Simplify the following Boolean expressions and draw their logic circuits: i) X = (A + B)(B + C) + (B + C) ii) Y = A . (A + B)(B + C) iii) Z = A . B + A . B . (B + C)

Here are the logic circuits for the given Boolean expressions, after simplification: i) X = (A + B)(B + C) + (B + C) Step 1: Simplify the expression. Let P = (B+C). The expression becomes X = (A+B)P + P. Using the Boolean algebra absorption law Q + QP = Q, where Q = P, we have: X = P Substitute P = (B+C) back: X = B + C Step 2: Construct the logic circuit for the simplified expression. The circuit for X = B + C consists of one OR gate. Inputs: B, C Gate: OR gate Output: X Circuit for i): ` B ---| |--- OR --- X C ---| ` ii) Y = A · (A + B)(B + C) Step 1: Simplify the expression. First, simplify A · (A+B). Using the Boolean algebra absorption law A · (A+B) = A. So, the expression becomes: Y = A · (B + C) Step 2: Construct the logic circuit for the simplified expression. The circuit for Y = A · (B + C) consists of one OR gate and one AND gate. Inputs for OR gate: B, C Output of OR gate: (B+C) Inputs for AND gate: A, output of OR gate Output of AND gate: Y Circuit for ii): ` B ---| |--- OR ---| C ---| |--- AND --- Y | A --------------| ` iii) Z = A · B + A · B · (B + C) Step 1: Simplify the expression. Let Q = A · B. The expression becomes Z = Q + Q · (B+C). Using the Boolean algebra absorption law Q + QP = Q, where Q = A · B, we have: Z = Q Substitute Q = A · B back: Z = A · B Step 2: Construct the logic circuit for the simplified expression. The circuit for Z = A · B consists of one AND gate. Inputs: A, B Gate: AND gate Output: Z Circuit for iii): ` A ---| |--- AND --- Z B ---| `