<PRE><B>Question 1191034</B>
Here&#39;s the simplification of the given Boolean functions using Boolean algebra:

**a. F(X, Y, Z) = (XY) + (X + Y + Z)&#39;X + YZ**

1. Distribute the X:
   F = XY + X(X + Y + Z)&#39; + YZ

2. Note that X(X + Y + Z)&#39; = X(X&#39;Y&#39;Z&#39;) = XX&#39;Y&#39;Z&#39; = 0 (because XX&#39; = 0)
   F = XY + 0 + YZ
   F = XY + YZ

**b. F(X, Y, Z) = (XY)&#39; + (X + Y + Z)&#39;**

1. Apply De Morgan&#39;s Law:
   F = X&#39; + Y&#39; + X&#39;Y&#39;Z&#39;

2. Notice that X&#39;Y&#39;Z&#39; is redundant as it&#39;s contained within X&#39; + Y&#39;
   F = X&#39; + Y&#39;

**c. F(X, Y, Z) = YZ + (X + Y)&#39; + (XYZ)&#39;**

1. Apply De Morgan&#39;s Law:
   F = YZ + X&#39;Y&#39; + X&#39; + Y&#39; + Z&#39;

2. Notice that X&#39;Y&#39; is redundant as it&#39;s contained in X&#39; and Y&#39;.
   F = YZ + X&#39; + Y&#39; + Z&#39;

**d. F(X, Y, Z) = (X + Y + Z)&#39;(X + Y)**

1. Apply De Morgan&#39;s Law:
   F = (X&#39;Y&#39;Z&#39;)(X + Y)

2. Distribute X&#39;Y&#39;Z&#39;:
   F = X&#39;Y&#39;Z&#39;X + X&#39;Y&#39;Z&#39;Y

3. Note that X&#39;Y&#39;Z&#39;X = XX&#39;Y&#39;Z&#39; = 0 and X&#39;Y&#39;Z&#39;Y = X&#39;YY&#39;Z&#39; = 0 (since XX&#39; = 0 and YY&#39; = 0)
   F = 0 + 0
   F = 0
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