Question

2х-1 --------+1<0 2х-3

Answer 1

(2x-1)÷(2x-3)+1<0
(2x-1)÷(2x-3)+1=-1
(2x-1)÷(2x-3)=-1-1
(2x-1)÷(2x-3)=-2
(Значит (2х-1) больше чем (2х-3) в 2 раза)
x=2,5

Answer 2

To solve the inequality $\frac{2x - 1}{x + 1} < 0$, we can follow these steps:

Step 1: Find the critical points where the expression in the inequality becomes zero or undefined. In this case, the critical point is when the denominator $(x + 1)$ becomes zero: $x + 1 = 0$ $x = -1$

Step 2: Create a number line with the critical point $x = -1$ and any other relevant points.

Number Line: _____|_______| -2 -1 0 1

Step 3: Choose a test point in each interval created by the critical points. Test points can be any value within each interval.

Test Point 1 (interval: -∞ to -1): Choose $x = -2$ Test Point 2 (interval: -1 to 0): Choose $x = -0.5$ Test Point 3 (interval: 0 to 1): Choose $x = 0.5$ Test Point 4 (interval: 1 to +∞): Choose $x = 2$

Step 4: Plug in each test point into the original inequality and determine the sign of the expression $\frac{2x - 1}{x + 1}$.

For Test Point 1 ($x = -2$): $\frac{2(-2) - 1}{(-2) + 1} = \frac{-5}{-1} = 5$ (Positive)

For Test Point 2 ($x = -0.5$): $\frac{2(-0.5) - 1}{(-0.5) + 1} = \frac{-2}{0.5} = -4$ (Negative)

For Test Point 3 ($x = 0.5$): $\frac{2(0.5) - 1}{(0.5) + 1} = 0$ (Zero)

For Test Point 4 ($x = 2$): $\frac{2(2) - 1}{2 + 1} = \frac{3}{3} = 1$ (Positive)

Step 5: Determine the sign of the expression for each interval based on the test points:

$\frac{2x - 1}{x + 1}$ is positive in the interval $(-2, -1)$ and $(1, \infty)$. $\frac{2x - 1}{x + 1}$ is negative in the interval $(-1, 0)$. $\frac{2x - 1}{x + 1}$ is zero at $x = 0$.

Step 6: Identify the solution set where $\frac{2x - 1}{x + 1} < 0$, which means the expression is negative.

The solution set for $\frac{2x - 1}{x + 1} < 0$ is $x \in (-1, 0)$.