The image contains two Boolean algebra expressions. Let’s break them down and simplify each.

Expression 1:

$X = \overline{A} \cdot B + A \cdot (\overline{A} + C)$ We will simplify this step by step:

1. Apply the Distributive Property to the second term: $A \cdot (\overline{A} + C) = A \cdot \overline{A} + A \cdot C$ Since $A \cdot \overline{A} = 0$, we get: $A \cdot C$ So the expression becomes: $X = \overline{A} \cdot B + A \cdot C$

This is the simplified form for expression 1.

Expression 2:

$X = \overline{(A \cdot B \cdot C + D)} \cdot (A \cdot B)$ We will simplify this step by step:

1. Use De Morgan's law to break the negation of the sum: $\overline{(A \cdot B \cdot C + D)} = \overline{A \cdot B \cdot C} \cdot \overline{D}$ Using De Morgan’s law on the first term $\overline{A \cdot B \cdot C}$, we get: $\overline{A \cdot B \cdot C} = \overline{A} + \overline{B} + \overline{C}$ So the expression becomes: $X = (\overline{A} + \overline{B} + \overline{C}) \cdot \overline{D} \cdot (A \cdot B)$
2. Distribute $A \cdot B$ across the terms: [ X = (A \cdot B \cdot \overline{A}) + (A \cdot B \cdot \overline{B}) + (A