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1/(|x|-3) le1/2...

`1/(|x|-3) le1/2`

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To solve the inequality \( \frac{1}{|x| - 3} \leq \frac{1}{2} \), we will consider two cases based on the definition of the absolute value. ### Step 1: Case 1 - \( x \geq 0 \) In this case, \( |x| = x \). Therefore, the inequality becomes: \[ \frac{1}{x - 3} \leq \frac{1}{2} \] ...
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