{"id":6580,"date":"2011-03-09T18:15:54","date_gmt":"2011-03-09T18:15:54","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6580"},"modified":"2022-01-03T00:36:25","modified_gmt":"2022-01-03T00:36:25","slug":"utilizando-o-m-m-c","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6580","title":{"rendered":"Utilizando o m.m.c."},"content":{"rendered":"<p><ul id='GTTabs_ul_6580' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6580' class='GTTabs_curr'><a  id=\"6580_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6580' ><a  id=\"6580_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_6580'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Utilizando o m.m.c.,<\/p>\n<ol>\n<li>escreve por ordem crescente as frac\u00e7\u00f5es \\[\\begin{matrix}<br \/>\n\\frac{7}{6}, &amp; \\frac{5}{9}, &amp; \\frac{19}{24}\u00a0 \\\\<br \/>\n\\end{matrix}\\]<\/li>\n<li>calcula \\[\\frac{5}{12}+\\frac{4}{9}-\\frac{3}{20}\\]<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6580' onClick='GTTabs_show(1,6580)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6580'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Como:<br \/>\n$6=2\\times 3$<br \/>\n$9={{3}^{2}}$<br \/>\n$24={{2}^{3}}\\times 3$<\/p>\n<p>Ent\u00e3o, $m.m.c.(6,9,24)={{2}^{3}}\\times {{3}^{2}}=8\\times 9=72$.<\/p>\n<p>Podemos agora escrever fra\u00e7\u00f5es equivalentes \u00e0s dadas com igual denominador, para as comparar com facilidade:<br \/>\n\\[\\frac{7}{\\underset{(12)}{\\mathop{6}}\\,}=\\frac{84}{72}\\]<br \/>\n\\[\\frac{5}{\\underset{(8)}{\\mathop{9}}\\,}=\\frac{40}{72}\\]<br \/>\n\\[\\frac{19}{\\underset{(3)}{\\mathop{24}}\\,}=\\frac{57}{72}\\]<br \/>\nLogo, \\[\\frac{5}{9}&lt;\\frac{19}{24}&lt;\\frac{7}{6}\\]<br \/>\n\u00ad<\/p>\n<\/li>\n<li>Como:<br \/>\n$12={{2}^{2}}\\times 3$<br \/>\n$9={{3}^{2}}$<br \/>\n$20={{2}^{2}}\\times 5$<\/p>\n<p>Ent\u00e3o, $m.m.c.(9,12,20)={{2}^{2}}\\times {{3}^{2}}\\times 5=4\\times 9\\times 5=180$.<\/p>\n<p>Assim, temos:<br \/>\n\\[\\frac{5}{\\underset{(15)}{\\mathop{12}}\\,}+\\frac{4}{\\underset{(20)}{\\mathop{9}}\\,}-\\frac{3}{\\underset{(9)}{\\mathop{20}}\\,}=\\frac{75}{180}+\\frac{80}{180}-\\frac{27}{180}=\\frac{155}{180}-\\frac{27}{180}=\\frac{128}{180}=\\frac{64}{90}=\\frac{32}{45}\\]<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6580' onClick='GTTabs_show(0,6580)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Utilizando o m.m.c., escreve por ordem crescente as frac\u00e7\u00f5es \\[\\begin{matrix} \\frac{7}{6}, &amp; \\frac{5}{9}, &amp; \\frac{19}{24}\u00a0 \\\\ \\end{matrix}\\] calcula \\[\\frac{5}{12}+\\frac{4}{9}-\\frac{3}{20}\\] Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":19188,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,137,97],"tags":[140],"series":[],"class_list":["post-6580","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-ainda-os-numeros","category-aplicando","tag-minimo-multiplo-comum"],"views":1810,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat74.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6580","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=6580"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6580\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19188"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6580"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6580"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6580"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6580"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}