Question

2<|x+1|<5 пжжжжжжжжжжддддж​

Answer 1

1) x+1<0, x<-1

на этом интервале выражение а модуле отрицательно, значит

-x-1>2 и -x-1<5

-x>3 и -x<6

x<-3 и x>-6. получаем, x принадлежит (-6;-3) и x<-1, то есть

x=(-6;-3)

2) x+1>=0, x>=-1

здесь модуль раскрывается со знеаком +

x+1>2 и x+1<5

x>1 и x<4 получаем х пинеадлежит (1;4) и x>=-1,

x=(1;4)

3) объединяя полученные результаты,

x прианадлежит (-6;-3) в объединении с (1;4)

Answer 2

To solve the inequality 2 < |x + 1| < 5, we need to break it down into two separate inequalities and solve each part individually. Remember that the absolute value of a number is always non-negative.

First, let's deal with the left side of the inequality:

1. 2 < |x + 1|

To remove the absolute value, we consider two cases:

Case 1: x + 1 is positive (x + 1 > 0) 2 < x + 1

Subtract 1 from both sides: 2 - 1 < x + 1 - 1 1 < x

Case 2: x + 1 is negative (x + 1 < 0) 2 < -(x + 1)

Multiply both sides by -1 (and reverse the inequality): -2 > x + 1

Subtract 1 from both sides: -2 - 1 > x -3 > x

So, the first part of the inequality is: -3 < x < 1

Next, let's deal with the right side of the inequality:

2. |x + 1| < 5

Since the absolute value of any number is always less than 5, this inequality is satisfied for all real numbers.

Combining the results from both parts, the final solution is:

-3 < x < 1

In interval notation, this can be expressed as: (-3, 1).