8. Find all real values of $x$ satisfying the inequality
$$
\sqrt{\left(\frac{1}{2-x}+1\right)^{2}} \geq 2 \text {. }
$$
8. $[1,2) \cup(2,7 / 3]$
We recall that $\sqrt{a^{2}}=|a|$ for any $a \in \mathbb{R}$. Thus, the given inequality is equivalent to
$$
\left|\frac{3-x}{2-x}\right| \geq 2,
$$
which is further equivalent to the following compound inequality:
$$
\frac{3-x}{2-x} \geq 2 \quad \text { or } \quad \frac{3-x}{2-x} \leq-2 .
$$

We solve the first inequality in $(\star)$.
$$
\frac{3-x}{2-x} \geq 2 \quad \Longrightarrow \quad \frac{3-x}{2-x}-2 \geq 0 \quad \Longrightarrow \quad \frac{x-1}{2-x} \geq 0
$$

The last inequality gives $x=1$ and $x=2$ as critical numbers.

Thus, the solution set of the first inequality in $(*)$ is $[1,2)$.
Solving the second inequality in $(\star)$, we get the interval $(2,7 / 3]$. Thus, the solution set of the original inequality is $[1,2) \cup(2,7 / 3]$.