{"id":123523,"date":"2019-03-07T20:31:34","date_gmt":"2019-03-07T23:31:34","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=123523"},"modified":"2024-06-10T04:20:08","modified_gmt":"2024-06-10T07:20:08","slug":"problema-para-ajudar-na-escola-expressoes-equivalentes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-expressoes-equivalentes\/","title":{"rendered":".Problema para ajudar na escola: Express\u00f5es equivalentes?"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=220113\" 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=\"g1n\" style=\"line-height:37px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#00BF00\"> (A partir do 9\u00ba ano do E. F. &#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: 1px; margin-bottom: 20px;border: 0 none\"\/>\n<p style=\"margin:10px;\">\n<p>(<strong>UNIFOR <\/strong>\u2013 CE) Sejam [tex]a[\/tex] e [tex]b[\/tex] n\u00fameros reais tais que [tex]a \\ne \\pm b[\/tex].<br \/>\nA express\u00e3o<br \/>\n[tex]\\quad  \\qquad \\dfrac{a^2+ab}{a^2-b^2}- \\dfrac{a^2+ab}{a^2+2ab+b^2 }[\/tex]<br \/>\nequivale \u00e0 express\u00e3o<br \/>\n[tex]\\qquad \\qquad \\dfrac{2}{a-b}[\/tex] ?\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"cb36 g1\" style=\"margin:30px;line-height:30px; width:70%; margin-left:auto; margin-right:auto; \">\n<p class=\"c3 ti3\" style=\"line-height:25px;margin:10px\"> Ajuda<\/p>\n<p style=\"margin:10px\">\n<p>Para resolver este problema, faremos uso dos seguintes produtos not\u00e1veis importantes:<\/p>\n<ul> [tex]\\textcolor{#800000}{(01)}\\quad (a +b)^2 = a^2 +2ab +b^2 \\, ; \\, \\forall  \\, a, \\, b \\, \\in \\mathbb{R}[\/tex].<br \/>\n[tex]\\textcolor{#800000}{(02)}\\quad  a^2 -b^2= (a+b) \\, (a-b) \\, ; \\, \\forall \\, a, \\, b\\in \\mathbb{R}.[\/tex]<\/ul>\n<table style=\" height:auto; margin-left: auto; margin-right:auto; width:99%;\">\n<tr>\n<td class=\"bf\" style=\"width:11%;\"><img loading=\"lazy\" decoding=\"async\" class=\" wp-image-1001 alignright\" style=\"border: 0px;\" alt=\"explicador_p\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"80\" height=\"100\" \/><\/td>\n<td class=\"bf\" style=\"width:85%;\">\n<p class=\"fala bg33 cb33\" style=\"width:89%; line-height: 20px; font-size:small\" > Para relembrar estes e outros produtos not\u00e1veis, voc\u00eas podem conferir a <a href=\"http:\/\/clubes.obmep.org.br\/blog\/sala-de-estudos-malabarismos-aritmeticos-e-algebricos\/\" target=\"_blank\"><strong>Sala de Estudos Malabarismos Aritm\u00e9ticos e Alg\u00e9bricos<\/strong><\/a>.  <\/p>\n<\/td>\n<\/tr>\n<\/table>\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:30px;\">\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: 1px; margin-bottom: 20px;border: 0 none\"\/>\n<p style=\"margin:10px\">\n<p>Sejam [tex]a[\/tex] e [tex]b[\/tex] n\u00fameros reais tais que [tex]a \\ne \\pm b[\/tex] e observe a seguinte sequ\u00eancia de igualdades equivalentes:<br \/>\n[tex]\\qquad \\begin{align*}\\boxed{\\dfrac{a^2+ab}{a^2-b^2}- \\dfrac{a^2+ab}{a^2+2ab+b^2 }}&#038;=\\dfrac{a^2+ab}{a^2-b^2}- \\dfrac{a\\,\\left(a+b\\right)}{a^2+2ab+b^2}\\\\<br \/>\n&#038;\\stackrel{\\textcolor{#800000}{(01)}}{=}\\dfrac{a^2+ab}{a^2-b^2}- \\dfrac{a\\,\\left(a+b\\right)}{(a+b)^2}\\\\<br \/>\n&#038;=\\dfrac{a^2+ab}{a^2-b^2}- \\dfrac{a\\,\\cancel{\\left(a+b\\right)}}{(a+b)\\,\\cancel{\\left(a+b\\right)}}\\\\<br \/>\n&#038; \\stackrel{\\textcolor{#800000}{(02)}}{=}\\dfrac{a^2+ab}{(a+b)\\,(a-b)}- \\dfrac{a}{(a+b)}\\\\<br \/>\n&#038; =\\dfrac{a^2+ab}{(a+b)\\,(a-b)}- \\dfrac{a\\,(a-b)}{(a+b)\\,(a-b)}\\\\<br \/>\n&#038; =\\dfrac{a^2+ab}{(a+b)\\,(a-b)}- \\dfrac{a^2-ab}{(a+b)\\,(a-b)}\\\\<br \/>\n&#038;\\stackrel{\\textcolor{#800000}{(02)}}{=} \\dfrac{a^2+ab}{a^2-b^2}- \\dfrac{a^2-ab}{a^2-b^2}\\\\<br \/>\n&#038;= \\dfrac{a^2+ab-(a^2-ab)}{a^2-b^2}\\\\<br \/>\n&#038;= \\dfrac{a^2+ab-a^2+ab}{a^2-b^2}\\\\<br \/>\n&#038;= \\dfrac{\\cancel{a^2}+ab-\\cancel{a^2}+ab}{a^2-b^2}\\\\<br \/>\n&#038;= \\boxed{\\dfrac{2ab}{a^2-b^2}}\\\\<br \/>\n\\end{align*}[\/tex]<br \/>\nPor essa sequ\u00eancia, conclu\u00edmos que [tex]\\dfrac{a^2+ab}{a^2-b^2}- \\dfrac{a^2+ab}{a^2+2ab+b^2 }=\\dfrac{2ab}{a^2-b^2} \\, .[\/tex]<br \/>\nPara que a express\u00e3o [tex]\\dfrac{a^2+ab}{a^2-b^2}- \\dfrac{a^2+ab}{a^2+2ab+b^2 }[\/tex] seja equivalente \u00e0 express\u00e3o [tex] \\dfrac{2}{a-b}[\/tex] para quaisquer n\u00fameros reais [tex]a[\/tex] e [tex]b[\/tex] tais que [tex]a \\ne \\pm b[\/tex], devemos ter [tex]\\dfrac{2}{a-b}=\\dfrac{2ab}{a^2-b^2}[\/tex] e, assim:<br \/>\n[tex]\\qquad \\dfrac{2}{a-b}=\\dfrac{2ab}{a^2-b^2}[\/tex]<br \/>\n[tex]\\qquad \\dfrac{2}{a-b}=\\dfrac{2ab}{(a+b)\\,(a-b)}[\/tex]<br \/>\n[tex]\\qquad \\dfrac{\\bcancel{2}}{\\cancel{a-b}}=\\dfrac{\\bcancel{2}ab}{(a+b)\\,\\cancel{(a-b)}}[\/tex]<br \/>\n[tex]\\qquad 1=\\dfrac{ab}{a+b}[\/tex]<br \/>\n[tex]\\qquad ab=a+b.[\/tex]<br \/>\nDessa forma, para que a equival\u00eancia das express\u00f5es ocorra, dever\u00edamos ter n\u00fameros reais [tex]a[\/tex] e [tex]b[\/tex], [tex]a \\ne \\pm b[\/tex], cuja soma fosse igual ao produto. Portanto, para n\u00fameros reais [tex]a[\/tex] e [tex]b[\/tex] <strong>quaisquer<\/strong>, com [tex]a \\ne \\pm b[\/tex], n\u00e3o temos a equival\u00eancia proposta.<\/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\/12\/Express\u00f5es-equivalentes-.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Express\u00f5es equivalentes.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 do 9\u00ba ano do E. F. &#8211; N\u00edvel de dificuldade: M\u00e9dio) (UNIFOR \u2013 CE) Sejam [tex]a[\/tex] e [tex]b[\/tex] n\u00fameros reais tais que [tex]a \\ne \\pm b[\/tex]. A express\u00e3o [tex]\\quad \\qquad \\dfrac{a^2+ab}{a^2-b^2}- \\dfrac{a^2+ab}{a^2+2ab+b^2 }[\/tex] equivale \u00e0 express\u00e3o [tex]\\qquad \\qquad \\dfrac{2}{a-b}[\/tex] ? Ajuda Para resolver este &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-expressoes-equivalentes\/\">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-123523","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/123523","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=123523"}],"version-history":[{"count":24,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/123523\/revisions"}],"predecessor-version":[{"id":220115,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/123523\/revisions\/220115"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=123523"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}