## Task B-1.4.

Determine all real numbers $a$ for which the equation

$$
\frac{a}{1-\frac{1}{1+\frac{1}{x}}}=2023 x
$$

has no solutions.
## Solution.

The conditions for the solution of the equation are $x \neq 0, x \neq -1$.

By simplifying the denominator of the fraction on the left side of the equation step by step, we get:

$\frac{a}{1-\frac{1}{1+\frac{1}{x}}}=2023 x$

$\frac{a}{1-\frac{1}{\frac{x+1}{x}}}=2023 x$

$\frac{a}{1-\frac{x}{x+1}}=2023 x$

![](https://cdn.mathpix.com/cropped/2024_05_30_5ec97620b8cc23e4b297g-03.jpg?height=149&width=848&top_left_y=1439&top_left_x=181)

Then, $a x + a = 2023 x$, which simplifies to

$x(2023 - a) = a$.

For $a = 2023$, the equation becomes $0 \cdot x = 2023$, so the equation has no solution.

For $a \neq 2023$, the solution to the original equation is $x = \frac{a}{2023 - a} \left(^{*}\right)$, but the original equation will not have a solution if the obtained solution $x$ does not satisfy the initial conditions.

The value of $a$ for which this can happen can be found by substituting $x = 0$ and $x = -1$ into $\left(^{*}\right)$.

In the first case, we get $a = 0$, and the second case is impossible.

Therefore, the given equation will not have a solution if $a = 0$ and if $a = 2023$.

Note: The student does not need to write the initial conditions if they have verified the result at the end and written the correct solution. If they have not written the conditions and only have the solution $a = 2023$, they can receive a maximum of 4 points.