2. In the set of real numbers, solve the equation

$$
(5-2 x)^{2}+\left(\frac{1}{2-x}-1\right)^{2}=9
$$
Solution. Clearly, $x \neq 2$. We introduce the substitution $2-x=t$. Now $5-2 x=2 t+1$, so we sequentially obtain the equations

$$
\begin{aligned}
& (1+2 t)^{2}+\left(\frac{1}{t}-1\right)^{2}=9 \\
& 4 t^{2}+4 t+1+\frac{1}{t^{2}}-\frac{2}{t}+1=9
\end{aligned}
$$

$$
\begin{aligned}
& 4 t^{2}-4+\frac{1}{t^{2}}+4 t-\frac{2}{t}=3 \\
& \left(2 t-\frac{1}{t}\right)^{2}+2\left(2 t-\frac{1}{t}\right)=3
\end{aligned}
$$

We introduce the substitution $2 t-\frac{1}{t}=z$ and obtain the equation $z^{2}+2 z-3=0$ whose solutions are 1 and -3. From $2 t-\frac{1}{t}=1$ we get $2 t^{2}-t-1=0$ and obtain the solutions $t_{1}=1$ and $t_{2}=-\frac{1}{2}$. From $2 t-\frac{1}{t}=-3$ we get $2 t^{2}+3 t-1=0$ and obtain the solutions $t_{3 / 4}=\frac{-3 \pm \sqrt{17}}{4}$. Finally, by substituting into $x=2-t$ we obtain the solutions $x_{1}=1$, $x_{2}=\frac{5}{2}$ and $x_{3 / 4}=\frac{11 \pm \sqrt{17}}{4}$.