Question

Решить систему неравенств: 1) 3 – х ≤ 2 2) х² – 1 ≥ 0 3) 9 – х² ≤ 0 2х + 1 ≤ 4 х > 2 х + 5 < 0

Answer 1

Вот, надеюсь сможешь разобрать.

Answer 2

1. 3 - x ≤ 2

Subtracting 3 from both sides, we get:

• x ≤ -1

So the solution for this inequality is:

x ∈ (-∞, -1]

2. x² - 1 ≥ 0

We can factor this inequality as:

(x - 1)(x + 1) ≥ 0

The critical values of x are -1 and 1, so we can make a sign table:

x	x - 1	x + 1	(x - 1)(x + 1)
< -1	-	-	+
-1	-	0	0
-1<x<1	-	+	-
1	0	+	0
> 1	+	+	+

The solution is:

x ∈ (-∞, -1] ∪ [1, +∞)

3. 9 - x² ≤ 0

Adding x² to both sides, we get:

x² ≥ 9

Taking the square root of both sides, we get:

|x| ≥ 3

So the solution is:

x ∈ (-∞, -3] ∪ [3, +∞)

4. 2x + 1 ≤ 4x

Subtracting 2x from both sides, we get:

1 ≤ 2x

Dividing both sides by 2, we get:

1/2 ≤ x

So the solution is:

x ∈ [1/2, +∞)

5. x + 5 < 0

Subtracting 5 from both sides, we get:

x < -5

So the solution is:

x ∈ (-∞, -5)