Question

Решить неравенство6 (3x -2) (7-8x) (4-10x) <0

Answer 1

Решение во вложении.

Answer 2

To solve the inequality $(3x - 2)(7 - 8x)(4 - 10x) < 0$, we need to find the intervals for $x$ where the expression is negative.

Step 1: Find the critical points where the expression becomes zero. To do this, set each factor equal to zero and solve for $x$:

$3x - 2 = 0 \Rightarrow x = \frac{2}{3}$

$7 - 8x = 0 \Rightarrow x = \frac{7}{8}$

$4 - 10x = 0 \Rightarrow x = \frac{4}{10} = \frac{2}{5}$

Step 2: Determine the intervals created by the critical points and test each interval to see whether the expression is positive or negative. We can create a sign chart:

$$\begin{array}{c|cccccccc} \textrm{Interval} & (-\infty, \frac{2}{5}) & \left(\frac{2}{5}, \frac{2}{3}\right) & \left(\frac{2}{3}, \frac{7}{8}\right) & \left(\frac{7}{8}, \infty\right) \\ \hline 3x - 2 & - & - & + & + \\ 7 - 8x & + & + & + & - \\ 4 - 10x & + & + & - & - \\ \hline \textrm{Expression} & - & + & - & + \end{array}$$

Step 3: Analyze the sign chart to find the intervals where the expression $(3x - 2)(7 - 8x)(4 - 10x)$ is negative ($< 0$).

The inequality is satisfied in two intervals:

1. $\frac{2}{5} < x < \frac{2}{3}$
2. $x > \frac{7}{8}$

So the solution to the inequality is $\frac{2}{5} < x < \frac{2}{3}$ and $x > \frac{7}{8}$.