{"id":132461,"date":"2019-07-17T21:19:31","date_gmt":"2019-07-18T00:19:31","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=132461"},"modified":"2023-06-02T03:32:13","modified_gmt":"2023-06-02T06:32:13","slug":"problemao-uma-bela-equacao","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-uma-bela-equacao\/","title":{"rendered":".Problem\u00e3o: Ali Bab\u00e3o e a d\u00e9cima de suas 40 equa\u00e7\u00f5es – Uma bela equa\u00e7\u00e3o"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 8\u00ba ano do E. F.)<\/span><\/p>\n

\n

\n

Resolva a equa\u00e7\u00e3o [tex]\\quad \\boxed{3=\\dfrac{1}{1 – \\dfrac{1}{1 + \\dfrac{1}{1 – \\dfrac{1}{x}}}}} \\, [\/tex].<\/p>\n

\n<\/div>\n

\n

\n

Solu\u00e7\u00e3o <\/p>\n

\n

Veja que
\n[tex]\\qquad \\qquad \\boxed{1 – \\dfrac{1}{x}}=\\boxed{\\dfrac{x}{x}-\\dfrac{1}{x}=\\dfrac{x-1}{x}}[\/tex],
\nassim, segue que:
\n[tex]\\qquad \\qquad 3=\\dfrac{1}{1 – \\dfrac{1}{1 + \\dfrac{1}{\\boxed{1 – \\dfrac{1}{x}}}}}=\\dfrac{1}{1 – \\dfrac{1}{1 + \\dfrac{1}{\\boxed{\\dfrac{x-1}{x}}}}}\\quad [\/tex].
\nComo
\n[tex]\\qquad \\qquad \\boxed{\\dfrac{1}{\\dfrac{x-1}{x}}}=\\boxed{\\dfrac{x}{x-1}}[\/tex],
\nvem que:
\n[tex]\\qquad \\qquad 3=\\dfrac{1}{1 – \\dfrac{1}{1 + \\boxed{\\dfrac{1}{\\dfrac{x-1}{x}}}}}=\\dfrac{1}{1 – \\dfrac{1}{1 +\\boxed{\\dfrac{x}{{x-1}}}}}\\quad [\/tex].
\nObserve agora que
\n[tex]\\qquad \\qquad \\boxed{1 + \\dfrac{x}{x-1}}=\\dfrac{x-1}{x-1}+\\dfrac{x}{x-1}=\\boxed{\\dfrac{2x-1}{x-1}}[\/tex] ,
\ndonde
\n[tex]\\qquad \\qquad 3=\\dfrac{1}{1 – \\dfrac{1}{\\boxed{1 +\\dfrac{x}{{x-1}}}}}=\\dfrac{1}{1 – \\dfrac{1}{\\boxed{\\dfrac{2x-1}{{x-1}}}}}\\quad [\/tex].
\nProsseguindo com "redu\u00e7\u00f5es ao mesmo denominador" e "invers\u00e3o de fra\u00e7\u00f5es", obtemos:
\n[tex]\\qquad \\qquad 3=\\dfrac{1}{1 -\\dfrac{1}{\\dfrac{2x-1}{{x-1}}}}[\/tex]<\/p>\n

\n

[tex]\\qquad \\qquad 3=\\dfrac{1}{1 – \\dfrac{x-1}{2x-1}}[\/tex]<\/p>\n

\n

[tex]\\qquad \\qquad 3=\\dfrac{1}{\\dfrac{x}{2x-1}}[\/tex]<\/p>\n

\n

[tex]\\qquad \\qquad 3=\\dfrac{2x-1}{x}[\/tex]
\n[tex]\\qquad \\qquad 3x\\ =\\ 2x-1[\/tex]
\n[tex]\\qquad \\qquad x =-1.[\/tex]
\nPortanto, a equa\u00e7\u00e3o tem uma \u00fanica raiz: [tex] \\, \\fcolorbox{black}{#eee0e5}{$x =-1$} \\, .[\/tex]<\/p>\n

\n

Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n

\n","protected":false},"excerpt":{"rendered":"

Problema (Indicado a partir do 8\u00ba ano do E. F.) Resolva a equa\u00e7\u00e3o [tex]\\quad \\boxed{3=\\dfrac{1}{1 – \\dfrac{1}{1 + \\dfrac{1}{1 – \\dfrac{1}{x}}}}} \\, [\/tex]. Solu\u00e7\u00e3o Veja que [tex]\\qquad \\qquad \\boxed{1 – \\dfrac{1}{x}}=\\boxed{\\dfrac{x}{x}-\\dfrac{1}{x}=\\dfrac{x-1}{x}}[\/tex], assim, segue que: [tex]\\qquad \\qquad 3=\\dfrac{1}{1 – \\dfrac{1}{1 + \\dfrac{1}{\\boxed{1 – \\dfrac{1}{x}}}}}=\\dfrac{1}{1 – \\dfrac{1}{1 + \\dfrac{1}{\\boxed{\\dfrac{x-1}{x}}}}}\\quad [\/tex]. Como [tex]\\qquad \\qquad \\boxed{\\dfrac{1}{\\dfrac{x-1}{x}}}=\\boxed{\\dfrac{x}{x-1}}[\/tex], vem que: [tex]\\qquad … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":734,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-132461","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/132461","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\/734"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=132461"}],"version-history":[{"count":7,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/132461\/revisions"}],"predecessor-version":[{"id":192897,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/132461\/revisions\/192897"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=132461"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}