1. **State the problem:** Solve the inequality $$\frac{2x}{(x+2)(x-2)(x-1)} < 0.$$\n\n2. **Identify critical points:** The expression is undefined or zero at points where the numerator or denominator is zero. These are $$x=0, x=-2, x=2, x=1.$$\n\n3. **Determine intervals:** The critical points divide the real line into intervals: $$(-\infty,-2), (-2,0), (0,1), (1,2), (2,\infty).$$\n\n4. **Analyze sign on each interval:** The numerator is $$2x$$, so sign depends on $$x$$. The denominator is product of three factors: $$(x+2), (x-2), (x-1).$$\n\n5. **Sign of numerator:** Positive if $$x>0$$, negative if $$x<0$$.\n\n6. **Sign of denominator:** Evaluate sign of each factor on intervals:\n- For $$x<-2$$: $$(x+2)<0, (x-2)<0, (x-1)<0$$ so denominator sign is $$(-)(-)(-) = -$$.\n- For $$-20, (x-2)<0, (x-1)<0$$ so denominator sign is $$(+)(-)(-) = +$$.\n- For $$00, (x-2)<0, (x-1)<0$$ same as above, denominator is $$+$$.\n- For $$10, (x-2)<0, (x-1)>0$$ denominator sign is $$(+)(-)(+) = -$$.\n- For $$x>2$$: all factors positive, denominator sign is $$+$$.\n\n7. **Combine numerator and denominator signs:**\n- $$x<-2$$: numerator negative, denominator negative, fraction positive.\n- $$-22$$: numerator positive, denominator positive, fraction positive.\n\n8. **Inequality $$<0$$ holds where fraction is negative:**\n$$-2 < x < 0$$ and $$1 < x < 2.$$\n\n9. **Check points where denominator zero:** excluded from solution. Numerator zero at $$x=0$$, fraction zero, but inequality is strict $$<0$$, so exclude $$x=0$$.\n\n**Final solution:** $$\boxed{(-2,0) \cup (1,2)}.$$