Question

(x-5)(3x-9)(7 - x) < 0

Answer 1

Ответ:

(х-5)(3х-9)(7-х)<0

х-5<0

х<5

3х-8<0

х<3

7-х<0

-х<7(*-1)

х>7

Answer 2

To solve the inequality $(x-5)(3x-9)(7 - x) < 0$, we need to find the intervals where the expression is negative.

First, let's find the critical points by setting each factor equal to zero:

$x - 5 = 0$ => $x = 5$

$3x - 9 = 0$ => $x = 3$

$7 - x = 0$ => $x = 7$

These critical points divide the number line into four intervals: $(-\infty, 3)$, $(3, 5)$, $(5, 7)$, and $(7, +\infty)$.

Now, we'll choose test points from each interval and evaluate the expression $(x-5)(3x-9)(7 - x)$ to determine the sign in each interval.

For $(-\infty, 3)$, we can choose $x = 0$:

$(0-5)(3(0)-9)(7 - 0) = (-5)(-9)(7) > 0$

The expression is positive in this interval.

For $(3, 5)$, we can choose $x = 4$:

$(4-5)(3(4)-9)(7 - 4) = (-1)(3)(3) < 0$

The expression is negative in this interval.

For $(5, 7)$, we can choose $x = 6$:

$(6-5)(3(6)-9)(7 - 6) = (1)(9)(1) > 0$

The expression is positive in this interval.

For $(7, +\infty)$, we can choose $x = 8$:

$(8-5)(3(8)-9)(7 - 8) = (3)(15)(-1) < 0$

The expression is negative in this interval.

Therefore, the solution to the inequality $(x-5)(3x-9)(7 - x) < 0$ is $x \in (3, 5) \cup (7, +\infty)$.