Solve ∣x∣−3∣x∣+2<0\frac{|x|-3}{|x|+2} < 0∣x∣+2∣x∣−3<0.
(-3, 3)
(-2, 2)
(-3, 3) \setminus {0}
(-\infty, -3) \cup (3, \infty)