utilisando todos los datos de cada ejercicio realiza los diagramas de venn de cada ejercicio ya estan resueltos 1. Ac∩BA^c \cap BAc∩B o Ac=U−A={1,3,5,7,9,10,11,12,13,14,15,16,17,18}A^c = U - A = \{ 1, 3, 5, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 \}Ac=U−A={1,3,5,7,9,10,11,12,13,14,15,16,17,18} o Resultado: Ac∩B={1,3,12}A^c \cap B = \{ 1, 3, 12 \}Ac∩B={1,3,12} 2. A∪BcA \cup B^cA∪Bc o Bc=U−B={0,5,7,8,9,10,11,13,14,15,16,17,18}B^c = U - B = \{ 0, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18 \}Bc=U−B={0,5,7,8,9,10,11,13,14,15,16,17,18} o Resultado: A∪Bc={0,2,4,5,6,7,8,9,10,11,13,14,15,16,17,18}A \cup B^c = \{ 0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18 \}A∪Bc={0,2,4,5,6,7,8,9,10,11,13,14,15,16,17,18} 3. Bc∩EcB^c \cap E^cBc∩Ec o Ec=U−E={10,11,12,13,14,15,16,17,18}E^c = U - E = \{ 10, 11, 12, 13, 14, 15, 16, 17, 18 \}Ec=U−E={10,11,12,13,14,15,16,17,18} o Resultado: Bc∩Ec={10,11,13,14,15,16,17,18}B^c \cap E^c = \{ 10, 11, 13, 14, 15, 16, 17, 18 \}Bc∩Ec={10,11,13,14,15,16,17,18} 4. Ac−GA^c - GAc−G o Resultado: Ac−G={1,3,5,7,9,10,11,13,15,17}A^c - G = \{ 1, 3, 5, 7, 9, 10, 11, 13, 15, 17 \}Ac−G={1,3,5,7,9,10,11,13,15,17} 5. (A∪B)c(A \cup B)^c(A∪B)c o A∪B={0,1,2,3,4,6,8,12}A \cup B = \{ 0, 1, 2, 3, 4, 6, 8, 12 \}A∪B={0,1,2,3,4,6,8,12} o Resultado: (A∪B)c={5,7,9,10,11,13,14,15,16,17,18}(A \cup B)^c = \{ 5, 7, 9, 10, 11, 13, 14, 15, 16, 17, 18 \}(A∪B)c={5,7,9,10,11,13,14,15,16,17,18} 6. (A∩B)c(A \cap B)^c(A∩B)c o A∩B={2,4,6}A \cap B = \{ 2, 4, 6 \}A∩B={2,4,6} o Resultado: (A∩B)c={0,1,3,5,7,8,9,10,11,12,13,14,15,16,17,18}(A \cap B)^c = \{ 0, 1, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 \}(A∩B)c={0,1,3,5,7,8,9,10,11,12,13,14,15,16,17,18} 7. (A∪F)∩C(A \cup F) \cap C(A∪F)∩C o A∪F={0,2,4,6,8,14,15,16,17,18}A \cup F = \{ 0, 2, 4, 6, 8, 14, 15, 16, 17, 18 \}A∪F={0,2,4,6,8,14,15,16,17,18} o Resultado: (A∪F)∩C={0,2,4}(A \cup F) \cap C = \{ 0, 2, 4 \}(A∪F)∩C={0,2,4} 8. B∪(F−G)B \cup (F - G)B∪(F−G) o F−G={15,17}F - G = \{ 15, 17 \}F−G={15,17} o Resultado: B∪(F−G)={1,2,3,4,6,12,15,17}B \cup (F - G) = \{ 1, 2, 3, 4, 6, 12, 15, 17 \}B∪(F−G)={1,2,3,4,6,12,15,17}