{"id":104461,"date":"2018-06-06T23:05:13","date_gmt":"2018-06-07T02:05:13","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=104461"},"modified":"2024-04-08T07:35:06","modified_gmt":"2024-04-08T10:35:06","slug":"problema-para-ajudar-na-escola-ali-babao-e-a-quarta-de-suas-40-equacoes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-ali-babao-e-a-quarta-de-suas-40-equacoes\/","title":{"rendered":".Problema para ajudar na escola: Ali Bab\u00e3o e a quarta de suas 40 equa\u00e7\u00f5es"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=216019\" rel=\"noopener\" target=\"_blank\">&#9999; <span style=\"color:#A0522D;font-size:14px;\">Link do problema para dispositivos da <em>Apple<\/em>.<\/span><\/a><\/strong><\/p>\n<div class=\"g1\" style=\"line-height:27px;\">\n<p class=\"tit\"  style=\"line-height:20px;margin:10px\">Problema <span style=\"font-size:12px; color:#0000FF\"><br \/>\n(A partir da 2\u00aa s\u00e9rie do E. M. &#8211; <span style=\"font-weight: normal;\"> N\u00edvel de dificuldade: M\u00e9dio<\/span>)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p>Resolva a equa\u00e7\u00e3o<br \/>\n[tex]\\qquad \\qquad \\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}=1.[\/tex]<\/p>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1\" 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<ul>\n<li>Elevando ambos os membros da equa\u00e7\u00e3o proposta ao quadrado, obtemos:<br \/>\n[tex]\\quad \\left(\\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}\\right)^2=1^2[\/tex]<br \/>\n[tex]\\quad 3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}=1[\/tex]<br \/>\n[tex]\\quad 3-1=\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}[\/tex]<br \/>\n[tex]\\quad 2=\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}\\,.\\qquad \\textcolor{#800000}{(i)}[\/tex]<\/li>\n<li>Agora, ao elevarmos os membros da equa\u00e7\u00e3o [tex]\\textcolor{#800000}{(i)}[\/tex] ao quadrado, segue que:<br \/>\n[tex]\\quad 2^2=\\left(\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}\\right)^2[\/tex]<br \/>\n[tex]\\quad 4=3+\\sqrt{x-\\sqrt{2x+1}}[\/tex]<br \/>\n[tex]\\quad 4-3=\\sqrt{x-\\sqrt{2x+1}}[\/tex]<br \/>\n[tex]\\quad 1=\\sqrt{x-\\sqrt{2x+1}} \\, .\\qquad \\textcolor{#800000}{(ii)}[\/tex]<\/li>\n<li>Elevando os membros da equa\u00e7\u00e3o [tex]\\textcolor{#800000}{(ii)}[\/tex] ao quadrado, temos que:<br \/>\n[tex]\\quad 1^2=\\left(\\sqrt{x-\\sqrt{2x+1}}\\right)^2[\/tex]<\/li>\n<p>[tex]\\quad 1=x-\\sqrt{2x+1}[\/tex]<\/li>\n<p>[tex]\\quad \\sqrt{2x+1}=x-1 \\, .\\qquad \\textcolor{#800000}{(iii)}[\/tex]<\/li>\n<li>Elevando os membros, agora, da equa\u00e7\u00e3o [tex]\\textcolor{#800000}{(iii)}[\/tex] ao quadrado:<br \/>\n[tex]\\quad \\left(\\sqrt{2x+1}\\right)^2=\\left(x-1\\right)^2[\/tex]<br \/>\n[tex]\\quad 2x+1=x^2-2x+1[\/tex]<br \/>\n[tex]\\quad x^2-4x=0[\/tex]<br \/>\n[tex]\\quad x(x-4)=0[\/tex]<br \/>\n[tex]\\quad x=0 \\, [\/tex] ou [tex]x= \\, 4 \\, .[\/tex]<\/li>\n<\/ul>\n<div class=\"cb36 g1\" style=\"margin:30px;line-height:30px; width:80%; margin-left:auto; margin-right:auto; \">\nCabe aqui uma palavrinha com rela\u00e7\u00e3o \u00e0s consequ\u00eancias do &quot; algebrismo &quot; que fizemos ao partir da equa\u00e7\u00e3o [tex]\\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}=1[\/tex] e chegar na equa\u00e7\u00e3o [tex]2x+1=x^2-2x+1[\/tex], cujas ra\u00edzes s\u00e3o [tex]0 \\, [\/tex] e [tex] \\, 4 \\, .[\/tex]<\/p>\n<ul>\n<li>Observe que se [tex]z=5[\/tex], ent\u00e3o podemos concluir que [tex]z^2=25[\/tex], uma vez que <\/li>\n<p>[tex]\\quad z^2=z\\cdot z=5\\cdot 5=25 \\, .[\/tex]<br \/>\nNo entanto, se [tex]z^2=25[\/tex] e se n\u00e3o tivermos outras informa\u00e7\u00f5es sobre [tex]z[\/tex], ent\u00e3o n\u00e3o podemos simplesmente concluir que [tex]z=5[\/tex], pois poder\u00edamos ter [tex]z=-5.[\/tex]<br \/>\nEm termos de s\u00edmbolos, estamos afirmando que<br \/>\n[tex]\\qquad z=5 \\Rightarrow z^2=25 [\/tex],<br \/>\nmas que<br \/>\n[tex]\\qquad z^2=25 \\nRightarrow z=5[\/tex],<br \/>\ne, com isso,<br \/>\n[tex]\\qquad z^2=25 \\nLeftrightarrow z=5[\/tex],<br \/>\no que quer dizer que <strong>as duas equa\u00e7\u00f5es n\u00e3o s\u00e3o equivalentes<\/strong>.<br \/>\n<span class=\"c3\"><strong>E se duas equa\u00e7\u00f5es n\u00e3o s\u00e3o equivalentes, elas n\u00e3o t\u00eam necessariamente as mesmas solu\u00e7\u00f5es!<\/strong><\/span><\/ul>\n<\/div>\n<p>Observe, ent\u00e3o, a sequ\u00eancia l\u00f3gica dos procedimentos de elevarmos ao quadrado ambos os termos da nossa sucess\u00e3o de igualdades:<br \/>\n[tex]\\qquad \\qquad \\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}=1 \\Longrightarrow  2=\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}} \\Longrightarrow \\\\<br \/>\n \\qquad \\qquad \\Longrightarrow 1=\\sqrt{x-\\sqrt{2x+1}}\\Longrightarrow \\sqrt{2x+1}=x-1 \\Longrightarrow   x^2-4x=0 \\, .[\/tex]<br \/>\nResumidamente, podemos concluir, somente,  que<br \/>\n[tex]\\qquad \\qquad \\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}=1 \\Longrightarrow   x^2-4x=0 \\, [\/tex]<br \/>\ne isso significa que todas as ra\u00edzes de [tex]\\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}=1 [\/tex] s\u00e3o ra\u00edzes de [tex]x^2-4x=0 \\, [\/tex], mas as ra\u00edzes de [tex]x^2-4x=0 \\, [\/tex] n\u00e3o s\u00e3o necessariamente  ra\u00edzes de [tex] \\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}=1 [\/tex].<br \/>\nDessa forma, devemos testar as duas ra\u00edzes da equa\u00e7\u00e3o [tex]x^2-4x=0 \\, [\/tex], para verificarmos se, ao fazermos sucessivas opera\u00e7\u00f5es de &#8220;elevar ao quadrado&#8221;, n\u00e3o ficamos com ra\u00edzes a mais.<br \/>\nVamos l\u00e1:<\/p>\n<ul>\n<li>Se [tex]x=0[\/tex] ent\u00e3o:<br \/>\n[tex]\\quad \\begin{align*} \\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}&#038;=\\sqrt{3-\\sqrt{3+\\sqrt{0-\\sqrt{2\\times 0+1}}}}\\\\<br \/>\n&#038;=\\sqrt{3-\\sqrt{3+\\sqrt{-1}}}\\ne 1 \\end{align*}[\/tex]<br \/>\numa vez que [tex] \\sqrt{-1}[\/tex] n\u00e3o \u00e9 um n\u00famero real.<br \/>\nAssim, <strong>[tex]x=0[\/tex] n\u00e3o \u00e9 solu\u00e7\u00e3o da equa\u00e7\u00e3o original.<\/strong><\/li>\n<li>Se [tex]x=4[\/tex]  ent\u00e3o:<br \/>\n[tex]\\quad \\begin{align*} \\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}&#038;=\\sqrt{3-\\sqrt{3+\\sqrt{4-\\sqrt{2\\times 4+1}}}}\\\\<br \/>\n&#038;=\\sqrt{3-\\sqrt{3+\\sqrt{4-3}}}=\\sqrt{3-\\sqrt{4}}=\\sqrt{1}=1 \\end{align*}[\/tex]<br \/>\ne, assim, <strong>[tex]x=4[\/tex] \u00e9 solu\u00e7\u00e3o da equa\u00e7\u00e3o original<\/strong>.<\/li>\n<\/ul>\n<p>Finalizando, a equa\u00e7\u00e3o [tex] \\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}=1 \\, [\/tex]tem uma \u00fanica ra\u00edz: [tex] \\, \\fcolorbox{black}{#eee0e5}{$x=4$} \\, .[\/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:20px\">\n<table style=\" height:auto; margin-left: auto; margin-right: auto;width:640px;\">\n<tr>\n<td class=\"bf\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1001 alignright\" style=\"border: 0px;\" alt=\"\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"80\" height=\"115\" \/><\/td>\n<td class=\"bf\">\n<p class=\"fala\" style=\"width:520px;background-color: #FFF5EE; padding:10px; font-family: calibri; font-size: 13px;color: #1C1C1C; line-height: 20px; \"> Se for conveniente, voc\u00ea pode obter um arquivo PDF desta p\u00e1gina, com o problema e a solu\u00e7\u00e3o, clicando no bot\u00e3o abaixo.<br \/>\nVoc\u00ea pode abrir o arquivo diretamente no seu navegador (<em>Chrome<\/em>, <em>Edge<\/em>, <em>Firefox<\/em>, <em>Safari<\/em>, entre outros), mas tamb\u00e9m pode utilizar o software gratuito <em><strong>Adobe Acrobat Reader<\/strong><\/em>.<br \/>\nCaso o dispositivo que voc\u00ea est\u00e1 utilizando n\u00e3o tenha o <strong><em>Acrobat Reader<\/em><\/strong> instalado, \u00e9 s\u00f3 clicar <a href=\"http:\/\/get.adobe.com\/br\/reader\/\" rel=\"noopener\" target=\"_blank\"><strong>AQUI<\/strong><\/a> para fazer o download adequado ao seu dispositivo.<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<div style=\" float:right;  width: 100px;\">\n<p style=\"text-align: center; border-style: dotted; border-color :#FFFAFA; border-width :thin; width:90px;  height:40px;  padding: 5px; background-color: #800000;\">\n<a href=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/Ali-Bab\u00e3o-e-a-quarta-de-suas-40-equa\u00e7\u00f5es.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Ali Bab\u00e3o e a quarta de suas 40 equa\u00e7\u00f5es.pdf\"><span style=\"color: #FFFAFA; font-family: georgia; font-size: 16px;\"><strong>Arquivo <\/strong><\/span><\/a>\n <\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema (A partir da 2\u00aa s\u00e9rie do E. M. &#8211; N\u00edvel de dificuldade: M\u00e9dio) Resolva a equa\u00e7\u00e3o [tex]\\qquad \\qquad \\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}=1.[\/tex] Solu\u00e7\u00e3o Elevando ambos os membros da equa\u00e7\u00e3o proposta ao quadrado, obtemos: [tex]\\quad \\left(\\sqrt{3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}}\\right)^2=1^2[\/tex] [tex]\\quad 3-\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}=1[\/tex] [tex]\\quad 3-1=\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}[\/tex] [tex]\\quad 2=\\sqrt{3+\\sqrt{x-\\sqrt{2x+1}}}\\,.\\qquad \\textcolor{#800000}{(i)}[\/tex] Agora, ao elevarmos os membros da equa\u00e7\u00e3o &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-ali-babao-e-a-quarta-de-suas-40-equacoes\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-104461","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/104461","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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=104461"}],"version-history":[{"count":16,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/104461\/revisions"}],"predecessor-version":[{"id":216021,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/104461\/revisions\/216021"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=104461"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}