{"id":150063,"date":"2020-04-07T01:04:30","date_gmt":"2020-04-07T04:04:30","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=150063"},"modified":"2024-06-24T22:47:51","modified_gmt":"2024-06-25T01:47:51","slug":"problema-para-ajudar-na-escola-muitas-raizes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-muitas-raizes\/","title":{"rendered":".Problema para ajudar na escola: Muitas ra\u00edzes…"},"content":{"rendered":"

✏ Link do problema para dispositivos da Apple<\/em>.<\/span><\/a><\/strong><\/p>\n

\n

Problema
\n (A partir do 9\u00ba ano do E. F. – N\u00edvel de dificuldade: Dif\u00edcil<\/span>) <\/span><\/p>\n

\n

(ONEM<\/strong>, 2005 – Adaptado)<\/span> Seja [tex]x[\/tex] um n\u00famero real maior do que [tex] 1[\/tex].
\n Determine o n\u00famero natural n\u00e3o nulo [tex]n[\/tex] tal que <\/p>\n

[tex] \\sqrt[4]{\\dfrac{\\sqrt[3]{\\dfrac{\\sqrt{\\dfrac{1}{x}}}{x}}}{x}}=\\sqrt{\\sqrt[3]{\\sqrt[4]{\\left(\\dfrac{1}{x}\\right)^n}}}[\/tex].<\/p>\n<\/div>\n

\n

\"explicador_p\" <\/p>\n

Lembretes <\/p>\n

\n

S\u00e3o bem conhecidas as seguintes defini\u00e7\u00f5es envolvendo pot\u00eancias e radicais:
\n►<\/span> Pot\u00eancia de expoente inteiro negativo:<\/strong> Se [tex]a[\/tex] \u00e9 um n\u00famero real n\u00e3o nulo e [tex]n[\/tex] \u00e9 um n\u00famero natural, ent\u00e3o a pot\u00eancia [tex]a^{-n}[\/tex] \u00e9 assim definida:
\n[tex]\\qquad \\qquad a^{\\color{red}{-n}}=\\dfrac{1}{a^{\\color{red}{n}}}\\,[\/tex]<\/span>.
\n►<\/span> Pot\u00eancia de expoente racional:<\/strong> Se [tex]a[\/tex] \u00e9 um n\u00famero real positivo, [tex]n[\/tex] \u00e9 um n\u00famero natural n\u00e3o nulo e [tex]m[\/tex] \u00e9 um n\u00famero inteiro, ent\u00e3o:
\n[tex]\\qquad \\qquad a^\\frac{\\color{blue}{m}}{\\color{red}{n}}=\\sqrt[\\color{red}{n}]{a^{\\color{blue}{m}}}\\,[\/tex]<\/span>.
\nAs propriedades de pot\u00eancias e ra\u00edzes nos permitem estender essas duas defini\u00e7\u00f5es para expoentes n\u00e3o necessariamente inteiros.<\/p>\n

(1)<\/strong><\/span> Pot\u00eancia de expoente negativo:<\/strong> Se [tex]a[\/tex] \u00e9 um n\u00famero real n\u00e3o nulo e [tex]n[\/tex] \u00e9 um n\u00famero racional positivo, ent\u00e3o:
\n[tex]\\qquad \\qquad a^{\\color{red}{-n}}=\\left(\\dfrac{1}{a}\\right)^{\\color{red}{n}}=\\dfrac{1}{a^{\\color{red}{n}}}[\/tex]<\/span><\/p>\n

\n

(2)<\/strong><\/span> Pot\u00eancia de expoente racional:<\/strong> Se [tex]a[\/tex] \u00e9 um n\u00famero real positivo, [tex]n[\/tex] \u00e9 um n\u00famero natural n\u00e3o nulo e [tex]m[\/tex] \u00e9 um n\u00famero racional, ent\u00e3o:
\n[tex]\\qquad \\qquad \\sqrt[\\color{red}{n}]{a^{\\color{blue}{m}}}=a^\\frac{\\color{blue}{m}}{\\color{red}{n}}[\/tex]<\/span><\/p>\n

\n

Podemos estender as propriedades de pot\u00eancias e ra\u00edzes com expoentes inteiros para pot\u00eancias e ra\u00edzes com expoentes racionais.
\nParticularmente nos interessam duas propriedades:
\n (3)<\/strong><\/span> Quociente de pot\u00eancias com a mesma base:<\/strong> Se [tex]a[\/tex] \u00e9 um n\u00famero real positivo e [tex]m[\/tex] e [tex]n[\/tex] s\u00e3o n\u00fameros racionais, ent\u00e3o:
\n[tex]\\qquad \\qquad \\dfrac{a^{\\color{blue}{m}}}{a^{\\color{red}{n}}}=a^{{\\color{blue}{m}}-{\\color{red}{n}}}[\/tex]<\/span><\/p>\n

\n

(4)<\/strong><\/span> Pot\u00eancia de pot\u00eancia:<\/strong> Se [tex]a[\/tex] \u00e9 um n\u00famero real positivo e [tex]m[\/tex] e [tex]n[\/tex] s\u00e3o n\u00fameros racionais, ent\u00e3o:
\n[tex]\\qquad \\qquad \\left(a^{\\color{red}{n}}\\right)^{\\color{blue}{m}}=a^{\\textcolor{red}{n} \\,\\cdot \\,\\color{blue}{m}}[\/tex]<\/span><\/p>\n

\n

(5)<\/strong><\/span> Igualdade de pot\u00eancias de mesma base:<\/strong> Se [tex]a[\/tex] \u00e9 um n\u00famero real positivo e diferente de [tex]1[\/tex] e [tex]m[\/tex] e [tex]n[\/tex] s\u00e3o n\u00fameros reais, ent\u00e3o:
\n[tex]\\qquad \\qquad a^n=a^m \\iff m=n[\/tex]<\/span><\/p>\n

\n<\/div>\n

\n

\n

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

\n

Para facilitar a aplica\u00e7\u00e3o dos Lembretes<\/strong><\/span>, vamos simplificar separadamente cada um dos lados da igualdade apresentada no problema.
\nVeja que
\n[tex] \\qquad \\boxed{\\sqrt[4]{\\dfrac{\\sqrt[3]{\\dfrac{\\sqrt{\\dfrac{1}{x}}}{x}}}{x}}}=\\sqrt[4]{\\dfrac{\\sqrt[3]{\\dfrac{\\sqrt{\\textcolor{red}{\\dfrac{1}{x^1}}}}{x}}}{x}}\\stackrel{\\textcolor{#800000}{(1)}}{=}\\sqrt[4]{\\dfrac{\\sqrt[3]{\\dfrac{\\sqrt{\\textcolor{red}{x^{-1}}}}{x}}}{x}}=[\/tex]
\n[tex] \\qquad=\\sqrt[4]{\\dfrac{\\sqrt[3]{\\dfrac{\\textcolor{#C72EE2}{\\sqrt[2]{x^{-1}}}}{x}}}{x}}\\stackrel{\\textcolor{#800000}{(2)}}{=}\\sqrt[4]{\\dfrac{\\sqrt[3]{\\dfrac{\\textcolor{#C72EE2}{x^{\\frac{-1}{2}}}}{x}}}{x}}=\\sqrt[4]{\\dfrac{\\sqrt[3]{\\textcolor{#42BA0D}{\\dfrac{x^{\\frac{-1}{2}}}{x}}}}{x}}\\stackrel{\\textcolor{#800000}{(3)}}{=}\\sqrt[4]{\\dfrac{\\sqrt[3]{\\textcolor{#42BA0D}{x^{\\frac{-1}{2}-1}}}}{x}}[\/tex]
\n[tex] \\qquad=\\sqrt[4]{\\dfrac{\\sqrt[3]{\\textcolor{#42BA0D}{x^{-\\frac{3}{2}}}}}{x}}=\\sqrt[4]{\\dfrac{\\textcolor{#FF00FF}{\\sqrt[3]{x^{-\\frac{3}{2}}}}}{x}}\\stackrel{\\textcolor{#800000}{(2)}}{=}\\sqrt[4]{\\dfrac{\\textcolor{#FF00FF}{x^{\\frac{-\\frac{3}{2}}{3}}}}{x}}=\\sqrt[4]{\\dfrac{\\textcolor{#FF00FF}{x^{-\\frac{1}{2}}}}{x}}=[\/tex]
\n[tex] \\qquad=\\sqrt[4]{\\textcolor{#F87217}{\\dfrac{x^{-\\frac{1}{2}}}{x^{1}}}}\\stackrel{\\textcolor{#800000}{(3)}}{=}\\sqrt[4]{\\textcolor{#F87217}{x^{-\\frac{1}{2}-1}}}=\\sqrt[4]{x^{-\\frac{3}{2}}}\\stackrel{\\textcolor{#800000}{(2)}}{=}x^{\\frac{-\\frac{3}{2}}{4}}=\\boxed{x^{-\\frac{3}{8}}}[\/tex]<\/span>
\ne
\n[tex] \\qquad \\qquad \\boxed{\\sqrt{\\sqrt[3]{\\sqrt[4]{\\left(\\dfrac{1}{x}\\right)^n}}}}=\\sqrt{\\sqrt[3]{\\textcolor{red}{\\sqrt[4]{\\left(\\dfrac{1}{x}\\right)^n}}}}\\stackrel{\\textcolor{#800000}{(2)}}{=}\\sqrt{\\sqrt[3]{\\textcolor{red}{\\left(\\dfrac{1}{x}\\right)^{\\frac{n}{4}}}}}=[\/tex]
\n[tex] \\qquad=\\sqrt{\\textcolor{#169DF7}{\\sqrt[3]{\\left(\\dfrac{1}{x}\\right)^{\\frac{n}{4}}}}}\\stackrel{\\textcolor{#800000}{(2)}}{=}\\sqrt{\\textcolor{#169DF7}{\\left(\\dfrac{1}{x}\\right)^{\\frac{\\frac{n}{4}}{3}}}}=\\sqrt{\\textcolor{#169DF7}{\\left(\\dfrac{1}{x}\\right)^{\\frac{n}{12}}}}[\/tex]
\n[tex] \\qquad=\\sqrt[2]{\\left(\\dfrac{1}{x}\\right)^{\\frac{n}{12}}}\\stackrel{\\textcolor{#800000}{(2)}}{=}\\left(\\dfrac{1}{x}\\right)^{\\frac{\\frac{n}{12}}{2}}=[\/tex]
\n[tex] \\qquad=\\left(\\dfrac{1}{x^1}\\right)^{\\frac{n}{24}}\\stackrel{\\textcolor{#800000}{(1)}}{=}\\left(x^{-1}\\right)^{\\frac{n}{24}}\\stackrel{\\textcolor{#800000}{(4)}}{=}\\boxed{x^{-\\frac{n}{24}}}[\/tex]<\/span>.
\nDessa forma, como <\/p>\n

[tex] \\qquad \\sqrt[4]{\\dfrac{\\sqrt[3]{\\dfrac{\\sqrt{\\dfrac{1}{x}}}{x}}}{x}}=\\sqrt{\\sqrt[3]{\\sqrt[4]{\\left(\\dfrac{1}{x}\\right)^n}}}\\,\\,[\/tex],<\/p>\n

segue que:
\n[tex] \\qquad x^{-\\frac{3}{8}}= x^{-\\frac{n}{24}}[\/tex].<\/span>
\nMas sabemos que [tex]x[\/tex] \u00e9 um n\u00famero real maior do que [tex] 1[\/tex], logo, utilizando o Lembrete (5)<\/strong><\/span>, segue dessa \u00faltima igualdade que :
\n[tex]\\qquad \\qquad -\\dfrac{3}{8}=-\\dfrac{n}{24}\\\\
\n\\qquad \\qquad \\dfrac{3}{8}=\\dfrac{n}{24}\\\\
\n\\qquad \\qquad n= 24 \\cdot \\dfrac{3}{8}\\\\
\n\\qquad \\qquad \\fcolorbox{black}{#eee0e5}{$n=9 $}\\,.[\/tex]<\/p>\n

\n

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

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

Se for conveniente, voc\u00ea pode obter um arquivo PDF desta p\u00e1gina, com o problema e a solu\u00e7\u00e3o, clicando no bot\u00e3o abaixo.
\nVoc\u00ea pode abrir o arquivo diretamente no seu navegador (Chrome<\/em>, Edge<\/em>, Firefox<\/em>, Safari<\/em>, entre outros), mas tamb\u00e9m pode utilizar o software gratuito Adobe Acrobat Reader<\/strong><\/em>.
\nCaso o dispositivo que voc\u00ea est\u00e1 utilizando n\u00e3o tenha o Acrobat Reader<\/em><\/strong> instalado, \u00e9 s\u00f3 clicar AQUI<\/strong><\/a> para fazer o download adequado ao seu dispositivo.<\/p>\n<\/td>\n<\/tr>\n<\/table>\n

\n

\nArquivo <\/strong><\/span><\/a>\n <\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

✏ Link do problema para dispositivos da Apple. Problema (A partir do 9\u00ba ano do E. F. – N\u00edvel de dificuldade: Dif\u00edcil) (ONEM, 2005 – Adaptado) Seja [tex]x[\/tex] um n\u00famero real maior do que [tex] 1[\/tex]. Determine o n\u00famero natural n\u00e3o nulo [tex]n[\/tex] tal que [tex] \\sqrt[4]{\\dfrac{\\sqrt[3]{\\dfrac{\\sqrt{\\dfrac{1}{x}}}{x}}}{x}}=\\sqrt{\\sqrt[3]{\\sqrt[4]{\\left(\\dfrac{1}{x}\\right)^n}}}[\/tex]. Lembretes S\u00e3o bem conhecidas as seguintes defini\u00e7\u00f5es envolvendo … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-150063","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/150063","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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=150063"}],"version-history":[{"count":20,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/150063\/revisions"}],"predecessor-version":[{"id":220759,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/150063\/revisions\/220759"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=150063"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}