1. In the field of real numbers, solve the equation $2^{|x+1|}-2^{x}=1+\left|2^{x}-1\right|$.
1. We divide the set of real numbers according to the zeros of the expressions. That is, according to the numbers for which the absolute values in the given equation are zero. The expression \( |x+1| \) corresponds to \( x = -1 \) and the expression \( \left|2^{x}-1\right| \) corresponds to \( x = 0 \). This gives us the following three cases:
1) In the case \( x \leq -1 \), we have \( |x+1| = -(x+1) \) and \( \left|2^{x}-1\right| = -\left(2^{x}-1\right) = 1 - 2^{x} \). The equation from the problem then becomes

\[
2^{-(x+1)} - 2^{x} = 1 + 1 - 2^{x} \quad \text{or} \quad 2^{-(x+1)} = 2
\]

and has the unique solution \( x = -2 \). By substituting back, we can verify that this is indeed a solution to the given equation.

2) In the case \( -1 < x \leq 0 \), we have \( |x+1| = x+1 \) and \( \left|2^{x}-1\right| = 1 - 2^{x} \). The equation from the problem then becomes

\[
2^{x+1} - 2^{x} = 1 + 1 - 2^{x} \quad \text{or} \quad 2^{x+1} = 2
\]

and has the unique solution \( x = 0 \). By substituting back, we can verify that this is indeed a solution to the given equation.

3) In the case \( x > 0 \), we have \( |x+1| = x+1 \) and \( \left|2^{x}-1\right| = 2^{x} - 1 \). The equation from the problem then becomes

\[
2^{x+1} - 2^{x} = 1 + 2^{x} - 1 \quad \text{or} \quad 2^{x+1} - 2^{x} = 2^{x}
\]

which simplifies to \( 2^{x+1} = 2 \cdot 2^{x} \), which is always true for \( x > 0 \). Therefore, all \( x > 0 \) are solutions to the given equation.

Answer. The solutions to the given equation are \( x = -2 \) and any \( x \geq 0 \).

Note. Instead of verifying the solutions by substituting them back into the given equation, it is sufficient to check that the found solutions fall within the investigated interval.

Alternative Solution. We consider two cases:

If \( x+1 \geq 0 \), then \( |x+1| = x+1 \), so the given equation can be simplified to

\[
2^{x+1} - 2^{x} = 1 + \left|2^{x} - 1\right| \quad \text{or} \quad 2^{x} - 1 = \left|2^{x} - 1\right|.
\]

This is satisfied if and only if \( 2^{x} - 1 \geq 0 \), which holds if and only if \( x \geq 0 \). In this case, the solutions are all \( x \) such that \( x+1 \geq 0 \) and \( x \geq 0 \), i.e., all non-negative \( x \).

If, on the other hand, \( x+1 < 0 \), then \( |x+1| = -(x+1) \), so the given equation can be simplified to

\[
2^{-(x+1)} - 2^{x} = 1 + \left|2^{x} - 1\right| \quad \text{or} \quad 2^{-(x+1)} - 2^{x} = 1 + 1 - 2^{x} \quad \text{or} \quad 2^{-(x+1)} = 2.
\]

This has the unique solution \( x = -2 \).

Grading. For the first solution, award 2 points for the first case \( x \leq -1 \), 3 points for the second case \( -1 < x \leq 0 \), and 1 point for the last case \( -1 < x \leq 0 \). If the solution only considers two cases as in the alternative solution, award 3 points for each part. For guessing all solutions, award 1 point, and for guessing only some (even infinitely many), do not award any points.