Question

(16-x^2)(x^2+4)(x^2+x+1)(x^2-x-3)<=0

Answer 1

Ответ:

я конечно не знаю разберешься ли ты в этих записях или нет

Объяснение:

если поставишь лучший будет круто

Answer 2

(16-x²)(x²+4)(x²+x+1)(x²-x-3) ≤0

1) x²+4 >0 при любом х ( 4>0 ,x²>0), корней нет

2) x²+x+1>0 при любом х , тк D=1-4=-3<0 ,y=x²+x+1  выше оси ох, корней нет

3)x²-x-3=0 ,D=1+12=13 ,x₁=(1-√13)/2 ,   x₂= (1+√13)/2

(4-х)(4+х)(x²+4)(x²+x+1)(х-(1+√13)/2)(х- (1-√13)/2 )≤0

по методу интервалов.

-                  +                        _                          +

/ / / / / /[-4]_____[(1-√13)/2]/ / / / / / [(1+√13)/2]_____

Ответх∈[-∞ ; -4] ,[ (1-√13)/2 ;  (1+√13)/2 ]                                            

Answer 3

To find the solution to the inequality $(16 - x^2)(x^2 + 4)(x^2 + x + 1)(x^2 - x - 3) \leq 0$, we need to determine the intervals on the real number line where this inequality is satisfied.

First, let's find the critical values by setting each factor equal to zero and solving for (x:

1. $16 - x^2 = 0$ when $x = 4$ or $x = -4$.
2. $x^2 + 4 = 0$ has no real solutions.
3. $x^2 + x + 1 = 0$ has no real solutions.
4. $x^2 - x - 3 = 0$ when $x = 3$ or $x = -1$.

Now, we'll create a sign chart using these critical values to test the sign of the expression $(16 - x^2)(x^2 + 4)(x^2 + x + 1)(x^2 - x - 3)$ within the intervals separated by the critical values.

We have the following intervals:

1. $(-\infty, -4)$
2. $(-4, -1)$
3. $(-1, 3)$
4. $(3, 4)$
5. $(4, \infty)$

Now, let's test the sign of each factor within these intervals:

1. $(16 - x^2)$ is positive for $x < -4$ and $x > 4$ and negative for $-4 < x < 4$.
2. $(x^2 + 4)$ is always positive for real $x$.
3. $(x^2 + x + 1)$ is always positive for real $x$.
4. $(x^2 - x - 3)$ is positive for $-1 < x < 3$ and negative for $x < -1$ and $x > 3$.

Now, let's combine the signs of all factors in each interval:

1. $(16 - x^2)(x^2 + 4)(x^2 + x + 1)(x^2 - x - 3)$ is negative in the interval $(-4, -1)$.
2. $(16 - x^2)(x^2 + 4)(x^2 + x + 1)(x^2 - x - 3)$ is positive in the interval $(-1, 3)$.

Since the inequality is less than or equal to zero ($\leq 0$), we are interested in the intervals where the expression is non-positive (negative or zero). Therefore, the solution to the inequality is $x \in [-4, -1] \cup [4, \infty)$. These are the values of $x$ that make the given expression non-positive.