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This image contains Boolean algebra expressions that need to be simplified. Let's go through each expression one by one and simplify them.

### 1. \( AB' + BC + C'A \)
   **Simplification**:
   - There's no further simplification possible, as these terms are already in simplified form. They have no common factors to factor out.
   
   **Final Expression**: \( AB' + BC + C'A \)

### 2. \( A'(BC + AB + BA') \)
   **Simplification**:
   - First, simplify inside the parentheses:
     \[
     AB + BA' = A(B + B') = A(1) = A
     \]
     So, the expression becomes:
     \[
     A'(BC + A)
     \]
   - Distribute \( A' \) across:
     \[
     A'BC + A'A = A'BC
     \]
     (since \( A'A = 0 \)).
   
   **Final Expression**: \( A'BC \)

### 3. \( ABC + AB + A \)
   **Simplification**:
   - Factor out \( A \):
     \[
     A(BC + B + 1)
     \]
     - Notice that \( B + 1 = 1 \), so the expression inside the parentheses simplifies to 1:
     \[
     A \times 1 = A
     \]
   
   **Final Expression**: \( A \)

### 4. \( (A' + AB)(A'B) \)
   **Simplification**:
   - Apply distribution:
     \[
     (A' + AB)(A'B) = A'A'B + ABA'B = A'B + 0 = A'B
     \]
   
   **Final Expression**: \( A'B \)

### 5. \( BC + AD + ABCD + ADC + A' \)
   **Simplification**:
   - Combine similar terms. Group \( AD \) and \( ADC \):
     \[
     AD + ADC = AD(C + 1) = AD
     \]
     - So the expression becomes:
     \[
     BC + AD + ABCD + A'
     \]
     - Notice that \( ABCD \) is already covered by \( AD \), so:
     \[
     BC + AD + A'
     \]
   
   **Final Expression**: \( BC + AD + A' \)

### Summary of Simplifications:
1. \( AB' + BC + C'A \)
2. \( A'BC \)
3. \( A \)
4. \( A'B \)
5. \( BC + AD + A' \)

Let me know if you would like more details or explanations on any of these steps!

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Here are five related questions:
1. How do you use De Morgan’s law in Boolean algebra simplification?
2. What is the role of the identity and null laws in Boolean algebra?
3. How does the distributive law affect the simplification of Boolean expressions?
4. Why does \( B + 1 \) always equal 1 in Boolean algebra?
5. Can you give examples of where Boolean algebra is used in digital circuit design?

**Tip**: When simplifying Boolean expressions, always look for opportunities to factor and apply fundamental laws like absorption or distribution for quicker results.

Sederhanakan ekspresi logika dibawah dengan Aljabar Boolean: AB' + BC + C'A; A'(BC + AB + BA'); ABC + AB + A; (A' + AB)(A'B); BC + AD + ABCD + ADC + A'

Simplify Boolean expressions such as AB' + BC + C'A and more using key Boolean algebra rules like Distributive and Absorption Laws. Suitable for students studying Computer Science or Electronics at the undergraduate level.

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