{"id":117358,"date":"2018-11-30T22:29:11","date_gmt":"2018-12-01T00:29:11","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=117358"},"modified":"2023-05-13T04:20:07","modified_gmt":"2023-05-13T07:20:07","slug":"problema-determine-o-valor-de-x","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-determine-o-valor-de-x\/","title":{"rendered":".Problema: Determine o valor de x"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n

\nDetermine o valor de [tex]x[\/tex] sabendo que [tex]x=\\sqrt[3]{25\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\cdots}}}}}[\/tex].<\/p>\n

\n<\/div>\n

\n

\n

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

\n

Observe que
\n[tex]\\qquad \\sqrt[3]{25}=25^{\\frac{1}{3}}[\/tex];
\n[tex]\\qquad \\sqrt[3]{\\sqrt[3]{25}}=25^{\\frac{1}{3}\\times\\frac{1}{3}}=25^{\\frac{1}{9}}[\/tex];
\n[tex]\\qquad \\sqrt[3]{\\sqrt[3]{\\sqrt[3]{25}}}=25^{\\frac{1}{3}\\times\\frac{1}{3}\\times\\frac{1}{3}}=25^{\\frac{1}{27}}[\/tex];
\ne assim sucessivamente.<\/p>\n

Podemos ent\u00e3o escrever [tex]x[\/tex] como
\n[tex]\\qquad \\begin{align*}x&=\\sqrt[3]{25\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\cdots}}}}}\\\\
\n&=\\sqrt[3]{25}\\sqrt[3]{\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\cdots}}}}}\\\\
\n&=\\sqrt[3]{25}\\sqrt[3]{\\sqrt[3]{{25}}}\\sqrt[3]{\\sqrt[3]{\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\cdots}}}}}\\\\
\n&=\\sqrt[3]{25}\\sqrt[3]{\\sqrt[3]{{25}}}\\sqrt[3]{\\sqrt[3]{\\sqrt[3]{{25}}}}\\sqrt[3]{\\sqrt[3]{\\sqrt[3]{\\sqrt[3]{{25}\\sqrt[3]{{25}\\cdots}}}}}\\\\
\n&\\vdots\\\\
\n&=25^{\\frac{1}{3}}\\times 25^{\\frac{1}{9}}\\times 25^{\\frac{1}{27}}\\times 25^{\\frac{1}{81}}\\cdots\\end{align*}[\/tex]<\/p>\n

\n

Dessa forma, temos que
\n[tex]\\qquad x=25^{\\frac{1}{3}+\\frac{1}{9}+\\frac{1}{27}+\\frac{1}{81}+\\cdots}. \\qquad \\qquad \\textcolor{#800000}{(i)}[\/tex]
\nObserve que o expoente [tex]{\\frac{1}{3}+\\frac{1}{9}+\\frac{1}{27}+\\frac{1}{81}+\\cdots}[\/tex] \u00e9 a soma dos termos de uma Progress\u00e3o Geom\u00e9trica (PG) infinita, em que o primeiro valor \u00e9 [tex]a_1=\\dfrac{1}{3}[\/tex] e a raz\u00e3o \u00e9 [tex]q=\\dfrac{1}{3}.[\/tex] Como [tex]|q|\\lt 1[\/tex], podemos utilizar a f\u00f3rmula da soma de infinitos termos de uma PG, [tex]\\boxed{S_\\infty=\\frac{a_1}{1-q}}[\/tex] e, no nosso caso, obter:<\/p>\n

\n

[tex]\\qquad S_\\infty=\\dfrac{\\frac{1}{3}}{1-\\frac{1}{3}}[\/tex]<\/p>\n

\n

[tex]\\qquad S_\\infty=\\dfrac{\\frac{1}{3}}{\\frac{2}{3}}[\/tex]<\/p>\n

\n

[tex]\\qquad S_\\infty=\\dfrac{1}{3}\\times\\dfrac{3}{2}[\/tex]<\/p>\n

\n

[tex]\\qquad S_\\infty=\\dfrac{1}{2}[\/tex].<\/p>\n

Assim, por [tex]\\textcolor{#800000}{(i)}[\/tex], [tex] \\, \\fcolorbox{black}{#eee0e5}{$x=25^\\frac{1}{2}=\\sqrt{25}=5$} \\, .[\/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 2\u00ba ano do E. M.) Determine o valor de [tex]x[\/tex] sabendo que [tex]x=\\sqrt[3]{25\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\cdots}}}}}[\/tex]. Solu\u00e7\u00e3o Observe que [tex]\\qquad \\sqrt[3]{25}=25^{\\frac{1}{3}}[\/tex]; [tex]\\qquad \\sqrt[3]{\\sqrt[3]{25}}=25^{\\frac{1}{3}\\times\\frac{1}{3}}=25^{\\frac{1}{9}}[\/tex]; [tex]\\qquad \\sqrt[3]{\\sqrt[3]{\\sqrt[3]{25}}}=25^{\\frac{1}{3}\\times\\frac{1}{3}\\times\\frac{1}{3}}=25^{\\frac{1}{27}}[\/tex]; e assim sucessivamente. Podemos ent\u00e3o escrever [tex]x[\/tex] como [tex]\\qquad \\begin{align*}x&=\\sqrt[3]{25\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\cdots}}}}}\\\\ &=\\sqrt[3]{25}\\sqrt[3]{\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\cdots}}}}}\\\\ &=\\sqrt[3]{25}\\sqrt[3]{\\sqrt[3]{{25}}}\\sqrt[3]{\\sqrt[3]{\\sqrt[3]{{25}\\sqrt[3]{{25}\\sqrt[3]{{25}\\cdots}}}}}\\\\ &=\\sqrt[3]{25}\\sqrt[3]{\\sqrt[3]{{25}}}\\sqrt[3]{\\sqrt[3]{\\sqrt[3]{{25}}}}\\sqrt[3]{\\sqrt[3]{\\sqrt[3]{\\sqrt[3]{{25}\\sqrt[3]{{25}\\cdots}}}}}\\\\ &\\vdots\\\\ &=25^{\\frac{1}{3}}\\times 25^{\\frac{1}{9}}\\times 25^{\\frac{1}{27}}\\times 25^{\\frac{1}{81}}\\cdots\\end{align*}[\/tex] Dessa forma, temos que [tex]\\qquad x=25^{\\frac{1}{3}+\\frac{1}{9}+\\frac{1}{27}+\\frac{1}{81}+\\cdots}. \\qquad \\qquad \\textcolor{#800000}{(i)}[\/tex] Observe … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":620,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-117358","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/117358","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\/620"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=117358"}],"version-history":[{"count":21,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/117358\/revisions"}],"predecessor-version":[{"id":119238,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/117358\/revisions\/119238"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=117358"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}