\n| \\( – 0,\\left( {56} \\right) = – \\frac{{56}}{{99}}\\)<\/td>\n | \\(3,\\left( {12} \\right) = \\frac{{103}}{{33}}\\)<\/td>\n | \\(0,4\\left( 5 \\right) = \\frac{{41}}{{90}}\\)<\/td>\n | \\(2,3\\left( 7 \\right) = \\frac{{214}}{{90}}\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \u00a0<\/p>\n C\u00e1lculos relativos \u00e0s fra\u00e7\u00f5es irredut\u00edveis das d\u00edzimas infinitas peri\u00f3dicas<\/h6>\nVamos apresentar o racioc\u00ednio e os c\u00e1lculos detalhados apenas em rela\u00e7\u00e3o \u00e0s tr\u00eas primeiras d\u00edzimas infinitas peri\u00f3dicas. Em rela\u00e7\u00e3o \u00e0s restantes tr\u00eas d\u00edzimas infinitas peri\u00f3dicas, os c\u00e1lculos apresentados ser\u00e3o breves, mas de maior exig\u00eancia de c\u00e1lculo mental (serve de exerc\u00edcio!).<\/p>\n \\[3,\\left( 4 \\right) = \\frac{{31}}{9}\\]<\/p>\n Designando a d\u00edzima por \\(x\\), vem: \\(x = 3,\\left( {4} \\right)\\).
Logo, multiplicando por \\(10\\) os dois membros da igualdade anterior, vem: \\(10x = 34,\\left( {4} \\right)\\).
Subtraindo, membro a membro, as duas equa\u00e7\u00f5es anteriores, temos:
\\[\\begin{array}{*{20}{r}}{}&{10x}& = &{34,\\left( {4} \\right)}\\\\ – &x& = &{3,\\left( {4} \\right)}\\\\\\hline{}&{9x}& = &{31\\quad\\;\\,\\,}\\end{array}\\]
Donde, \\(x = \\frac{{31}}{{99}}\\).
Portanto, \\(3,\\left( {4} \\right) = \\frac{{31}}{{9}}\\).<\/p>\n \\[1,\\left( 2 \\right) = \\frac{{11}}{9}\\]<\/p>\n Designando a d\u00edzima por \\(x\\), vem: \\(x = 1,\\left( {2} \\right)\\).
Logo, multiplicando por \\(10\\) os dois membros da igualdade anterior, vem: \\(10x = 12,\\left( {2} \\right)\\).
Subtraindo, membro a membro, as duas equa\u00e7\u00f5es anteriores, temos:
\\[\\begin{array}{*{20}{r}}{}&{10x}& = &{12,\\left( {2} \\right)}\\\\ – &x& = &{1,\\left( {2} \\right)}\\\\\\hline{}&{9x}& = &{11\\quad\\;\\,\\,}\\end{array}\\]
Donde, \\(x = \\frac{{11}}{{9}}\\).
Portanto, \\(1,\\left( {2} \\right) = \\frac{{11}}{{9}}\\).<\/p>\n \\[ – 0,\\left( {56} \\right) = – \\frac{{56}}{{99}}\\]<\/p>\n Designando a d\u00edzima por \\(x\\), vem: \\(x = 0,\\left( {56} \\right)\\).
Logo, multiplicando por \\(100\\) os dois membros da igualdade anterior, vem: \\(100x = 56,\\left( {56} \\right)\\).
Subtraindo, membro a membro, as duas equa\u00e7\u00f5es anteriores, temos:
\\[\\begin{array}{*{20}{r}}{}&{100x}& = &{56,\\left( {56} \\right)}\\\\ – &x& = &{0,\\left( {56} \\right)}\\\\\\hline{}&{99x}& = &{56\\quad\\;\\;\\;\\;}\\end{array}\\]
Donde, \\(x = \\frac{{56}}{{99}}\\).
Portanto, \\( – 0,\\left( {56} \\right) = – \\frac{{56}}{{99}}\\).<\/p>\n \\[3,\\left( {12} \\right) = \\frac{{103}}{{33}}\\]<\/p>\n \\[\\begin{array}{*{20}{c}}{100x – x = 312,\\left( {12} \\right) – 3,\\left( {12} \\right)}& \\Leftrightarrow &{99x = 309}& \\Leftrightarrow &{x = \\frac{{103}}{{33}}}\\end{array}\\]<\/p>\n \u00a0<\/p>\n \\[0,4\\left( 5 \\right) = \\frac{{41}}{{90}}\\]<\/p>\n \\[\\begin{array}{*{20}{c}}{100x – 10x = 45,\\left( 5 \\right) – 4,\\left( 5 \\right)}& \\Leftrightarrow &{90x = 41}& \\Leftrightarrow &{x = \\frac{{41}}{{90}}}\\end{array}\\]<\/p>\n \u00a0<\/p>\n \\[2,3\\left( 7 \\right) = \\frac{{214}}{{90}}\\]<\/p>\n \\[\\begin{array}{*{20}{c}}{100x – 10x = 237,\\left( 7 \\right) – 23,\\left( 7 \\right)}& \\Leftrightarrow &{90x = 214}& \\Leftrightarrow &{x = \\frac{{214}}{{90}}}\\end{array}\\]<\/p>\n << Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"Enunciado Resolu\u00e7\u00e3o Enunciado Escreve cada um dos n\u00fameros sob a forma de fra\u00e7\u00e3o irredut\u00edvel: \\( – 1,6\\) \\(1,25\\) \\( – 3,6\\) \\(10,5\\) \\(0,2\\) \\( – 0,49\\) \\(3,\\left( 4 \\right)\\) \\(1,\\left( 2 \\right)\\) \\( –...<\/p>\n","protected":false},"author":1,"featured_media":19171,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,664],"tags":[424,665,666],"series":[],"class_list":["post-22473","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-numeros-reais","tag-8-o-ano","tag-dizima-finita","tag-dizima-infinita-periodica"],"views":295,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat62.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/22473","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=22473"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/22473\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=22473"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=22473"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=22473"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=22473"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}} |