{"id":13737,"date":"2014-02-03T22:24:00","date_gmt":"2014-02-04T00:24:00","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=13737"},"modified":"2023-01-21T23:52:21","modified_gmt":"2023-01-22T02:52:21","slug":"problema-olimpico-nivel-a-um-problema-com-fracoes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-olimpico-nivel-a-um-problema-com-fracoes\/","title":{"rendered":".Problema Ol\u00edmpico – N\u00edvel A: Um problema com fra\u00e7\u00f5es"},"content":{"rendered":"
\n

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

\n

Se [tex]~\\dfrac{1}{a+1}=2[\/tex], [tex]~\\dfrac{1}{b+2}=3~[\/tex] e [tex]~\\dfrac{1}{c+3}=6[\/tex], calcular [tex]\\dfrac{1}{a+b+c}[\/tex].<\/p>\n

<\/span><\/div>\n

 <\/p>\n

\n

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

\n

Temos que
\n[tex]\\quad\\quad\\quad \\dfrac{x}{y}=z \\iff zy=x, \\forall \\ y\\ne0[\/tex].
\nLogo:
\n[tex]\\quad\\quad\\,\\,\\bullet\\,\\,\\dfrac{1}{a+1}=2\\Rightarrow 2(a+1)=2\\Rightarrow a= – \\dfrac{1}{2}[\/tex];
\n 
\n[tex]\\quad\\quad\\,\\,\\bullet \\,\\,\\dfrac{1}{b+2}= 3\\Rightarrow 3(b+2)=1\\Rightarrow b= – \\dfrac{5}{3}[\/tex];
\n 
\n[tex]\\quad\\quad\\,\\,\\bullet \\,\\,\\dfrac{1}{c+3}=6\\Rightarrow 6(c+3)=1\\Rightarrow c= – \\dfrac{17}{6}[\/tex];
\n 
\nde sorte que
\n[tex]\\quad\\quad\\quad \\dfrac{1}{a+b+c}=\\dfrac{1}{-\\dfrac{1}{2}-\\dfrac{5}{3}-\\dfrac{17}{6}}= \\dfrac{1}{\\dfrac{-3-10-17}{6}}[\/tex].
\nAssim, [tex]\\boxed{ \\dfrac{1}{a+b+c}= -\\dfrac{1}{5}}.[\/tex]<\/span><\/p>\n

\n

Solu\u00e7\u00e3o elaborada pela aluna do PIC-OBMEP Miriam de C\u00e1ssia Souza<\/strong>.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Problema Se [tex]~\\dfrac{1}{a+1}=2[\/tex], [tex]~\\dfrac{1}{b+2}=3~[\/tex] e [tex]~\\dfrac{1}{c+3}=6[\/tex], calcular [tex]\\dfrac{1}{a+b+c}[\/tex].   Solu\u00e7\u00e3o Temos que [tex]\\quad\\quad\\quad \\dfrac{x}{y}=z \\iff zy=x, \\forall \\ y\\ne0[\/tex]. Logo: [tex]\\quad\\quad\\,\\,\\bullet\\,\\,\\dfrac{1}{a+1}=2\\Rightarrow 2(a+1)=2\\Rightarrow a= – \\dfrac{1}{2}[\/tex];   [tex]\\quad\\quad\\,\\,\\bullet \\,\\,\\dfrac{1}{b+2}= 3\\Rightarrow 3(b+2)=1\\Rightarrow b= – \\dfrac{5}{3}[\/tex];   [tex]\\quad\\quad\\,\\,\\bullet \\,\\,\\dfrac{1}{c+3}=6\\Rightarrow 6(c+3)=1\\Rightarrow c= – \\dfrac{17}{6}[\/tex];   de sorte que [tex]\\quad\\quad\\quad \\dfrac{1}{a+b+c}=\\dfrac{1}{-\\dfrac{1}{2}-\\dfrac{5}{3}-\\dfrac{17}{6}}= \\dfrac{1}{\\dfrac{-3-10-17}{6}}[\/tex]. Assim, [tex]\\boxed{ \\dfrac{1}{a+b+c}= -\\dfrac{1}{5}}.[\/tex] Solu\u00e7\u00e3o elaborada pela aluna … <\/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-13737","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/13737","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=13737"}],"version-history":[{"count":25,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/13737\/revisions"}],"predecessor-version":[{"id":183324,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/13737\/revisions\/183324"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=13737"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}