Task B-2.3. Solve the equation

$$
x^{2}+\left(\frac{x}{x-1}\right)^{2}=8
$$
## Solution.

Let's introduce the substitution $t=x-1$. Then the equation transforms into

$$
\begin{aligned}
& (t+1)^{2}+\left(\frac{t+1}{t}\right)^{2}=8 \\
& t^{2}+2 t+1+1+\frac{2}{t}+\frac{1}{t^{2}}=8 \\
& t^{2}+2+\frac{1}{t^{2}}+2 \cdot\left(t+\frac{1}{t}\right)-8=0 \\
& \left(t+\frac{1}{t}\right)^{2}+2 \cdot\left(t+\frac{1}{t}\right)-8=0
\end{aligned}
$$

Solving this equation, we get

$$
t+\frac{1}{t}=2 \quad \text { or } \quad t+\frac{1}{t}=-4
$$

This leads to $t^{2}-2 t+1=0$ or $t^{2}+4 t+1=0$, that is,

$$
\begin{aligned}
& t_{1}=1 \quad \Rightarrow \quad x_{1}=2 \\
& t_{2}=-2-\sqrt{3} \quad \Rightarrow \quad x_{2}=-1-\sqrt{3} \\
& t_{3}=-2+\sqrt{3} \quad \Rightarrow \quad x_{3}=-1+\sqrt{3}
\end{aligned}
$$

The solutions are all $x \in\{2,-1-\sqrt{3},-1+\sqrt{3}\}$.