{"id":128413,"date":"2019-06-05T22:23:20","date_gmt":"2019-06-06T01:23:20","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=128413"},"modified":"2023-05-31T18:38:29","modified_gmt":"2023-05-31T21:38:29","slug":"probleminha-proporcoes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/probleminha-proporcoes\/","title":{"rendered":".Probleminha: Propor\u00e7\u00f5es"},"content":{"rendered":"<p style=\"margin:30px\">\n<div class=\"g1\" style=\"line-height:25px;\">\n<p class=\"tit\"style=\"line-height:20px; margin-bottom: 10px;\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 8\u00ba ano do E. F.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p>Sejam [tex]a[\/tex], [tex]b[\/tex] e [tex]c[\/tex] n\u00fameros reais n\u00e3o nulos.<br \/>\nSabendo que as propor\u00e7\u00f5es [tex]\\boxed{\\dfrac{a+b}{c}=\\dfrac{b+c}{a}=\\dfrac{a+c}{b}}[\/tex] s\u00e3o v\u00e1lidas, qual o valor num\u00e9rico da express\u00e3o [tex] \\, \\boxed{\\dfrac{a+b}{c}} \\, [\/tex]?<\/p>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"cb36 g1n\" style=\"margin:25px;line-height:30px; width:85%; margin-left:auto; margin-right:auto; \"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" style=\"border:none\"  alt=\"explicador_p\" width=\"90\" height=\"108\" class=\"alignleft size-full wp-image-1001\" \/> <\/p>\n<p class=\"c3 ti3\"style=\"line-height:20px; margin-bottom: 10px;\"> Lembrete<\/p>\n<p style=\"margin:10px\">\n<p>Em uma propor\u00e7\u00e3o, a soma (ou a diferen\u00e7a) dos antecedentes est\u00e1 para a soma (ou a diferen\u00e7a) dos consequentes, assim como cada antecedente est\u00e1 para o seu respectivo consequente.<br \/>\nEm s\u00edmbolos:<\/p>\n<p style=\"text-align:center; margin-bottom:20px;\">[tex] \\dfrac{x}{y}=\\dfrac{z}{t}  \\,  \\Rightarrow  \\, \\begin{cases}<br \/>\n \\dfrac{x\\pm z}{y\\pm t}=\\dfrac{x}{y}\\\\<br \/>\n \\,  \\, \\\\<br \/>\n \\dfrac{x\\pm z}{y\\pm t}=\\dfrac{z}{t}\\\\<br \/>\n\\end{cases}[\/tex].\n<\/p>\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1\" style=\"line-height:33px;\">\n<p class=\"tit\" style=\"line-height:30px; margin-bottom: 10px;\">Solu\u00e7\u00e3o <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 30px;border: 0 none\"\/>\n<ul>\n<li>Se [tex]a+b+c\\ne 0[\/tex], a partir das propor\u00e7\u00f5es [tex]\\dfrac{a+b}{c}=\\dfrac{b+c}{a}=\\dfrac{a+c}{b}[\/tex] e da propriedade citada no Lembrete, podemos garantir que:<\/li>\n<p>[tex]\\qquad \\qquad \\dfrac{a+b}{c}=\\dfrac{b+c}{a}=\\dfrac{a+c}{b}=\\dfrac{2\\cdot (a+b+c)}{a+b+c}.\\\\ \\,  \\, [\/tex]<br \/>\nPortanto,<br \/>\n[tex]\\qquad \\qquad \\boxed{\\dfrac{a+b}{c}}=\\dfrac{2\\cdot (a+b+c)}{a+b+c}=\\boxed{2}.[\/tex]<\/p>\n<li>Agora, se [tex]a+b+c= 0[\/tex], temos que [tex]a+b=-c[\/tex]. Logo, <\/li>\n<p> [tex]\\qquad \\qquad \\boxed{\\dfrac{a+b}{c}}=\\dfrac{-c}{c}=\\boxed{-1}.[\/tex]<\/p>\n<p>Portanto,  [tex] \\, \\fcolorbox{black}{#eee0e5}{$\\dfrac{a+b}{c}=-1$} \\, [\/tex] ou [tex] \\, \\fcolorbox{black}{#eee0e5}{$\\dfrac{a+b}{c}=2$} \\, [\/tex].<\/ul>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:10px\">\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 8\u00ba ano do E. F.) Sejam [tex]a[\/tex], [tex]b[\/tex] e [tex]c[\/tex] n\u00fameros reais n\u00e3o nulos. Sabendo que as propor\u00e7\u00f5es [tex]\\boxed{\\dfrac{a+b}{c}=\\dfrac{b+c}{a}=\\dfrac{a+c}{b}}[\/tex] s\u00e3o v\u00e1lidas, qual o valor num\u00e9rico da express\u00e3o [tex] \\, \\boxed{\\dfrac{a+b}{c}} \\, [\/tex]? Lembrete Em uma propor\u00e7\u00e3o, a soma (ou a diferen\u00e7a) dos antecedentes est\u00e1 para a soma (ou a diferen\u00e7a) &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/probleminha-proporcoes\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":668,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-128413","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/128413","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/668"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=128413"}],"version-history":[{"count":9,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/128413\/revisions"}],"predecessor-version":[{"id":192598,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/128413\/revisions\/192598"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=128413"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}