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Ix+1I=3 ⇔ x+1=3 ∨ x+1= -3
x=3-1 ∨ x= -3-1
x=2 ∨ x= -4
Pierwiastkiem rownania jest : -4 ; 2
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Ix-4I < 7 ⇔ x-4 < 7 ∧ x-4 > -7
x < 7+4 ∧ x > -7+4
x < 11 ∧ x > -3
x ∈ ( -3 ; 11)
|x + 1| = 3
x + 1 = 3 lub x + 1 = -3
x = 2 lub x = -4
|x - 4| < 7
x - 4 < 7 i x - 4 > -7
x < 11 i x > -3
x ∈ (-3,11)