<\/p>\n
Problema <\/p>\n Solu\u00e7\u00e3o 1 <\/strong><\/span><\/p>\n Solu\u00e7\u00e3o enviada pelo Clube Eureka!<\/strong>.<\/p>\n<\/div>\n <\/p>\n Solu\u00e7\u00e3o 2 <\/strong><\/span><\/p>\n Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n <\/p>\n Participou da discuss\u00e3o do problema o Clube Eureka!<\/strong>.<\/span><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":" Problema (Indicado a partir do 8\u00ba ano do E. F.) Qual \u00e9 a maior fra\u00e7\u00e3o: [tex]\\, \\dfrac{2005}{2006}\\, [\/tex] ou [tex]\\, \\dfrac{2006}{2007}\\, [\/tex]? Observa\u00e7\u00e3o: Resolva o problema sem efetuar as divis\u00f5es! Solu\u00e7\u00e3o 1 Se [tex] x [\/tex] \u00e9 um n\u00famero natural n\u00e3o nulo, segue que [tex]\\qquad\\qquad \\dfrac{x-1}{x}-\\dfrac{x}{x+1}=\\dfrac{x^2-1-x^2}{x^2+x}=\\dfrac{-1}{x^2+x}[\/tex]. Assim, se [tex]x=2006[\/tex], temos … <\/p>\n
\n (Indicado a partir do 8\u00ba ano do E. F.)<\/span><\/span><\/strong><\/p>\n
\n Qual \u00e9 a maior fra\u00e7\u00e3o: [tex]\\, \\dfrac{2005}{2006}\\, [\/tex] ou [tex]\\, \\dfrac{2006}{2007}\\, [\/tex]?
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\nObserva\u00e7\u00e3o:<\/strong> Resolva o problema sem efetuar as divis\u00f5es!\n<\/div>\n
\nSe [tex] x [\/tex] \u00e9 um n\u00famero natural n\u00e3o nulo, segue que
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\n[tex]\\qquad\\qquad \\dfrac{x-1}{x}-\\dfrac{x}{x+1}=\\dfrac{x^2-1-x^2}{x^2+x}=\\dfrac{-1}{x^2+x}[\/tex].
\n
\nAssim, se [tex]x=2006[\/tex], temos que:
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\n[tex]\\qquad \\dfrac{2006-1}{2006}-\\dfrac{2006}{2006+1}=\\dfrac{-1}{2006^2+2006}\\\\
\n~~\\\\
\n\\qquad \\dfrac{2005}{2006}-\\dfrac{2006}{2007}=\\dfrac{-1}{2006^2+2006}.
\n[\/tex]
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\nCom isso, notamos que [tex]\\dfrac{2005}{2006}-\\dfrac{2006}{2007}\\lt 0[\/tex] e, portanto, [tex]\\dfrac{2006}{2007}[\/tex] \u00e9 a maior dentre as fra\u00e7\u00f5es dadas.<\/p>\n
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\nApresentamos uma generaliza\u00e7\u00e3o deste problema e sua resolu\u00e7\u00e3o:<\/p>\n\n
\nRealizando a subtra\u00e7\u00e3o das fra\u00e7\u00f5es, temos
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\n[tex]\\qquad \\begin{align*} \\dfrac{n+1}{n+2}-\\dfrac{n}{n+1} &= \\dfrac{(n+1) \\cdot (n+1)}{(n+2) \\cdot (n+1)}-\\dfrac{n \\cdot (n+2)}{(n+1) \\cdot (n+2)}\\\\
\n&= \\dfrac{n^2 + 2n + 1}{(n+2) \\cdot (n+1)}-\\dfrac{n^2 + 2n}{(n+1) \\cdot (n+2)}\\\\
\n&= \\dfrac{n^2 + 2n + 1-n^2-2n}{(n+1) \\cdot (n+2)}\\\\
\n&= \\dfrac{1}{(n+1) \\cdot (n+2)} \\gt 0.\\end{align*}[\/tex]
\n
\nMas se [tex]\\dfrac{n+1}{n+2} – \\dfrac{n}{n+1} > 0[\/tex], podemos concluir que [tex]\\dfrac{n+1}{n+2} > \\dfrac{n}{n+1}[\/tex].
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\nDe modo particular, para [tex]n = 2005[\/tex], temos que [tex]\\fcolorbox{black}{#eee0e5}{$\\dfrac{2006}{2007} \\gt \\dfrac{2005}{2006}$}\\,.[\/tex]<\/p>\n
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