{"id":22369,"date":"2022-10-18T08:18:16","date_gmt":"2022-10-18T07:18:16","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=22369"},"modified":"2022-10-19T01:57:34","modified_gmt":"2022-10-19T00:57:34","slug":"representa-na-forma-de-fracao","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=22369","title":{"rendered":"Representa na forma de fra\u00e7\u00e3o"},"content":{"rendered":"\n<p><ul id='GTTabs_ul_22369' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_22369' class='GTTabs_curr'><a  id=\"22369_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_22369' ><a  id=\"22369_1\" onMouseOver=\"GTTabsShowLinks(' Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'> Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_22369'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Representa na forma de fra\u00e7\u00e3o os n\u00fameros racionais dados pelas seguintes d\u00edzimas infinitas peri\u00f3dicas:<\/p>\n<ol>\n<li>\\(3,\\left( 4 \\right)\\)<\/li>\n<li>\\(1,\\left( {45} \\right)\\)<\/li>\n<li>\\(7,226\\left( {72} \\right)\\)<\/li>\n<li>\\(0,5\\left( 9 \\right)\\)<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_22369' onClick='GTTabs_show(1,22369)'> Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_22369'>\n<span class='GTTabs_titles'><b> Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Designando a d\u00edzima por \\(x\\), vem: \\(x = 3,\\left( 4 \\right)\\).<br \/>Logo, multiplicando por \\(10\\) os dois membros da igualdade anterior, temos: \\(10x = 34,\\left( 4 \\right)\\).<br \/>Subtraindo, membro a membro, as duas equa\u00e7\u00f5es anteriores, temos:<br \/>\\[\\begin{array}{*{20}{r}}{}&amp;{10x}&amp; = &amp;{34,\\left( 4 \\right)}\\\\ &#8211; &amp;x&amp; = &amp;{3,\\left( 4 \\right)}\\\\\\hline{}&amp;{9x}&amp; = &amp;{31\\quad\\;\\,\\,}\\end{array}\\]<br \/>Donde, \\(x = \\frac{{31}}{9}\\).<br \/>Portanto, \\(3,\\left( 4 \\right) = \\frac{{31}}{9}\\).<br \/><br \/><\/li>\n<li>Designando a d\u00edzima por \\(x\\), vem: \\(x = 1,\\left( {45} \\right)\\).<br \/>Logo, multiplicando por \\(100\\) os dois membros da igualdade anterior, temos: \\(100x = 145,\\left( {45} \\right)\\).<br \/>Subtraindo, membro a membro, as duas equa\u00e7\u00f5es anteriores, temos:<br \/>\\[\\begin{array}{*{20}{r}}{}&amp;{100x}&amp; = &amp;{145,\\left( {45} \\right)}\\\\ &#8211; &amp;x&amp; = &amp;{1,\\left( {45} \\right)}\\\\\\hline{}&amp;{99x}&amp; = &amp;{144\\quad\\;\\;\\;\\;\\,}\\end{array}\\]<br \/>Donde, \\(x = \\frac{{144}}{{99}} = \\frac{{16}}{{11}}\\).<br \/>Portanto, \\(1,\\left( {45} \\right) = \\frac{{16}}{{11}}\\).<br \/><br \/><\/li>\n<li>Designando a d\u00edzima por \\(x\\), vem: \\(x = 7,226\\left( {72} \\right)\\).<br \/>Logo, multiplicando por \\(1000\\)\u00a0 e por \\(100000\\) os dois membros da igualdade anterior, temos: \\(1000x = 7226,\\left( {72} \\right)\\) e \\(100000x = 722672,\\left( {72} \\right)\\), respetivamente.<br \/>Subtraindo, membro a membro, as duas equa\u00e7\u00f5es anteriores, temos:<br \/>\\[\\begin{array}{*{20}{r}}{}&amp;{100000x}&amp; = &amp;{722672,\\left( {72} \\right)}\\\\ &#8211; &amp;{1000x}&amp; = &amp;{7226,\\left( {72} \\right)}\\\\\\hline{}&amp;{99000x}&amp; = &amp;{715446\\quad\\quad\\,}\\end{array}\\]<br \/>Donde, \\(x = \\frac{{715446}}{{99000}} = \\frac{{79494}}{{11000}} = \\frac{{39747}}{{5500}}\\).<br \/>Portanto, \\(7,226\\left( {72} \\right) = \\frac{{39747}}{{5500}}\\).<br \/><br \/><\/li>\n<li>Designando a d\u00edzima por \\(x\\), vem: \\(x = 0,5\\left( 9 \\right)\\).<br \/>Logo, multiplicando por \\(10\\) e por \\(100\\) os dois membros da igualdade anterior, temos: \\(10x = 5,\\left( 9 \\right)\\) e \\(100x = 59,\\left( 9 \\right)\\), respetivamente.<br \/>Subtraindo, membro a membro, as duas equa\u00e7\u00f5es anteriores, temos:<br \/>\\[\\begin{array}{*{20}{r}}{}&amp;{100x}&amp; = &amp;{59,\\left( 9 \\right)}\\\\ &#8211; &amp;{10x}&amp; = &amp;{5,\\left( 9 \\right)}\\\\\\hline{}&amp;{90x}&amp; = &amp;{54\\quad\\;\\;\\,}\\end{array}\\]<br \/>Donde, \\(x = \\frac{{54}}{{90}} = \\frac{6}{{10}} = \\frac{3}{5} = 0,6\\).<br \/>Portanto, \\(0,5\\left( 9 \\right) = \\frac{3}{5}\\).<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_22369' onClick='GTTabs_show(0,22369)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Representa na forma de fra\u00e7\u00e3o os n\u00fameros racionais dados pelas seguintes d\u00edzimas infinitas peri\u00f3dicas: \\(3,\\left( 4 \\right)\\) \\(1,\\left( {45} \\right)\\) \\(7,226\\left( {72} \\right)\\) \\(0,5\\left( 9 \\right)\\) Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":22368,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,664],"tags":[424,261,666],"series":[],"class_list":["post-22369","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","tag-dizima-infinita-periodica"],"views":274,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/10\/Mat269.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/22369","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=22369"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/22369\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/22368"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=22369"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=22369"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=22369"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=22369"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}