{"id":102524,"date":"2018-05-05T19:00:04","date_gmt":"2018-05-05T22:00:04","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=102524"},"modified":"2023-04-20T22:02:22","modified_gmt":"2023-04-21T01:02:22","slug":"problemao-sistema-de-equacoes-2","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-sistema-de-equacoes-2\/","title":{"rendered":".Problem\u00e3o: Sistema de Equa\u00e7\u00f5es (2)"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\" style=\"line-height:30px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 1\u00ba ano do E. M.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nDetermine as solu\u00e7\u00f5es reais do seguinte sistema de equa\u00e7\u00f5es:<\/p>\n<p>[tex]\\qquad \\qquad  \\left\\{ \\begin{array}{lll} x+y &#038;=&#038; 3 \\\\ x^3+y^3 &#038; =&#038; 18  \\end{array} \\right.[\/tex] .<\/p>\n<p style=\"margin:20px\">\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: 10px; margin-bottom: 20px;border: 0 none\"\/>\n\tUsando o produto not\u00e1vel &#8220;cubo perfeito&#8221; vemos que<br \/>\n[tex]\\qquad \\qquad (x+y)^3=x^3+3x^2y+3xy^2+y^3=x^3+y^3+3xy(x+y),[\/tex]<br \/>\nou seja,<br \/>\n[tex]\\qquad \\qquad (x+y)^3=x^3+y^3+3xy(x+y).\\qquad \\textcolor{#800000}{(i)}[\/tex]<br \/>\nUtilizando as equa\u00e7\u00f5es do sistema na identidade [tex] \\textcolor{#800000}{(i)}[\/tex], obtemos que [tex] 3^3=18+9xy \\,  [\/tex], donde [tex]xy=1.[\/tex]<br \/>\nDe [tex]x+y=3[\/tex] segue que [tex]\\boxed{y=3-x}[\/tex]; assim, substituindo essa rela\u00e7\u00e3o em [tex]\\boxed{xy=1}[\/tex] obtemos que<br \/>\n[tex]\\qquad 3x-x^2=1 [\/tex]<br \/>\n[tex]\\qquad x^2-3x+1=0.\\qquad \\textcolor{#800000}{(ii)}[\/tex]<br \/>\nNote que as solu\u00e7\u00f5es da equa\u00e7\u00e3o do segundo grau [tex] \\textcolor{#800000}{(ii)}[\/tex] s\u00e3o<br \/>\n[tex]\\qquad \\qquad x_1=\\frac{3+\\sqrt{5}}{2}\\qquad [\/tex] e [tex]\\qquad x_2=\\frac{3-\\sqrt{5}}{2}[\/tex]<br \/>\ne, que  a partir dessas solu\u00e7\u00f5es, obtemos respectivamente os valores<br \/>\n[tex]\\qquad \\qquad y_1=\\frac{3-\\sqrt{5}}{2}\\qquad [\/tex] e [tex]\\qquad y_2=\\frac{3+\\sqrt{5}}{2}[\/tex],<br \/>\npara [tex]y[\/tex].<br \/>\nPortanto, as solu\u00e7\u00f5es para o sistema de equa\u00e7\u00f5es s\u00e3o:<br \/>\n[tex]\\qquad \\qquad \\fcolorbox{black}{#eee0e5}{$ \\,  x=\\frac{3-\\sqrt{5}}{2}\\quad  \\text{ e }\\quad y=\\frac{3+\\sqrt{5}}{2}$} \\,  \\, [\/tex] ou [tex]\\quad \\fcolorbox{black}{#eee0e5}{$ \\,  x=\\frac{3+\\sqrt{5}}{2}\\quad  \\text{ e }\\quad y=\\frac{3-\\sqrt{5}}{2}$} \\,  \\, .[\/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","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 1\u00ba ano do E. M.) Determine as solu\u00e7\u00f5es reais do seguinte sistema de equa\u00e7\u00f5es: [tex]\\qquad \\qquad \\left\\{ \\begin{array}{lll} x+y &#038;=&#038; 3 \\\\ x^3+y^3 &#038; =&#038; 18 \\end{array} \\right.[\/tex] . Solu\u00e7\u00e3o Usando o produto not\u00e1vel &#8220;cubo perfeito&#8221; vemos que [tex]\\qquad \\qquad (x+y)^3=x^3+3x^2y+3xy^2+y^3=x^3+y^3+3xy(x+y),[\/tex] ou seja, [tex]\\qquad \\qquad (x+y)^3=x^3+y^3+3xy(x+y).\\qquad \\textcolor{#800000}{(i)}[\/tex] Utilizando as equa\u00e7\u00f5es &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-sistema-de-equacoes-2\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":667,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-102524","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/102524","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\/667"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=102524"}],"version-history":[{"count":50,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/102524\/revisions"}],"predecessor-version":[{"id":187659,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/102524\/revisions\/187659"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=102524"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}