Răspuns :
|x-3|•(|x+1| - 5) < 0
|x-3| ≠ 0 => x ≠ 3
|x-3|•(|x+1| - 5) < 0 / :|x-3| {|x-3| > 0}
|x+1| - 5 < 0
|x+1| < 5
-5 < x+1 < 5 \ (-1)
-6 < x < 4
=> S = {-5,-4,-3,-2,-1,0,1,2}
|x-3| ≠ 0 => x ≠ 3
|x-3|•(|x+1| - 5) < 0 / :|x-3| {|x-3| > 0}
|x+1| - 5 < 0
|x+1| < 5
-5 < x+1 < 5 \ (-1)
-6 < x < 4
=> S = {-5,-4,-3,-2,-1,0,1,2}