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How would one proceed in solving this difficult inequation with multiple absolute values? is there a way one should proceed ?

$$\frac{x}{||x|-2|} \le \frac{x-1}{|x-3|}$$

thanks

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1 Answer 1

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Hint

Discuss several cases: $$x\in(-\infty,-2)\cup(-2 ,0]\; ; \; x\in(0 ,2)\; ;\; x\in(2 ,3)\; ;\; x\in(3 ,+\infty)$$

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    $\begingroup$ @franck It might also help to cross multiply the fractions as well. $\endgroup$ Aug 14, 2013 at 20:03
  • $\begingroup$ @Ataraxia how come don't we consider the case from x ∈ (1,2) ? $\endgroup$
    – franck
    Aug 14, 2013 at 20:06
  • $\begingroup$ @franck The $x-1$ doesn't form its own separate case because it's not absolute valued in the equation. $\endgroup$ Aug 14, 2013 at 20:24
  • $\begingroup$ Thanks for the comment @AndréNicolas $\endgroup$
    – user63181
    Aug 14, 2013 at 20:55

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