## Task B-2.1.

Solve the equation $\sqrt{3-\frac{1}{2020 x}}=1-\frac{1}{2020 x}$.
## Solution.

Let's introduce the substitution $t=\frac{1}{2020 x}, x \neq 0$.

Then the equation becomes $\sqrt{3-t}=1-t$, with the conditions $3-t \geqslant 0, 1-t \geqslant 0$, i.e., $t \leqslant 1$.

After squaring, we get $3-t=1-2 t+t^{2}$, or $t^{2}-t-2=0$.

Now, $t=-1$ or $t=2$.

Due to the above condition, the only possible solution is $t=-1$.

This gives $\frac{1}{2020 x}=-1$, so $x=-\frac{1}{2020}$.

Let's verify the solution by substituting it back into the given equation. We get

$$
\sqrt{3-(-1)}=1-(-1)
$$

which is correct, so this is the solution to the given equation.

Note: If the student did not write the conditions, do not deduct points if they have verified both solutions.

If the student did not write the conditions and did not verify the solutions to determine which are correct, deduct 2 points.

If the student does not use the substitution, they will get the equation $2 \cdot 2020^{2} x^{2}+2020 x-1=0$.