{"id":262,"date":"2020-03-26T20:47:55","date_gmt":"2020-03-26T23:47:55","guid":{"rendered":"http:\/\/cinoto.com.br\/matematica\/?p=262"},"modified":"2020-03-26T20:47:55","modified_gmt":"2020-03-26T23:47:55","slug":"denotemos-por-nx-o-numero-de-elementos-de-um-conjunto-finito-x","status":"publish","type":"post","link":"http:\/\/cinoto.com.br\/matematica\/denotemos-por-nx-o-numero-de-elementos-de-um-conjunto-finito-x\/","title":{"rendered":"1) (ITA) Denotemos por n(x) o n\u00famero de elementos de um conjunto finito X. Sejam A, B e C conjuntos tais que n(A U B) = 8, n(A U C) = 9, n(B U C) = 10, N(A U B U C) = 11 e n(A \u2229 B \u2229 C) = 2. Ent\u00e3o n(A) + n(B) + n(C) \u00e9 igual a:"},"content":{"rendered":"\n

a)\u00a011 \u00a0 \u00a0\u00a0b)\u00a014 \u00a0 \u00a0\u00a0c)\u00a015\u00a0 \u00a0 \u00a0d)\u00a018 \u00a0 \u00a0\u00a0e)\u00a025<\/p>\n\n\n\n\n\n\n\n

Resolu\u00e7\u00e3o:<\/h2>\n\n\n\n

Em primeiro lugar, vamos convencionar o seguinte:<\/p>\n\n\n\n

(A U B) = A uni\u00e3o com B<\/p>\n\n\n\n

(A \u2229 B) = A intersec\u00e7\u00e3o com B<\/p>\n\n\n\n

n(A) = n\u00famero de elementos de A<\/p>\n\n\n\n

Para resolver esse exerc\u00edcio voc\u00ea deve saber que:<\/p>\n\n\n\n

n(A U B) = n(A) + n(B) – n(A \u2229 B)<\/p>\n\n\n\n

n(A U B U C) = n(A) + n(B) + n(C) – n(A \u2229 B) – n(A \u2229 C) – n(B \u2229 C) + n(A \u2229 B \u2229 C)<\/p>\n\n\n\n

Como temos que encontrar o valor de n(A) + n(B) + n(C), podemos usar a f\u00f3rmula acima, colocando os valores que foram dados:<\/p>\n\n\n\n

n(A U B U C) = n(A) + n(B) + n(C) – n(A \u2229 B) – n(A \u2229 C) – n(B \u2229 C) + n(A \u2229 B \u2229 C)<\/p>\n\n\n\n

11 = n(A) + n(B) + n(C) – n(A \u2229 B) – n(A \u2229 C) – n(B \u2229 C) + 2<\/p>\n\n\n\n

Mas precisamos descobrir os valores de n(A \u2229 B), n(A \u2229 C) e n(B \u2229 C). Ent\u00e3o vamos usar o seguinte:<\/p>\n\n\n\n

n(A U B) = n(A) + n(B) – n(A \u2229 B)<\/p>\n\n\n\n

8 = n(A) + n(B) – n(A \u2229 B) (i)<\/p>\n\n\n\n

n(A U C) = n(A) + n(C) – n(A \u2229 C)<\/p>\n\n\n\n

9 = n(A) + n(C) – n(A \u2229 C) (ii)<\/p>\n\n\n\n

n(B U C) = n(B) + n(C) – n(B \u2229 C)<\/p>\n\n\n\n

10 = n(B) + n(C) – n(B \u2229 C) (iii)<\/p>\n\n\n\n

E se somarmos as tr\u00eas equa\u00e7\u00f5es (i), (ii) e (iii), temos:<\/p>\n\n\n\n

8 + 9 + 10 = n(A) + n(B) – n(A \u2229 B) + n(A) + n(C) – n(A \u2229 C) + n(B) + n(C) – n(B \u2229 C)<\/p>\n\n\n\n

27 = 2.n(A) + 2.n(B) + 2.n(C) – n(A \u2229 C) – n(B \u2229 C) – n(A \u2229 B)<\/p>\n\n\n\n

27 – 2.n(A) – 2.n(B) – 2.n(C) = – n(A \u2229 C) – n(B \u2229 C) – n(A \u2229 B)<\/p>\n\n\n\n

E veja que o segundo membro \u00e9 justamente o que est\u00e1vamos procurando, ent\u00e3o vamos colocar isso na equa\u00e7\u00e3o que est\u00e1vamos trabalhando:<\/p>\n\n\n\n

11 = n(A) + n(B) + n(C) – n(A \u2229 B) – n(A \u2229 C) – n(B \u2229 C) + 2<\/p>\n\n\n\n

11 = n(A) + n(B) + n(C) + 27 – 2.n(A) – 2.n(B) – 2.n(C) + 2<\/p>\n\n\n\n

11 = n(A) + n(B) + n(C) + 29 – 2.n(A) – 2.n(B) – 2.n(C)<\/p>\n\n\n\n

11 = – n(A) – n(B) – n(C) + 29<\/p>\n\n\n\n

n(A) + n(B) + n(C) = 29 – 11<\/p>\n\n\n\n

n(A) + n(B) + n(C) = 18<\/p>\n","protected":false},"excerpt":{"rendered":"

a)\u00a011 \u00a0 \u00a0\u00a0b)\u00a014 \u00a0 \u00a0\u00a0c)\u00a015\u00a0 \u00a0 \u00a0d)\u00a018 \u00a0 \u00a0\u00a0e)\u00a025<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[30],"class_list":["post-262","post","type-post","status-publish","format-standard","hentry","category-conjuntos","tag-insano"],"_links":{"self":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/262","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/comments?post=262"}],"version-history":[{"count":1,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/262\/revisions"}],"predecessor-version":[{"id":263,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/262\/revisions\/263"}],"wp:attachment":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/media?parent=262"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/categories?post=262"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/tags?post=262"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}