{"id":89226,"date":"2017-11-13T22:23:46","date_gmt":"2017-11-14T00:23:46","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=89226"},"modified":"2023-04-14T02:41:32","modified_gmt":"2023-04-14T05:41:32","slug":"desafio-ali-babao-e-a-primeira-de-suas-40-equacoes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/desafio-ali-babao-e-a-primeira-de-suas-40-equacoes\/","title":{"rendered":".Desafio: Ali Bab\u00e3o e a primeira de suas 40 equa\u00e7\u00f5es"},"content":{"rendered":"<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:28px;\">\n<p class=\"tit\" style=\"line-height:20px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 9\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>Ali Bab\u00e3o \u00e9 um mestre que adora resolver equa\u00e7\u00f5es. Prop\u00f4s o seguinte problema para seus disc\u00edpulos matem\u00e1ticos:<\/p>\n<ul>\n<li>&#8220;Resolva a equa\u00e7\u00e3o [tex]\\boxed{(x^2-3x+1)^2-3(x^2-3x+1)+1=x}[\/tex] no conjunto dos n\u00fameros reais.&#8221;<\/li>\n<\/ul>\n<p>Qual a solu\u00e7\u00e3o correta?<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:35px;\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p> Adicionando [tex]\\textcolor{red}{x^2-3x+1}[\/tex] a ambos os lados da igualdade [tex](x^2-3x+1)^2-3(x^2-3x+1)+1=x \\, [\/tex], segue que:<br \/>\n[tex]\\quad(x^2-3x+1)^2-3(x^2-3x+1)+1\\textcolor{red}{+(x^2-3x+1)}=x\\textcolor{red}{+(x^2-3x+1)}  [\/tex]<br \/>\n[tex]\\quad (x^2-3x+1)^2-2(x^2-3x+1)+1=(x^2-3x+1)+x[\/tex]<br \/>\n[tex]\\quad (x^2-3x+1)^2-2(x^2-3x+1)+1=x^2-2x+1[\/tex]<br \/>\n[tex]\\quad \\left((x^2-3x+1)-1\\right)^2=(x-1)^2[\/tex]<br \/>\n[tex]\\quad \\left(x^2-3x\\right)^2=(x-1)^2[\/tex]<br \/>\n[tex]\\quad (x^2-3x)^2-(x-1)^2=0[\/tex]<br \/>\n[tex]\\quad ((x^2-3x)+(x-1))\\cdot ((x^2-3x)-(x-1))=0[\/tex]<br \/>\n[tex]\\quad (x^2-3x+x-1)\\cdot (x^2-3x-x+1)=0[\/tex]<br \/>\n[tex]\\quad (x^2-2x-1)\\cdot (x^2-4x+1)=0[\/tex].<br \/>\nAssim, temos que [tex]\\boxed{x^2-2x-1=0} \\, [\/tex] ou [tex] \\, \\boxed{x^2-4x+1=0} \\, [\/tex], donde obtemos que [tex] \\, \\fcolorbox{black}{#eee0e5}{$x=1\\pm \\sqrt{2}$} \\, [\/tex] ou [tex] \\, \\fcolorbox{black}{#eee0e5}{$x=2\\pm \\sqrt{3}$} \\, [\/tex].<\/p>\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:30px\">\n<table style=\" height:auto; margin-left: auto;margin-right:auto;width:562px;\">\n<tr>\n<td class=\"bf\"><img loading=\"lazy\" decoding=\"async\" class=\" wp-image-1001 alignrightt\" style=\"border: 0px;\" alt=\"explicador_p\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"90\" height=\"108\" \/><\/td>\n<td class=\"bf\">\n<p class=\"fala bg33 cb33\" style=\"width:430px; line-height: 20px; font-size:medium\" > Olhando a igualdade [tex]\\boxed{\\left(x^2-3x\\right)^2=(x-1)^2} [\/tex], resista \u00e0 tenta\u00e7\u00e3o de &#8220;cortar os quadrados&#8221; e concluir que [tex]\\boxed{x^2-3x=x-1} [\/tex].<br \/>\n<span class=\"c3\" style=\"padding-left:30px;\"><strong>Esse procedimento n\u00e3o \u00e9 correto!<\/strong><\/span><\/p>\n<\/td>\n<\/tr>\n<\/table>\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 9\u00ba ano do E. F.) Ali Bab\u00e3o \u00e9 um mestre que adora resolver equa\u00e7\u00f5es. Prop\u00f4s o seguinte problema para seus disc\u00edpulos matem\u00e1ticos: &#8220;Resolva a equa\u00e7\u00e3o [tex]\\boxed{(x^2-3x+1)^2-3(x^2-3x+1)+1=x}[\/tex] no conjunto dos n\u00fameros reais.&#8221; Qual a solu\u00e7\u00e3o correta? Solu\u00e7\u00e3o Adicionando [tex]\\textcolor{red}{x^2-3x+1}[\/tex] a ambos os lados da igualdade [tex](x^2-3x+1)^2-3(x^2-3x+1)+1=x \\, [\/tex], segue que: [tex]\\quad(x^2-3x+1)^2-3(x^2-3x+1)+1\\textcolor{red}{+(x^2-3x+1)}=x\\textcolor{red}{+(x^2-3x+1)} &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/desafio-ali-babao-e-a-primeira-de-suas-40-equacoes\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":164,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-89226","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/89226","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\/164"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=89226"}],"version-history":[{"count":31,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/89226\/revisions"}],"predecessor-version":[{"id":186907,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/89226\/revisions\/186907"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=89226"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}