{"id":107086,"date":"2018-07-14T00:35:15","date_gmt":"2018-07-14T03:35:15","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=107086"},"modified":"2023-05-03T02:53:06","modified_gmt":"2023-05-03T05:53:06","slug":"problema-radicais-e-fracoes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-radicais-e-fracoes\/","title":{"rendered":".Problema: Radicais e Fra\u00e7\u00f5es"},"content":{"rendered":"

\n

\n

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

\n

Calcule o valor da express\u00e3o num\u00e9rica abaixo:<\/p>\n

[tex]\\dfrac{1}{\\sqrt{1}+\\sqrt{2}}+\\dfrac{1}{\\sqrt{2}+\\sqrt{3}}+\\dfrac{1}{\\sqrt{3}+\\sqrt{4}}+\\cdots +\\dfrac{1}{\\sqrt{98}+\\sqrt{99}}+\\dfrac{1}{\\sqrt{99}+\\sqrt{100}}\\,[\/tex]. <\/p>\n

\n<\/p><\/div>\n

\n

\n

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

\nAo racionalizar uma express\u00e3o do tipo [tex]\\dfrac{1}{\\sqrt{x}+\\sqrt{x+1}}[\/tex], obtemos:
\n[tex]\\qquad \\dfrac{1}{\\sqrt{x}+\\sqrt{x+1}} \\cdot \\dfrac{{\\sqrt{x}-\\sqrt{x+1}}}{\\sqrt{x}-\\sqrt{x+1}}= \\dfrac{{\\sqrt{x}-\\sqrt{x+1}}}{-1}= \\sqrt{x+1}-\\sqrt{x}[\/tex].
\nPortanto,
\n[tex]\\qquad \\dfrac{1}{\\sqrt{1}+\\sqrt{2}}= \\sqrt{2}-\\sqrt{1}[\/tex]
\n[tex]\\qquad \\dfrac{1}{\\sqrt{2}+\\sqrt{3}}= \\sqrt{3}-\\sqrt{2}[\/tex]
\n[tex]\\qquad \\dfrac{1}{\\sqrt{3}+\\sqrt{4}}= \\sqrt{4}-\\sqrt{3}[\/tex]
\n[tex]\\qquad \\vdots[\/tex]
\n[tex]\\qquad \\dfrac{1}{\\sqrt{98}+\\sqrt{99}}= \\sqrt{99}-\\sqrt{98}[\/tex]
\n[tex]\\qquad \\dfrac{1}{\\sqrt{99}+\\sqrt{100}}= \\sqrt{100}-\\sqrt{99} \\, .[\/tex]
\nSomando membro a membro todas as igualdades acima, teremos:
\n[tex]\\dfrac{1}{\\sqrt{1}+\\sqrt{2}}+\\dfrac{1}{\\sqrt{2}+\\sqrt{3}}+\\dfrac{1}{\\sqrt{3}+\\sqrt{4}}+\\cdots +\\dfrac{1}{\\sqrt{98}+\\sqrt{99}}+\\dfrac{1}{\\sqrt{99}+\\sqrt{100}}=[\/tex]
\n[tex]\\qquad =\\sqrt{2}-\\sqrt{1}+ \\sqrt{3}-\\sqrt{2}+ \\sqrt{4}-\\sqrt{3}+ \\cdots\\sqrt{99}-\\sqrt{98}+\\sqrt{100}-\\sqrt{99}=[\/tex]
\n[tex]\\qquad =\\bcancel{\\sqrt{2}}-\\sqrt{1}+\\bcancel{\\sqrt{3}}-\\bcancel{\\sqrt{2}}+\\bcancel{\\sqrt{4}}-\\bcancel{\\sqrt{3}}+ \\cdots\\\\
\n\\qquad \\quad \\cdots +\\bcancel{\\sqrt{99}}-\\bcancel{\\sqrt{98}}+ {\\sqrt{100}}-\\bcancel{\\sqrt{99}}=[\/tex]
\n[tex]\\qquad =\\sqrt{100}- \\sqrt{1}=10-1=9 \\, .[\/tex]
\nPortanto, [tex]\\fcolorbox{black}{#eee0e5}{$\\dfrac{1}{\\sqrt{1}+\\sqrt{2}}+\\dfrac{1}{\\sqrt{2}+\\sqrt{3}}+\\dfrac{1}{\\sqrt{3}+\\sqrt{4}}+\\cdots +\\dfrac{1}{\\sqrt{98}+\\sqrt{99}}+\\dfrac{1}{\\sqrt{99}+\\sqrt{100}}=9$} \\, .[\/tex] <\/p>\n
\n

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

Problema (Indicado a partir do 9\u00ba ano do E. F.) Calcule o valor da express\u00e3o num\u00e9rica abaixo: [tex]\\dfrac{1}{\\sqrt{1}+\\sqrt{2}}+\\dfrac{1}{\\sqrt{2}+\\sqrt{3}}+\\dfrac{1}{\\sqrt{3}+\\sqrt{4}}+\\cdots +\\dfrac{1}{\\sqrt{98}+\\sqrt{99}}+\\dfrac{1}{\\sqrt{99}+\\sqrt{100}}\\,[\/tex]. Solu\u00e7\u00e3o Ao racionalizar uma express\u00e3o do tipo [tex]\\dfrac{1}{\\sqrt{x}+\\sqrt{x+1}}[\/tex], obtemos: [tex]\\qquad \\dfrac{1}{\\sqrt{x}+\\sqrt{x+1}} \\cdot \\dfrac{{\\sqrt{x}-\\sqrt{x+1}}}{\\sqrt{x}-\\sqrt{x+1}}= \\dfrac{{\\sqrt{x}-\\sqrt{x+1}}}{-1}= \\sqrt{x+1}-\\sqrt{x}[\/tex]. Portanto, [tex]\\qquad \\dfrac{1}{\\sqrt{1}+\\sqrt{2}}= \\sqrt{2}-\\sqrt{1}[\/tex] [tex]\\qquad \\dfrac{1}{\\sqrt{2}+\\sqrt{3}}= \\sqrt{3}-\\sqrt{2}[\/tex] [tex]\\qquad \\dfrac{1}{\\sqrt{3}+\\sqrt{4}}= \\sqrt{4}-\\sqrt{3}[\/tex] [tex]\\qquad \\vdots[\/tex] [tex]\\qquad \\dfrac{1}{\\sqrt{98}+\\sqrt{99}}= \\sqrt{99}-\\sqrt{98}[\/tex] [tex]\\qquad \\dfrac{1}{\\sqrt{99}+\\sqrt{100}}= \\sqrt{100}-\\sqrt{99} \\, .[\/tex] Somando … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":668,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-107086","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/107086","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\/668"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=107086"}],"version-history":[{"count":24,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/107086\/revisions"}],"predecessor-version":[{"id":188683,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/107086\/revisions\/188683"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=107086"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}