Question

Решить неравенство (x-4)*(4x+1)*(4x+3)*(5x-2)<0

Answer 1

(x-4)*(4x+1)*(4x+3)*(5x-2) < 0
метод интервалов
х - 4 = 0     4х +1 = 0         4х +3 = 0       5х -2 = 0
х = 4            х = -1/4           х = -3/4            х = 2/5
-∞               -3/4            -1/4            2/5           4         +∞
           -                  -                 -               -           +         это знаки(х -4)
           -                  -                 +              +          +         это знаки (4х +1)  
           -                  +                +               +          +         это знаки( 4х +3)
           -                  -                  -               +          +         это знаки (5х -2)
                    IIIIIIIIIIIIIIIIIIII                IIIIIIIIIIIIIIII                это решение
Ответ: (-3/4; -1/4)∪(2/5; 4)
    

Answer 2

To solve the inequality $(x-4)(4x+1)(4x+3)(5x-2) < 0$, we need to find the intervals for which the expression is less than zero (i.e., negative).

To do this, we can use the concept of interval notation. We need to find the intervals on the number line where the expression $(x-4)(4x+1)(4x+3)(5x-2)$ is negative.

Step 1: Find the critical points by setting each factor to zero and solving for $x$:

1. $x - 4 = 0 \Rightarrow x = 4$
2. $4x + 1 = 0 \Rightarrow x = -\frac{1}{4}$
3. $4x + 3 = 0 \Rightarrow x = -\frac{3}{4}$
4. $5x - 2 = 0 \Rightarrow x = \frac{2}{5}$

Step 2: Use these critical points to create intervals on the number line and test points within each interval to determine the sign of $(x-4)(4x+1)(4x+3)(5x-2)$:

Let's construct the number line with the critical points:

$$\begin{array}{ccccccc} -\infty & \quad & -\frac{3}{4} & \quad & -\frac{1}{4} & \quad & 4 & \quad & \frac{2}{5} & \quad & \infty \\ \end{array}$$

Step 3: Choose a test point within each interval and check the sign of the expression $(x-4)(4x+1)(4x+3)(5x-2)$:

Test point in $(-\infty, -\frac{3}{4})$: $x = -1$ $((-1) - 4)(4(-1) + 1)(4(-1) + 3)(5(-1) - 2) = (-5)(-3)(-1)(-7) = 105 > 0$

Test point in $(- \frac{3}{4}, -\frac{1}{4})$: $x = -\frac{1}{2}$ $((-\frac{1}{2}) - 4)(4(-\frac{1}{2}) + 1)(4(-\frac{1}{2}) + 3)(5(-\frac{1}{2}) - 2) = (-\frac{9}{2})(-\frac{3}{2})(\frac{5}{2})(-\frac{9}{2}) = \frac{6075}{16} > 0$

Test point in $(- \frac{1}{4}, 4)$: $x = 0$ $(0 - 4)(4(0) + 1)(4(0) + 3)(5(0) - 2) = (-4)(1)(3)(-2) = 24 > 0$

Test point in $(4, \frac{2}{5})$: $x = 1$