1) IxI=7;2) Ix - 3I = 5;3) Ix - 1I = - 3;4) Ix + 7I = 12

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вопрос

To solve these equations, we need to separate the absolute value bars and look at the two possible cases for each equation.

1) |x| = 7
This equation has two possible solutions:
x = 7 or x = -7

2) |x - 3| = 5
This equation has two possible solutions based on the two possible cases:
x - 3 = 5 or x - 3 = -5
Solving each case:
Case 1: x - 3 = 5
x = 8
Case 2: x - 3 = -5
x = -2

3) |x - 1| = -3
This equation has no real solutions since the absolute value of a real number cannot be negative.

4) |x + 7| = 12
This equation has two possible solutions based on the two possible cases:
x + 7 = 12 or x + 7 = -12
Solving each case:
Case 1: x + 7 = 12
x = 5
Case 2: x + 7 = -12
x = -19

Therefore, the solutions to the given equations are:
1) x = 7 or x = -7
2) x = 8 or x = -2
3) No real solutions
4) x = 5 or x = -19