Solve for x the inequality: frac{|x-2|-1}{|x- | vedclass

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(D) Let $y = |x-2|$.Then the inequality becomes $\frac{y-1}{y-2} \leq 0$.The critical points are $y=1$ and $y=2$.For the fraction to be $\leq 0$,$y$ m

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$x \in (0, 1] \cup [3, 4)$

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(D) Let $y = |x-2|$.Then the inequality becomes $\frac{y-1}{y-2} \leq 0$.The critical points are $y=1$ and $y=2$.For the fraction to be $\leq 0$,$y$ must lie between the roots,including the numerator root but excluding the denominator root: $1 \leq y Substituting back $y = |x-2|$,we get $1 \leq |x-2| This splits into two parts:$1) |x-2| \geq 1 \implies x-2 \leq -1$ or $x-2 \geq 1 \implies x \leq 1$ or $x \geq 3$.$2) |x-2| Taking the intersection of these conditions:$(x \leq 1 	ext{ or } x \geq 3) \cap (0 Thus,$x \in (0, 1] \cup [3, 4)$.

Solve for $x$ the inequality: $\frac{|x-2|-1}{|x-2|-2} \leq 0$

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