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a) |x-3|≤4 ⇔ -4 ≤ x-3 ≤ 4
-4+3 ≤ x ≤ 4+3
-1 ≤ x ≤ 7
Odp x∈ <-1;7>
dla a>0, |w|>a ⇔ w>a lub w<-a więc
|x+2|>5 ⇔
x+2>5 lub x+2 < -5 tzn
x>3 lub x<-7
Odp. x∈(-∞;-7)U(3;+∞)