Answer
$(-\infty,-1)\cup(1,\infty)$.
Work Step by Step
Step 1. For $\frac{(x-2)^2}{x^2-1}\ge0\Longrightarrow \frac{(x-2)^2}{(x+1)(x-1)}\ge0$, identify boundary points as $x=-1,1,2$.
Step 2. Form intervals $(-\infty,-1),(-1,1),(1,2],[2,\infty)$.
Step 3. Choose test values for each interval $x=-2,0,1.5,3$.
Step 4. Test the inequality to get results: $True,\ False,\ True,\ True$
Step 5. We have the solution $(-\infty,-1)\cup(1,\infty)$.
