# Solution.

Let $f(x)=k=x-\frac{2}{x}, f(f(x))=f(k)=k-\frac{2}{k}, f(f(f(x)))=f(f(k))=f(k)-\frac{2}{f(k)}$. The equation $f(f(f(x)))=f(k)-\frac{2}{f(k)}=1 \Leftrightarrow f^{2}(k)-f(k)-2=0$ has two solutions $f_{1}(k)=-1$ and $f_{2}(k)=2$.

We obtain $\left[\begin{array}{l}f(k)=k-\frac{2}{k}=-1 \\ f(k)=k-\frac{2}{k}=2\end{array} \Leftrightarrow\left[\begin{array}{l}k^{2}+k-2=0 \\ k^{2}-2 k-2=0\end{array}\right.\right.$.

The system has four distinct roots $k_{1}, k_{2}, k_{3}, k_{4}$.

Notice that the equations $x-\frac{2}{x}=k \Leftrightarrow x^{2}-k_{i} x-2=0$ have two distinct roots and these equations with different $k_{i}$ do not have common roots.

(Schoolchildren should strictly prove this or find the roots $x$ explicitly).

Thus, the original equation has 8 distinct roots.

Answer. 8.

| Content of the criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 12 |
| The answer is correct. All main logical steps of the solution are provided. There are arithmetic errors or typos that did not affect the overall course of the solution. | +. | 10 |
| The answer is correct. All main logical steps of the solution are provided. There is no strict justification for some conclusions. | $\pm$ | 9 |
| All main logical steps of the solution are provided. The incorrect reasoning led to an incorrect answer. | $+/ 2$ | 6 |
| The answer is correct. The main logical steps of the solution are provided. There is no proof that the equations $x^{2}-k_{i} x-2=0$ for different $k_{i}$ do not have common roots. |  |  |
| The answer is correct. The solution is missing or incorrect. | $\mp$ | 2 |
| :---: | :---: | :---: |
| Some steps reflecting the general idea of the solution are provided. The answer is missing or incorrect. |  |  |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |
| Maximum score | 12 |  |
# Task 11.3. (12 points)

How many distinct roots does the equation $f(f(f(x)))=1$ have, if $f(x)=x-\frac{2}{x}$.