{"id":33097,"date":"2015-03-18T20:51:33","date_gmt":"2015-03-18T23:51:33","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=33097"},"modified":"2023-03-21T21:24:39","modified_gmt":"2023-03-22T00:24:39","slug":"problema-um-problema-astuto-de-algebra","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-um-problema-astuto-de-algebra\/","title":{"rendered":".Problema: Um problema astuto de \u00e1lgebra"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 7\u00ba ano do E. F.)<\/span><\/p>\n

\n

\n

Simplificar a express\u00e3o<\/p>\n

[tex]\\qquad \\qquad \\dfrac{a-b}{a+b}+\\dfrac{b-c}{b+c}+\\dfrac{c-a}{c+a}+\\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}, [\/tex]<\/p>\n

admitindo que nenhum denominador seja zero.\n<\/p><\/div>\n

\n

\n

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

\n

\n

Observe o seguinte desenvolvimento:
\n[tex]\\quad S=\\dfrac{a-b}{a+b}+\\dfrac{b-c}{b+c}+\\dfrac{c-a}{c+a}+\\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}\\\\
\n\\quad S=\\dfrac{(a-b)(b+c)+(b-c)(a+b)}{(a+b)(b+c)}+\\dfrac{c-a}{c+a} \\left(1+\\dfrac{(a-b)(b-c)}{(a+b)(b+c)} \\right)\\\\
\n\\quad S=\\dfrac{ab+\\cancel{ac}-\\cancel{b^2}-bc+ab+\\cancel{b^2}-\\cancel{ac}-bc}{(a+b)(b+c)}+\\\\
\n\\quad \\quad \\quad + \\dfrac{c-a}{c+a} \\left(\\dfrac{ab+\\cancel{ac}+\\cancel{b^2}+bc+ab-\\cancel{ac}-\\cancel{b^2}+bc}{(a+b)(b+c)} \\right)\\\\
\n\\quad S= \\dfrac{2b(a-c)}{(a+b)(b+c)}+\\dfrac{c-a}{\\cancel{c+a}} \\left(\\dfrac{2b\\cancel{(c+a)}}{(a+b)(b+c)}\\right)\\\\
\n\\quad S=\\dfrac{2b(a-c)}{(a+b)(b+c)}+\\dfrac{2b(c-a)}{(a+b)(b+c)}\\\\
\n\\quad S=\\dfrac{2b(a-c)}{(a+b)(b+c)}-\\dfrac{2b(a-c)}{(a+b)(b+c)}=0\\,.[\/tex]
\nPortanto,
\n[tex]\\qquad \\boxed{\\dfrac{a-b}{a+b}+\\dfrac{b-c}{b+c}+\\dfrac{c-a}{c+a}+\\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=0 }\\,.[\/tex]<\/p>\n

\n

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

\n

\n

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

\n

\n

Note que:
\n[tex]\\qquad \\dfrac{a-b}{a+b}+\\dfrac{b-c}{b+c}+\\dfrac{c-a}{c+a}+\\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=\\\\
\n\\qquad \\qquad \\dfrac{(a-b)(b+c)(c+a)+(b-c)(a+b)(c+a)}{(a+b)(b+c)(c+a)}+\\\\
\n\\qquad \\qquad +\\dfrac{(c-a)(a+b)(b+c)+(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=\\\\
\n\\qquad \\qquad \\dfrac{(c+a)[(a-b)(b+c)+(b-c)(a+b)]}{(a+b)(b+c)(c+a)}+\\\\
\n\\qquad \\qquad +\\dfrac{(c-a)[(a+b)(b+c)+(a-b)(b-c)]}{(a+b)(b+c)(c+a)}=\\\\
\n\\qquad \\qquad \\dfrac{(c+a)(ab\\cancel{-b^2}\\cancel{+ac}-bc+ab\\cancel{-ac}\\cancel{+b^2}-bc)}{(a+b)(b+c)(c+a)}+\\\\
\n\\qquad \\qquad +\\dfrac{(c-a)(ab\\bcancel{+b^2}\\bcancel{+ac}+bc+ab\\bcancel{-b^2}\\bcancel{-ac}+bc)}{(a+b)(b+c)(c+a)}=[\/tex]
\n[tex]\\qquad \\qquad \\dfrac{2b(c+a)(a-c)+2b(c-a)(a+c)}{(a+b)(b+c)(c+a)}=[\/tex]
\n[tex]\\qquad \\qquad \\dfrac{2b(a^2-c^2)+2b(c^2-a^2)}{(a+b)(b+c)(c+a)}=[\/tex]
\n[tex]\\qquad \\qquad \\dfrac{2b(a^2-a^2+c^2-c^2)}{(a+b)(b+c)(c+a)}=0[\/tex]
\nAssim,
\n[tex]\\qquad \\boxed{\\dfrac{a-b}{a+b}+\\dfrac{b-c}{b+c}+\\dfrac{c-a}{c+a}+\\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=0}\\,. [\/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 7\u00ba ano do E. F.) Simplificar a express\u00e3o [tex]\\qquad \\qquad \\dfrac{a-b}{a+b}+\\dfrac{b-c}{b+c}+\\dfrac{c-a}{c+a}+\\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}, [\/tex] admitindo que nenhum denominador seja zero. Solu\u00e7\u00e3o 1 Observe o seguinte desenvolvimento: [tex]\\quad S=\\dfrac{a-b}{a+b}+\\dfrac{b-c}{b+c}+\\dfrac{c-a}{c+a}+\\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}\\\\ \\quad S=\\dfrac{(a-b)(b+c)+(b-c)(a+b)}{(a+b)(b+c)}+\\dfrac{c-a}{c+a} \\left(1+\\dfrac{(a-b)(b-c)}{(a+b)(b+c)} \\right)\\\\ \\quad S=\\dfrac{ab+\\cancel{ac}-\\cancel{b^2}-bc+ab+\\cancel{b^2}-\\cancel{ac}-bc}{(a+b)(b+c)}+\\\\ \\quad \\quad \\quad + \\dfrac{c-a}{c+a} \\left(\\dfrac{ab+\\cancel{ac}+\\cancel{b^2}+bc+ab-\\cancel{ac}-\\cancel{b^2}+bc}{(a+b)(b+c)} \\right)\\\\ \\quad S= \\dfrac{2b(a-c)}{(a+b)(b+c)}+\\dfrac{c-a}{\\cancel{c+a}} \\left(\\dfrac{2b\\cancel{(c+a)}}{(a+b)(b+c)}\\right)\\\\ \\quad S=\\dfrac{2b(a-c)}{(a+b)(b+c)}+\\dfrac{2b(c-a)}{(a+b)(b+c)}\\\\ \\quad S=\\dfrac{2b(a-c)}{(a+b)(b+c)}-\\dfrac{2b(a-c)}{(a+b)(b+c)}=0\\,.[\/tex] Portanto, [tex]\\qquad \\boxed{\\dfrac{a-b}{a+b}+\\dfrac{b-c}{b+c}+\\dfrac{c-a}{c+a}+\\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=0 … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":164,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-33097","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/33097","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\/164"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=33097"}],"version-history":[{"count":22,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/33097\/revisions"}],"predecessor-version":[{"id":185038,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/33097\/revisions\/185038"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=33097"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}