{"id":188168,"date":"2023-05-02T22:52:51","date_gmt":"2023-05-03T01:52:51","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=188168"},"modified":"2023-05-21T00:39:54","modified_gmt":"2023-05-21T03:39:54","slug":"problemao-manipulacao-algebrica-s","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-manipulacao-algebrica-s\/","title":{"rendered":".Problem\u00e3o: Manipula\u00e7\u00e3o alg\u00e9brica"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=189935\" 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:25px;padding-bottom:15px\">\n<p class=\"tit\" style=\"line-height:25px;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\"\/>\nDados os n\u00fameros reais n\u00e3o nulos [tex]a,\\, b\\,[\/tex] e [tex]\\,c[\/tex], determine o valor da express\u00e3o<\/p>\n<p style=\"margin:20px\">\n[tex]\\qquad \\dfrac{a^5+b^5+c^5}{5} \\cdot \\dfrac{a^2+b^2+c^2}{a^2 \\cdot b^3 \\cdot c^2}[\/tex],\n<\/p>\n<p>de forma que se tenha [tex]a^2+b^2+c^2=a \\cdot b+a \\cdot c+b \\cdot c[\/tex].<\/p>\n<p style=\"font-size: 12px;text-align:right;line-height:1px;\">Extra\u00eddo de <strong> \u00c1lgebra \u2013 Cole\u00e7\u00e3o Ci\u00eancias e Humanas<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"g1n\" style=\"line-height:28px;\">\n<p class=\"tit\" style=\"line-height:28px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 1px; margin-bottom: 30px;border: 0 none\"\/>\n<ul>\n<li>Inicialmente, tomemos a express\u00e3o [tex]\\boxed{a^2+b^2+c^2=a \\cdot b + a \\cdot c + b \\cdot c}[\/tex] e vamos multiplic\u00e1-la por [tex]2[\/tex].<br \/>\nCom isso, obtemos a express\u00e3o<br \/>\n[tex]\\qquad \\qquad 2 \\cdot a^2+2 \\cdot b^2+2 \\cdot c^2=2 \\cdot a \\cdot b+2 \\cdot a \\cdot c+2 \\cdot b \\cdot c\\,[\/tex].<\/li>\n<li>Agora, vamos reescrever os termos do primeiro membro desta \u00faltima equa\u00e7\u00e3o da seguinte forma:<br \/>\n[tex]\\qquad \\qquad a^2+a^2+b^2+b^2+c^2+c^2=2 \\cdot a \\cdot b+2 \\cdot a \\cdot c+2 \\cdot b \\cdot c\\,[\/tex].<\/li>\n<li>Passando os termos do segundo para o primeiro membro e reagrupando, obtemos:<br \/>\n[tex]\\qquad \\qquad (a^2-2 \\cdot a \\cdot b+b^2)+(b^2\u20132 \\cdot b \\cdot c+c^2)+(a^2-2 \\cdot a \\cdot c+c^2)=0[\/tex].<\/li>\n<li>Perceba que cada express\u00e3o entre par\u00eanteses corresponde a um trin\u00f4mio quadrado perfeito. Logo, podemos reescrever a equa\u00e7\u00e3o assim:<br \/>\n[tex]\\qquad \\qquad (a-b)^2+(b-c)^2+(a-c)^2=0\\,.\t\\qquad \\textcolor{#800000}{(*)}[\/tex]<\/li>\n<li>Como o quadrado de todo n\u00famero real ou \u00e9 igual a zero ou \u00e9 maior que zero, para que a soma dos quadrados de [tex] \\textcolor{#800000}{(*)}[\/tex] seja zero, \u00e9 necess\u00e1rio que cada parcela seja zero. Desta forma, segue que: <\/li>\n<\/ul>\n<table class=\"bf\" style=\"width:40%; margin-left:50px; \">\n<tr class=\"bf\" >\n<td class=\"bf\" style=\"border-right:2px solid #800000\">\n[tex](a-b)^2=0[\/tex]<br \/>\n[tex]a-b=0[\/tex]<br \/>\n[tex]a=b[\/tex]<\/td>\n<td class=\"bf\" style=\"border-right:2px solid #800000\">\n[tex](b-c)^2=0[\/tex]<br \/>\n[tex]b-c=0[\/tex]<br \/>\n[tex]b=c[\/tex]<\/td>\n<td class=\"bf\">\n[tex](a-c)^2=0[\/tex]<br \/>\n[tex]a\u2013c=0[\/tex]<br \/>\n[tex]a=c[\/tex]<\/td>\n<\/tr>\n<\/table>\n<p>Sinteticamente, temos [tex]\\boxed{~a=b=c~}[\/tex] e, finalmente, podemos calcular o valor da express\u00e3o em quest\u00e3o:<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\quad\\begin{align}<br \/>\n\\dfrac{a^5+b^5+c^5}{5} \\cdot \\dfrac{a^2+b^2+c^2}{a^2 \\cdot b^3 \\cdot c^2}&#038;= \\dfrac{a^5+a^5+a^5}{5} \\cdot \\dfrac{a^2+a^2+a^2}{a^2 \\cdot a^3 \\cdot a^2} \\\\<br \/>\n&#038;=\\dfrac{3 \\cdot a^5}{5} \\cdot \\dfrac{3 \\cdot a^2}{a^7}\\\\<br \/>\n&#038;=\\dfrac{9 \\cdot a^7}{5 \\cdot a^7}\\\\<br \/>\n&#038;=\\dfrac{9}{5}\\,.<br \/>\n\\end{align}[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>Assim, [tex]\\,\\fcolorbox{black}{#eee0e5}{$\\dfrac{a^5+b^5+c^5}{5} \\cdot \\dfrac{a^2+b^2+c^2}{a^2 \\cdot b^3 \\cdot c^2}=\\dfrac{9}{5}$}[\/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\" >Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:10px\">\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema (Indicado a partir do 9\u00ba ano do E. F.) Dados os n\u00fameros reais n\u00e3o nulos [tex]a,\\, b\\,[\/tex] e [tex]\\,c[\/tex], determine o valor da express\u00e3o [tex]\\qquad \\dfrac{a^5+b^5+c^5}{5} \\cdot \\dfrac{a^2+b^2+c^2}{a^2 \\cdot b^3 \\cdot c^2}[\/tex], de forma que se tenha [tex]a^2+b^2+c^2=a \\cdot b+a \\cdot c+b \\cdot c[\/tex]. Extra\u00eddo de &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-manipulacao-algebrica-s\/\">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-188168","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/188168","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=188168"}],"version-history":[{"count":9,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/188168\/revisions"}],"predecessor-version":[{"id":191206,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/188168\/revisions\/191206"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=188168"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}