{"id":48246,"date":"2015-10-07T18:05:14","date_gmt":"2015-10-07T21:05:14","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=48246"},"modified":"2023-03-29T03:37:55","modified_gmt":"2023-03-29T06:37:55","slug":"problema-um-inteiro-bem-diferente","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-um-inteiro-bem-diferente\/","title":{"rendered":".Problema: Um inteiro bem diferente"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\">\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\"\/>\n<strong>Prove que <\/strong><br \/>\n[tex]\\qquad \\qquad \\dfrac{1}{\\sqrt{1} + \\sqrt{2}} + \\dfrac{1}{\\sqrt{2} + \\sqrt{3}} + &#8230; + \\dfrac{1}{\\sqrt{99} + \\sqrt{100}}[\/tex]<br \/>\n\u00e9 um n\u00famero inteiro.<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height: 50px;\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o 1<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nObserve que<br \/>\n[tex]\\qquad \\dfrac{1}{\\sqrt{1}+\\sqrt{2}} = \\dfrac{1}{\\sqrt{1}+\\sqrt{2}} \\cdot \\dfrac{\\sqrt{1}-\\sqrt{2}}{\\sqrt{1}-\\sqrt{2}} = \\dfrac{1-\\sqrt{2}}{-1} = \\sqrt{2}- 1[\/tex]<br \/>\n[tex]\\qquad \\dfrac{1}{\\sqrt{2}+\\sqrt{3}} = \\dfrac{1}{\\sqrt{2}+\\sqrt{3}} \\cdot \\dfrac{\\sqrt{2}-\\sqrt{3}}{\\sqrt{2}-\\sqrt{3}} = \\dfrac{\\sqrt{2}-\\sqrt{3}}{-1} = \\sqrt{3}-\\sqrt{2}[\/tex]<br \/>\n[tex]\\qquad \\dfrac{1}{\\sqrt{3}+\\sqrt{4}} = \\dfrac{1}{\\sqrt{3}+\\sqrt{4}} \\cdot \\dfrac{\\sqrt{3}-\\sqrt{4}}{\\sqrt{3}-\\sqrt{4}} = \\dfrac{\\sqrt{3}-\\sqrt{4}}{-1} = \\sqrt{4}-\\sqrt{3}[\/tex]<br \/>\n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.&nbsp;.&nbsp;.&nbsp;<br \/>\nDeste modo,<br \/>\n[tex]\\qquad \\dfrac{1}{\\sqrt{1} + \\sqrt{2}} + \\dfrac{1}{\\sqrt{2} + \\sqrt{3}} + \u2026 + \\dfrac{1}{\\sqrt{99} + \\sqrt{100}} =[\/tex]<br \/>\n[tex]\\qquad = \\sqrt{2}-1 + \\sqrt{3}-\\sqrt{2} + \\sqrt{4}-\\sqrt{3} + \\ldots + \\sqrt{99}-\\sqrt{98} +\\sqrt{100}-\\sqrt{99}[\/tex]<br \/>\n[tex]\\qquad = \\sqrt{100}-1 = 9.[\/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 pelo COM <strong> MIRIM APRENDIZ<\/strong>, com formata\u00e7\u00e3o dos Moderadores do Blog.<\/p>\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 2<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nSeja<br \/>\n[tex]\\qquad S = \\dfrac{1}{\\sqrt{1} + \\sqrt{2}} + \\dfrac{1}{\\sqrt{2} + \\sqrt{3}} + &#8230; + \\dfrac{1}{\\sqrt{99} + \\sqrt{100}}[\/tex]<br \/>\ne note que<br \/>\n[tex]\\qquad\\dfrac{1}{\\sqrt{n} + \\sqrt{n+1}} = \\dfrac{1}{\\sqrt{n} + \\sqrt{n+1}} \\cdot \\dfrac{\\sqrt{n+1}-\\sqrt{n}}{\\sqrt{n+1}-\\sqrt{n}} = \\sqrt{n+1}-\\sqrt{n}[\/tex].<br \/>\nDeste modo,<br \/>\n[tex]\\qquad S =\\left(\\sqrt{2}-\\sqrt{1}\\right) + \\left(\\sqrt{3}-\\sqrt{2}\\right) + \\left(\\sqrt{4}-\\sqrt{3}\\right) + &#8230; + \\left(\\sqrt{99}-\\sqrt{98}\\right) +\\left( \\sqrt{100}-\\sqrt{99}\\right)[\/tex],<br \/>\nou seja,<br \/>\n[tex]\\qquad  S = \\sqrt{100}-\\sqrt{1} = 10-1 = 9.[\/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:30px\">\n<div class=\"g3\">\nParticipou da discuss\u00e3o o Clube <strong>MIRIM APRENDIZ<\/strong>.<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 9\u00ba ano do E. F.) Prove que [tex]\\qquad \\qquad \\dfrac{1}{\\sqrt{1} + \\sqrt{2}} + \\dfrac{1}{\\sqrt{2} + \\sqrt{3}} + &#8230; + \\dfrac{1}{\\sqrt{99} + \\sqrt{100}}[\/tex] \u00e9 um n\u00famero inteiro. Solu\u00e7\u00e3o 1 Observe que [tex]\\qquad \\dfrac{1}{\\sqrt{1}+\\sqrt{2}} = \\dfrac{1}{\\sqrt{1}+\\sqrt{2}} \\cdot \\dfrac{\\sqrt{1}-\\sqrt{2}}{\\sqrt{1}-\\sqrt{2}} = \\dfrac{1-\\sqrt{2}}{-1} = \\sqrt{2}- 1[\/tex] [tex]\\qquad \\dfrac{1}{\\sqrt{2}+\\sqrt{3}} = \\dfrac{1}{\\sqrt{2}+\\sqrt{3}} \\cdot \\dfrac{\\sqrt{2}-\\sqrt{3}}{\\sqrt{2}-\\sqrt{3}} = \\dfrac{\\sqrt{2}-\\sqrt{3}}{-1} = &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-um-inteiro-bem-diferente\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":179,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-48246","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/48246","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\/179"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=48246"}],"version-history":[{"count":26,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/48246\/revisions"}],"predecessor-version":[{"id":185735,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/48246\/revisions\/185735"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=48246"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}