Carry out a sign study to solve the inequality (|x - 2| - x)/(x² - 4) ≤ 0.

Step-by-step solution

1. 1

   Fix the domain. The denominator factors as $x^{2}-4=(x-2)(x+2)$, so it vanishes at $x=\pm2$. Those two points must be excluded:

   $D=\mathbb{R}\setminus\{-2,\,2\}.$

   They will also be the natural break points of the sign chart.

2. 2

   Remove the absolute value by cases at $x=2$. For $x\ge2$, $|x-2|=x-2$ and the numerator is

   $x-2-x=-2,$

   a constant that is always negative. For $x<2$, $|x-2|=2-x$ and the numerator is

   $2-x-x=2-2x=2(1-x),$

   positive for $x<1$, zero at $x=1$, negative for $1<x<2$. So the numerator changes sign only at $x=1$.

3. 3

   Chart the denominator. $(x-2)(x+2)$ is positive outside $[-2,2]$ and negative inside:

   $x<-2:\;+\qquad -2<x<2:\;-\qquad x>2:\;+.$

4. 4

   Combine on the four intervals cut by $-2$, $1$ and $2$.

   • $(-\infty,-2)$: numerator $+$, denominator $+$ $\Rightarrow$ quotient $+$ — excluded.
   • $(-2,1)$: numerator $+$, denominator $-$ $\Rightarrow$ quotient $-$ — included.
   • $(1,2)$: numerator $-$, denominator $-$ $\Rightarrow$ quotient $+$ — excluded.
   • $(2,+\infty)$: numerator $-2$, denominator $+$ $\Rightarrow$ quotient $-$ — included.
5. 5

   Handle the boundary points and write the answer. At $x=1$ the numerator is $0$ and the denominator is $-3\ne0$, so the quotient is $0$ and the point satisfies $\le 0$; it is included. The points $x=\pm2$ are never allowed. Hence

   $(-2,\,1]\;\cup\;(2,\,+\infty).$

   Spot checks: $x=0$ gives $2/(-4)=-0.5\le0$ ✓, $x=1.1$ gives $-0.2/(-2.79)=+0.0717>0$ ✗, $x=3$ gives $-2/5=-0.4\le0$ ✓, $x=-3$ gives $8/5=+1.6>0$ ✗.