Question

(x+7)(x+1)(x-4)<0 решить

Answer 1

Решение на фото........

Answer 2

Заменяем знак "<" знаком "="

(x+7)(x+1)(x-4)=0

Произведение равно нулю, когда хотя бы один из множителей равен нулю:

$\begin{bmatrix}x+7=0 \\ x+1=0 \\ x-4=0\\\end{matrix} \ \Leftrightarrow \ \begin{bmatrix} x=-7\\ x=-1\\ x=4 \end{matrix}$

Находим решения методов интервалов:

----(-7)++++(-1)----(4)+++++>x

ОТВЕТ: х∈(-∞;-7) ∪ (-1;4)

Answer 3

To solve the inequality $(x+7)(x+1)(x-4) < 0$, we need to find the intervals where the expression on the left-hand side is negative.

To do this, we can use the concept of intervals and test the sign of the expression in each interval. The inequality will be satisfied when the expression is negative (less than zero).

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

1. $x + 7 = 0 \Rightarrow x = -7$
2. $x + 1 = 0 \Rightarrow x = -1$
3. $x - 4 = 0 \Rightarrow x = 4$

Step 2: Plot these critical points on a number line and divide it into intervals:

$-\infty \quad -7 \quad -1 \quad 4 \quad \infty$

Step 3: Test the sign of the expression $(x+7)(x+1)(x-4)$ in each interval:

• Interval 1: $-\infty < x < -7$ Pick a test point, e.g., $x = -8$ $((-8)+7)((-8)+1)((-8)-4) = (-1)(-7)(-12) = -84$ (negative)
• Interval 2: $-7 < x < -1$ Pick a test point, e.g., $x = -5$ $((-5)+7)((-5)+1)((-5)-4) = (2)(-4)(-9) = 72$ (positive)
• Interval 3: $-1 < x < 4$ Pick a test point, e.g., $x = 0$ $(0+7)((0)+1)((0)-4) = (7)(1)(-4) = -28$ (negative)
• Interval 4: $4 < x < \infty$ Pick a test point, e.g., $x = 5$ $(5+7)((5)+1)((5)-4) = (12)(6)(1) = 72$ (positive)

Step 4: Determine the solution based on the sign of the expression in each interval:

The inequality $(x+7)(x+1)(x-4) < 0$ is satisfied when the expression is negative, i.e., in Intervals 1 and 3.

So, the solution to the inequality is $x \in (-\infty, -7) \cup (-1, 4)$.