Diberikan S={1,2,3,4,5,6,7,8,9,10} A= {1,2,3,4,5} B=

a. {5, 6, 7, 8, 9, 10}, b. {4}, c. {4}

Pembahasan

Diberikan:
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
C = {3, 5, 6, 9}

Kita akan menentukan:

a. A^c ∪ (B∩C)
   Pertama, cari A^c (komplemen A terhadap S).
   A^c = S - A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {1, 2, 3, 4, 5}
   A^c = {6, 7, 8, 9, 10}

   Kedua, cari B∩C (irisan B dan C).
   B∩C = {4, 5, 6, 7, 8} ∩ {3, 5, 6, 9}
   B∩C = {5, 6}

   Ketiga, cari A^c ∪ (B∩C) (gabungan A^c dan B∩C).
   A^c ∪ (B∩C) = {6, 7, 8, 9, 10} ∪ {5, 6}
   A^c ∪ (B∩C) = {5, 6, 7, 8, 9, 10}

b. (A∩B) ∩ C^c
   Pertama, cari A∩B (irisan A dan B).
   A∩B = {1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8}
   A∩B = {4, 5}

   Kedua, cari C^c (komplemen C terhadap S).
   C^c = S - C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 5, 6, 9}
   C^c = {1, 2, 4, 7, 8, 10}

   Ketiga, cari (A∩B) ∩ C^c (irisan A∩B dan C^c).
   (A∩B) ∩ C^c = {4, 5} ∩ {1, 2, 4, 7, 8, 10}
   (A∩B) ∩ C^c = {4}

c. (B-C) ∩ A
   Pertama, cari B-C (selisih B dan C).
   B-C = {4, 5, 6, 7, 8} - {3, 5, 6, 9}
   B-C = {4, 7, 8}

   Kedua, cari (B-C) ∩ A (irisan B-C dan A).
   (B-C) ∩ A = {4, 7, 8} ∩ {1, 2, 3, 4, 5}
   (B-C) ∩ A = {4}

Hasil:
a. A^c ∪ (B∩C) = {5, 6, 7, 8, 9, 10}
b. (A∩B) ∩ C^c = {4}
c. (B-C) ∩ A = {4}