = 2$.

For the factor $4 - x^2$, we have: $4 - x^2 = 0$ Rearranging, we get: $x^2 - 4 = 0$ Factoring the quadratic, we get: $(x-2)(x+2) = 0$

So, the critical points for this factor are $x = -2$ and $x = 2$.

Step 2: Plot the critical points on a number line.

-2 0 1 2

Step 3: Choose test points within each interval and evaluate the sign of the expression $(x^2-3x+2)(x^3-2x^2)(4-x^2)$ for each test point.

Let's choose test points: -3, -1, 1.5, and 3.

For $x = -3$, the expression becomes: $(-)(-)(+)$ which is positive.

For $x = -1$, the expression becomes: $(+)(-)(+)$ which is negative.

For $x = 1.5$, the expression becomes: $(+)(-)(+)$ which is negative.

For $x = 3$, the expression becomes: $(+)(+)(+)$ which is positive.

Step 4: Determine the intervals where the expression is negative.

Based on the signs of the expression for the test points, we find that the intervals where the expression $(x^2-3x+2)(x^3-2x^2)(4-x^2) < 0$ are: -2 < x < 0 and 1 < x < 2.

Therefore, the solution to the inequality is $-2 < x < 0$ and $1 < x < 2$.