{"id":63878,"date":"2016-07-11T16:36:03","date_gmt":"2016-07-11T19:36:03","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=63878"},"modified":"2023-04-04T03:04:23","modified_gmt":"2023-04-04T06:04:23","slug":"problemao-triplas-de-racionais","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-triplas-de-racionais\/","title":{"rendered":".Problem\u00e3o: Triplas de Racionais"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n

\nEncontre triplas de n\u00fameros racionais [tex](a, b, c)[\/tex] tais que [tex]\\sqrt[3]{\\sqrt[3]{2}-1}=\\sqrt[3]{a}+\\sqrt[3]{b}+\\sqrt[3]{c}[\/tex].<\/p>\n

\n<\/div>\n

\n

\n

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

\nSejam [tex]x=\\sqrt[3]{\\sqrt[3]{2}-1}[\/tex] e [tex]y=\\sqrt[3]{2}[\/tex]. Ent\u00e3o [tex]y^3=2[\/tex] e [tex]x=\\sqrt[3]{y-1}[\/tex].
\n Note que [tex]1=y^3-1=(y-1)(y^2+y+1)[\/tex], logo<\/p>\n

[tex]\\qquad \\qquad y^2+y+1=\\dfrac{3y^2+3y+3}{3}=\\dfrac{y^3+3y^2+3y+1}{3}=\\dfrac{(y+1)^3}{3}[\/tex],<\/p>\n

o que implica em
\n[tex]\\qquad \\qquad x^3=y-1=\\dfrac{1}{y^2+y+1}=\\dfrac{3}{(y+1)^3}[\/tex]
\ndonde
\n [tex]\\qquad \\qquad x=\\dfrac{\\sqrt[3]{3}}{y+1}\t\\qquad {(I)}[\/tex].
\nPor outro lado,
\n[tex]\\qquad \\qquad 3=y^3+1=(y+1)(y^2-y+1) [\/tex]
\ne assim
\n[tex]\\qquad \\qquad \\dfrac{1}{y+1}=\\dfrac{y^2-y+1}{3}\t\\qquad{(II)}[\/tex].
\nCombinando as equa\u00e7\u00f5es [tex](I)[\/tex] e [tex](II)[\/tex], obtemos:<\/p>\n

[tex]\\qquad \\qquad x=\\sqrt[3]{\\dfrac{1}{9}}(\\sqrt[3]{4}-\\sqrt[3]{2}+1)=\\sqrt[3]{\\dfrac{4}{9}}-\\sqrt[3]{\\dfrac{2}{9}}+\\sqrt[3]{\\dfrac{1}{9}}[\/tex].<\/p>\n

Consequentemente, [tex](a, b, c)=\\left(\\dfrac{4}{9}, -\\dfrac{2}{9}, \\dfrac{1}{9}\\right)[\/tex].<\/p>\n

\n\n\n
\"\"<\/td>\n <\/p>\n

Uma tripla foi encontrada!
\nSer\u00e1 que existem outras satisfazendo as condi\u00e7\u00f5es do problemas?<\/span><\/p>\n

<\/td>\n<\/tr>\n<\/table>\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 9\u00ba ano do E. F.) Encontre triplas de n\u00fameros racionais [tex](a, b, c)[\/tex] tais que [tex]\\sqrt[3]{\\sqrt[3]{2}-1}=\\sqrt[3]{a}+\\sqrt[3]{b}+\\sqrt[3]{c}[\/tex]. Solu\u00e7\u00e3o Sejam [tex]x=\\sqrt[3]{\\sqrt[3]{2}-1}[\/tex] e [tex]y=\\sqrt[3]{2}[\/tex]. Ent\u00e3o [tex]y^3=2[\/tex] e [tex]x=\\sqrt[3]{y-1}[\/tex]. Note que [tex]1=y^3-1=(y-1)(y^2+y+1)[\/tex], logo [tex]\\qquad \\qquad y^2+y+1=\\dfrac{3y^2+3y+3}{3}=\\dfrac{y^3+3y^2+3y+1}{3}=\\dfrac{(y+1)^3}{3}[\/tex], o que implica em [tex]\\qquad \\qquad x^3=y-1=\\dfrac{1}{y^2+y+1}=\\dfrac{3}{(y+1)^3}[\/tex] donde [tex]\\qquad \\qquad x=\\dfrac{\\sqrt[3]{3}}{y+1} \\qquad {(I)}[\/tex]. Por outro lado, [tex]\\qquad \\qquad … <\/p>\n

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-63878","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/63878","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=63878"}],"version-history":[{"count":29,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/63878\/revisions"}],"predecessor-version":[{"id":186142,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/63878\/revisions\/186142"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=63878"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}