Question

Решите неравенства и запишите множество их решении 1) 4х +1| < 7;2) 2x + 3 = 4;(3) Ix +1] < 2, 5;0) 2x – 5 = 3;​

Answer 1

0) 2х-5=3

2х=5+3

2х=8

х=8:2

х=4

Проверка:2•4-5=3

Ответ:4.

Answer 2

1. 4x + 1 < 7 Subtract 1 from both sides: 4x < 6 Divide both sides by 4 (since the coefficient of x is 4 and it's positive): x < 6/4 Simplify the right side: x < 3/2

   4x + 1 > -7 Subtract 1 from both sides: 4x > -8 Divide both sides by 4: x > -2

   The solution set is (-2, 3/2).

2. 2x + 3 = 4 Subtract 3 from both sides: 2x = 1 Divide both sides by 2: x = 1/2

   The solution is x = 1/2.

3. |x + 1| < 2 There are two cases to consider:

   Case 1: x + 1 > 0 In this case, the absolute value becomes x + 1. x + 1 < 2 Subtract 1 from both sides: x < 1

   Case 2: x + 1 < 0 In this case, the absolute value becomes -(x + 1). -(x + 1) < 2 Multiply both sides by -1 (note the change in direction of the inequality): x + 1 > -2 Subtract 1 from both sides: x > -3

   Combining the solutions from both cases, the solution set is (-3, 1).

4. 2x - 5 = 3 Add 5 to both sides: 2x = 8 Divide both sides by 2: x = 4

   The solution is x = 4.