Math Problem Statement

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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'

Solution

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+CAAB' + 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+CAAB' + BC + C'A

2. A(BC+AB+BA)A'(BC + AB + BA')

Simplification:

  • First, simplify inside the parentheses: AB+BA=A(B+B)=A(1)=AAB + BA' = A(B + B') = A(1) = A So, the expression becomes: A(BC+A)A'(BC + A)
  • Distribute AA' across: ABC+AA=ABCA'BC + A'A = A'BC (since AA=0A'A = 0).

Final Expression: ABCA'BC

3. ABC+AB+AABC + AB + A

Simplification:

  • Factor out AA: A(BC+B+1)A(BC + B + 1)
    • Notice that B+1=1B + 1 = 1, so the expression inside the parentheses simplifies to 1: A×1=AA \times 1 = A

Final Expression: AA

4. (A+AB)(AB)(A' + AB)(A'B)

Simplification:

  • Apply distribution: (A+AB)(AB)=AAB+ABAB=AB+0=AB(A' + AB)(A'B) = A'A'B + ABA'B = A'B + 0 = A'B

Final Expression: ABA'B

5. BC+AD+ABCD+ADC+ABC + AD + ABCD + ADC + A'

Simplification:

  • Combine similar terms. Group ADAD and ADCADC: AD+ADC=AD(C+1)=ADAD + ADC = AD(C + 1) = AD
    • So the expression becomes: BC+AD+ABCD+ABC + AD + ABCD + A'
    • Notice that ABCDABCD is already covered by ADAD, so: BC+AD+ABC + AD + A'

Final Expression: BC+AD+ABC + AD + A'

Summary of Simplifications:

  1. AB+BC+CAAB' + BC + C'A
  2. ABCA'BC
  3. AA
  4. ABA'B
  5. BC+AD+ABC + AD + A'

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


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+1B + 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.

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Math Problem Analysis

Mathematical Concepts

Boolean Algebra
Logic Simplification

Formulas

Distributive Law
Identity Law
Absorption Law

Theorems

De Morgan’s Law
Absorption Theorem

Suitable Grade Level

Undergraduate Computer Science or Electronics