{"id":14245,"date":"2014-02-10T00:51:05","date_gmt":"2014-02-10T02:51:05","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=14245"},"modified":"2023-01-24T19:16:36","modified_gmt":"2023-01-24T22:16:36","slug":"problema-olimpico-nivel-b-operando-com-raizes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-olimpico-nivel-b-operando-com-raizes\/","title":{"rendered":".Problema Ol\u00edmpico – N\u00edvel B: Operando com ra\u00edzes"},"content":{"rendered":"
\n

Problema<\/strong><\/p>\n

\n

Determine o valor de [tex]x[\/tex], sabendo que [tex]x=\\sqrt{7+\\sqrt{24}}-\\sqrt{7-\\sqrt{24}}[\/tex].<\/p>\n<\/div>\n

 <\/p>\n

\n

Solu\u00e7\u00e3o 1<\/strong><\/p>\n

\n

Temos que [tex]x=\\sqrt{7+\\sqrt{24}}-\\sqrt{7-\\sqrt{24}}[\/tex]; mas observamos que
\n[tex]\\qquad \\quad \\sqrt{24}=\\sqrt{4\\cdot6}=2\\sqrt{6}[\/tex]
\ne, portanto,
\n[tex]\\qquad x=\\sqrt{7+\\sqrt{24}}-\\sqrt{7-\\sqrt{24}}=\\sqrt{7+2\\sqrt{6}}-\\sqrt{7-2\\sqrt{6}}[\/tex].
\nLembrando que
\n[tex]\\qquad (a\\pm b)^2=a^2\\pm 2ab+b^2[\/tex],
\npodemos observar que:
\n[tex]\\qquad 7+2\\sqrt{6}=6+2\\sqrt{6}+1=(\\sqrt{6})^2+2\\sqrt{6}+1=(\\sqrt{6}+1)^2[\/tex]
\n[tex]\\qquad 7-2\\sqrt{6}=6-2\\sqrt{6}+1=(\\sqrt{6})^2-2\\sqrt{6}+1=(\\sqrt{6}-1)^2[\/tex].
\nTemos tamb\u00e9m que
\n[tex]\\qquad x\\in\\mathbb{R}\\Rightarrow\\sqrt{x^2}=|x|[\/tex],
\nsendo [tex]|x|[\/tex] o valor absoluto de [tex]x[\/tex] (o valor de [tex]x[\/tex] sem sinal),
\nent\u00e3o podemos escrever
\n[tex]\\qquad \\begin{align*} x&=\\sqrt{7+2\\sqrt{6}}-\\sqrt{7-2\\sqrt{6}}\\\\
\n&=\\sqrt{(\\sqrt{6}+1)^2}-\\sqrt{(\\sqrt{6}-1)^2}\\\\
\n&=|\\sqrt{6}+1|-|\\sqrt{6}-1|.\\end{align*}[\/tex]
\nComo [tex]\\sqrt{6}>1[\/tex], ent\u00e3o [tex]\\sqrt{6}-1>0[\/tex] e com isso [tex]|\\sqrt{6}-1|=\\sqrt{6}-1[\/tex].
\nPortanto,
\n[tex]\\qquad \\begin{align*}x&=|\\sqrt{6}+1|-|\\sqrt{6}-1|\\\\
\n&=\\sqrt{6}+1-(\\sqrt{6}-1)\\\\
\n&=\\cancel{\\sqrt{6}}+1-\\cancel{\\sqrt{6}}+1=2 .\\end{align*}[\/tex]
\nAssim,
\n[tex]\\qquad \\fbox{x=2} \\, [\/tex].<\/p>\n

\n

Solu\u00e7\u00e3o elaborada pelo aluno do PIC-OBMEP Andr\u00e9 Oliveira Mendes<\/strong>, com contribui\u00e7\u00f5es dos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n

 <\/p>\n

\n

Solu\u00e7\u00e3o 2<\/strong><\/p>\n

\n

Sejam [tex]a=7+\\sqrt{24} \\, \\, [\/tex] e [tex] \\, \\, b= 7-\\sqrt{24} \\, \\, [\/tex] e note que [tex]a>b>0[\/tex].
\nComo [tex]x=\\sqrt{7+\\sqrt{24}}-\\sqrt{7-\\sqrt{24}}[\/tex], ent\u00e3o [tex]x=\\sqrt{a}-\\sqrt{b}>0[\/tex].
\nObserve que
\n[tex]\\qquad \\left(\\sqrt{a}-\\sqrt{b}\\right)^2=\\left(\\sqrt{a}\\right)^2-2\\sqrt{a}\\sqrt{b}+\\left(\\sqrt{b}\\right)^2=a-2\\sqrt{a}\\sqrt{b}+b\\;;[\/tex]
\nassim,
\n[tex]\\qquad x=\\sqrt{a}-\\sqrt{b}=\\sqrt{a-2\\sqrt{ab}+b}[\/tex].
\nVamos calcular
\n[tex]\\qquad ab=\\left(7+\\sqrt{24}\\right)\\cdot \\left(7-\\sqrt{24}\\right)[\/tex]
\ne, para tanto, observe que
\n[tex]\\qquad (m+n) \\, (m-n)=m^2-n^2[\/tex].
\nAssim
\n[tex]\\qquad ab=\\left(7+\\sqrt{24}\\right)\\cdot \\left(7-\\sqrt{24}\\right)=7^2-\\left(\\sqrt{24}\\right)^2=49-24=25[\/tex]
\ne, ent\u00e3o,
\n[tex]\\qquad x=\\sqrt{a}-\\sqrt{b}=\\sqrt{a-2\\sqrt{ab}+b}\\\\
\n\\qquad x=\\sqrt{7+\\sqrt{24}-2\\cdot \\sqrt{25}+7-\\sqrt{24}}\\\\
\n\\qquad x=\\sqrt{7+\\sqrt{24}-2\\cdot 5+7-\\sqrt{24}},[\/tex]
\ndonde
\n[tex]\\qquad x=\\sqrt{7+\\cancel{\\sqrt{24}}-10+7-\\cancel{\\sqrt{24}}}\\\\
\n\\qquad x=\\sqrt{7-10+7}=\\sqrt{14-10}=\\sqrt{4}.[\/tex]
\nAssim,
\n[tex]\\qquad \\fcolorbox{red}{yellow}{x=2}\\,.[\/tex]<\/p>\n

\n

Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Problema<\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-14245","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/14245","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=14245"}],"version-history":[{"count":83,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/14245\/revisions"}],"predecessor-version":[{"id":183371,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/14245\/revisions\/183371"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=14245"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}