{"id":39742,"date":"2015-06-10T22:15:11","date_gmt":"2015-06-11T01:15:11","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=39742"},"modified":"2023-03-22T23:26:20","modified_gmt":"2023-03-23T02:26:20","slug":"problemao-um-problema-complexo","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-um-problema-complexo\/","title":{"rendered":".Problem\u00e3o: Um problema complexo"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 10px;border: 0 none\"\/>\n<p style=\"margin:20px\">\n<p>Encontre todos os n\u00fameros complexos [tex]z[\/tex] tais que<br \/>\n[tex]\\qquad \\qquad (3z+1)(4z+1)(6z+1)(12z+1)=2[\/tex].\n\t<\/p><\/div>\n<p style=\"margin:40px\">\n<div class=\"g1n\">\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: 10px; margin-bottom: 10px;border: 0 none\"\/>\n<p style=\"margin:20px\">\n<p> Inicialmente, multiplique a equa\u00e7\u00e3o inicial por [tex]4[\/tex], [tex]3[\/tex] e [tex]2[\/tex]: <\/p>\n<p>[tex]\\qquad (3z+1)(4z+1)(6z+1)(12z+1)=2 \t\\qquad (\\times 4) (\\times 3) (\\times 2)[\/tex]<\/p>\n<p>[tex]\\qquad (12z+4)(12z+3)(12z+2)(12z+1)=48[\/tex].<\/p>\n<p>Agora, chame [tex]12z+1[\/tex] de [tex]x[\/tex] e, ent\u00e3o, a equa\u00e7\u00e3o fica assim:<\/p>\n<p>[tex]\\qquad (x+3)(x+2)(x+1)x=48[\/tex],<br \/>\n\tou ainda,<\/p>\n<p>[tex]\\qquad (x^2+3x)(x^2+3x+2)=48[\/tex].<\/p>\n<p>Fa\u00e7a, agora, a substitui\u00e7\u00e3o de [tex]x^2+3x[\/tex] por [tex]a[\/tex]. A equa\u00e7\u00e3o se torna:<\/p>\n<p>[tex]\\qquad a(a+2)=48[\/tex]<\/p>\n<p>e, dessa forma, [tex]a=-8[\/tex] ou [tex]a=6[\/tex].<\/p>\n<p>Substituindo esses valores de [tex]a[\/tex] na equa\u00e7\u00e3o [tex]x^2+3x=a[\/tex], temos que:<\/p>\n<p>[tex]\\qquad x^2+3x=-8~[\/tex] e, portanto, [tex]x=\\dfrac{-3\\pm i\\sqrt{23}}{2}[\/tex]<br \/>\nou<br \/>\n[tex]\\qquad x^2+3x=6~[\/tex] e, ent\u00e3o, [tex] x=\\dfrac{-3\\pm \\sqrt{33}}{2}[\/tex].<br \/>\nFinalmente, como [tex]x=12z+1[\/tex],<\/p>\n<table class=\"bf\" style=\"margin-left:5%; width:70%;\">\n<tr>\n<td class=\"bf\" style=\" line-height:30px;\">\n<ul>\n<li>para [tex]x=\\dfrac{-3\\pm i\\sqrt{23}}{2}\\\\<br \/>\n[\/tex], segue que:<br \/>\n[tex]\\qquad \\dfrac{-3\\pm i\\sqrt{23}}{2}=12z+1\\\\<br \/>\n\\qquad 12z=\\dfrac{-3\\pm i\\sqrt{23}}{2}-1\\\\<br \/>\n\\qquad 12z=\\dfrac{-5\\pm i\\sqrt{23}}{2}\\\\<br \/>\n\\qquad \\boxed{z=\\dfrac{-5\\pm i\\sqrt{23}}{24}_~}\\,.[\/tex]\n<\/li>\n<\/ul>\n<\/td>\n<td class=\"bf\" style=\"border-left: solid 2px  #800000\">\n<ul>\n<li>para [tex]x=\\dfrac{-3\\pm \\sqrt{33}}{2}\\\\<br \/>\n[\/tex], segue que:<br \/>\n[tex]\\qquad \\dfrac{-3\\pm \\sqrt{33}}{2}=12z+1\\\\<br \/>\n\\qquad 12z=\\dfrac{-3\\pm \\sqrt{33}}{2}-1\\\\<br \/>\n\\qquad 12z=\\dfrac{-5\\pm \\sqrt{33}}{2}\\\\<br \/>\n\\qquad \\boxed{z=\\dfrac{-5\\pm \\sqrt{33}}{24_~}}\\,.[\/tex]\n<\/li>\n<\/ul>\n<\/td>\n<\/tr>\n<\/table>\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 Encontre todos os n\u00fameros complexos [tex]z[\/tex] tais que [tex]\\qquad \\qquad (3z+1)(4z+1)(6z+1)(12z+1)=2[\/tex]. Solu\u00e7\u00e3o Inicialmente, multiplique a equa\u00e7\u00e3o inicial por [tex]4[\/tex], [tex]3[\/tex] e [tex]2[\/tex]: [tex]\\qquad (3z+1)(4z+1)(6z+1)(12z+1)=2 \\qquad (\\times 4) (\\times 3) (\\times 2)[\/tex] [tex]\\qquad (12z+4)(12z+3)(12z+2)(12z+1)=48[\/tex]. Agora, chame [tex]12z+1[\/tex] de [tex]x[\/tex] e, ent\u00e3o, a equa\u00e7\u00e3o fica assim: [tex]\\qquad (x+3)(x+2)(x+1)x=48[\/tex], ou ainda, [tex]\\qquad (x^2+3x)(x^2+3x+2)=48[\/tex]. Fa\u00e7a, agora, a substitui\u00e7\u00e3o &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-um-problema-complexo\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":164,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-39742","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/39742","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=39742"}],"version-history":[{"count":28,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/39742\/revisions"}],"predecessor-version":[{"id":185132,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/39742\/revisions\/185132"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=39742"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}