{"id":164685,"date":"2020-09-18T20:43:48","date_gmt":"2020-09-18T23:43:48","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=164685"},"modified":"2023-07-23T03:22:14","modified_gmt":"2023-07-23T06:22:14","slug":"problema-uma-expressao-em-si-mesma","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-uma-expressao-em-si-mesma\/","title":{"rendered":".Problema: Uma express\u00e3o em si mesma"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 8\u00ba ano do E. F.)<\/span><\/p>\n

\nNa express\u00e3o [tex]\\dfrac{x+2}{x-2}[\/tex], substitu\u00edmos cada [tex]x[\/tex] por [tex]\\dfrac{x+2}{x-2}[\/tex].
\nQual \u00e9 o valor resultante se substituirmos, na express\u00e3o obtida, [tex]x[\/tex] por [tex]\\dfrac{1}{3}[\/tex]?<\/strong><\/p>\n

Adaptado de UNMSM – 88<\/strong>.<\/p>\n<\/div>\n

\n

\n

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

\nSe na express\u00e3o [tex]\\dfrac{x+2}{x-2}[\/tex] substituirmos cada [tex]x[\/tex] por [tex]\\dfrac{x+2}{x-2}[\/tex], obtemos:
\n[tex]\\,\\\\
\n \\quad \\begin{align*}\\dfrac{\\dfrac{x+2}{x-2}+2}{\\dfrac{x+2}{x-2}-2}&=\\dfrac{\\dfrac{x+2}{x-2}+\\dfrac{2(x-2)}{x-2}}{\\dfrac{x+2}{x-2}-\\dfrac{2(x-2)}{x-2}}\\\\
\n\\,\\\\
\n&=\\dfrac{\\dfrac{x+2+2x-4}{x-2}}{\\dfrac{x+2-2x+4}{x-2}}\\\\
\n\\,\\\\
\n&=\\dfrac{\\dfrac{3x-2}{x-2}}{\\dfrac{6-x}{x-2}}\\\\
\n\\,\\\\
\n&=\\dfrac{3x-2}{\\cancel{x-2}}\\cdot \\dfrac{\\cancel{x-2}}{6-x}\\\\
\n\\,\\\\
\n&=\\boxed{\\dfrac{3x-2}{6-x}}\\,.\\\\
\n\\,
\n\\end{align*}[\/tex]
\nSubstituindo [tex]x=\\frac{1}{3}[\/tex] na express\u00e3o final obtida, segue que:<\/p>\n

[tex]\\quad \\dfrac{3x-2}{6-x}=\\dfrac{3\\cdot \\frac{1}{3}-2}{6-\\frac{1}{3}}=\\dfrac{-1}{\\frac{17}{3}}=-\\dfrac{3}{17}[\/tex]
\ne, assim, o valor obtido com a substitui\u00e7\u00e3o proposta foi [tex] \\,\\fcolorbox{black}{#eee0e5}{$-\\,\\dfrac{3}{17}$}\\,.[\/tex]<\/p>\n

\n

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

\n","protected":false},"excerpt":{"rendered":"

Problema (Indicado a partir do 8\u00ba ano do E. F.) Na express\u00e3o [tex]\\dfrac{x+2}{x-2}[\/tex], substitu\u00edmos cada [tex]x[\/tex] por [tex]\\dfrac{x+2}{x-2}[\/tex]. Qual \u00e9 o valor resultante se substituirmos, na express\u00e3o obtida, [tex]x[\/tex] por [tex]\\dfrac{1}{3}[\/tex]? Adaptado de UNMSM – 88. Solu\u00e7\u00e3o Se na express\u00e3o [tex]\\dfrac{x+2}{x-2}[\/tex] substituirmos cada [tex]x[\/tex] por [tex]\\dfrac{x+2}{x-2}[\/tex], obtemos: [tex]\\,\\\\ \\quad \\begin{align*}\\dfrac{\\dfrac{x+2}{x-2}+2}{\\dfrac{x+2}{x-2}-2}&=\\dfrac{\\dfrac{x+2}{x-2}+\\dfrac{2(x-2)}{x-2}}{\\dfrac{x+2}{x-2}-\\dfrac{2(x-2)}{x-2}}\\\\ \\,\\\\ &=\\dfrac{\\dfrac{x+2+2x-4}{x-2}}{\\dfrac{x+2-2x+4}{x-2}}\\\\ \\,\\\\ &=\\dfrac{\\dfrac{3x-2}{x-2}}{\\dfrac{6-x}{x-2}}\\\\ \\,\\\\ … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":255,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-164685","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/164685","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\/255"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=164685"}],"version-history":[{"count":22,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/164685\/revisions"}],"predecessor-version":[{"id":197736,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/164685\/revisions\/197736"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=164685"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}