Question

Решить неравенство1) |1+x|≤0,32) |2+x|<0,23) |x+0,5|<1,5

Answer 1

1) 1+x≤0.3 или 1+х≥-0.3
х≤-0.7 х≥-1.3
Ответ: [-1.3;-0.7]
2) 2+х<0.2 или 2+х>-0.2
х<-1.8 х>-2.2
Ответ: (-2.2;-1.8)
3) х+0.5<1.5 или х+0.5>-1.5
х<1 х>-2
Ответ: (-2;1)

Answer 2

To solve the given inequalities, we need to consider different cases based on the absolute value expressions. Let's break it down step by step:

1. |1+x| ≤ 0.3

For an absolute value expression to be less than or equal to a constant, the absolute value must be either zero or negative. Since the absolute value of any number is always non-negative, the only possibility is when the expression inside the absolute value is zero.

1 + x = 0

Solving for x:

x = -1

So, the solution for the first inequality is x = -1.

2. |2+x| < 0.2

Similarly, the absolute value must be less than 0.2, which means the expression inside the absolute value must be between -0.2 and 0.2 (exclusive).

-0.2 < 2 + x < 0.2

Subtract 2 from all parts:

-2.2 < x < -1.8

The solution for the second inequality is -2.2 < x < -1.8.

3. |x + 0.5| < 1.5

Here, the absolute value is less than 1.5, so the expression inside the absolute value must be between -1.5 and 1.5 (exclusive).

-1.5 < x + 0.5 < 1.5

Subtract 0.5 from all parts:

-2 < x < 1

The solution for the third inequality is -2 < x < 1.

So, to summarize:

1. x = -1
2. -2.2 < x < -1.8
3. -2 < x < 1