Question

(6x+18)(4-16x)(7x-21)(5-2x)≥0

Answer 1

Ответ: ..........

Удачи!!!

Answer 2

To solve the inequality $(6x+18)(4-16x)(7x-21)(5-2x) \geq 0$, we need to find the intervals for which the expression is non-negative.

First, let's factorize the expression:

$(6x+18)(4-16x)(7x-21)(5-2x) = 36(x+3)(2-x)(7x-21)(5-2x)$

We can see that the expression is zero when any of the factors are zero. Therefore, the critical values occur at $x = -3$, $x = 2$, $x = 3$, and $x = \frac{21}{7} = 3$.

We'll divide the number line into intervals based on these critical values and test the sign of the expression within each interval.

Interval 1: $(-\infty, -3)$ Let's pick a value in this interval, for example, $x = -4$. Plugging it into the expression:

$(36(-4+3)(2-(-4))(7(-4)-21)(5-2(-4)) = 36(-1)(6)(-28)(13) = -272,16 > 0$

Since the expression is negative in this interval, it doesn't satisfy the inequality.

Interval 2: $(-3, 2)$ Let's pick a value in this interval, for example, $x = 0$. Plugging it into the expression:

$(36(0+3)(2-(0))(7(0)-21)(5-2(0)) = 36(3)(2)(-21)(5) = -22680 < 0$

Since the expression is negative in this interval, it doesn't satisfy the inequality.

Interval 3: $(2, \frac{21}{7})$ Let's pick a value in this interval, for example, $x = 3$. Plugging it into the expression:

$(36(3+3)(2-(3))(7(3)-21)(5-2(3)) = 36(6)(-1)(0)(-1) = 0$

Since the expression is zero in this interval, it satisfies the inequality.

Interval 4: $(\frac{21}{7}, \infty)$ Let's pick a value in this interval, for example, $x = 4$. Plugging it into the expression:

$(36(4+3)(2-(4))(7(4)-21)(5-2(4)) = 36(7)(-2)(7)(-3) = 31752 > 0$

Since the expression is positive in this interval, it satisfies the inequality.

Therefore, the solution to the inequality $(6x+18)(4-16x)(7x-21)(5-2x) \geq 0$ is:

$\frac{21}{7} \leq x < 2$ or $x \geq 4$