{"id":226164,"date":"2025-08-28T17:06:28","date_gmt":"2025-08-28T20:06:28","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=226164"},"modified":"2025-08-28T19:03:50","modified_gmt":"2025-08-28T22:03:50","slug":"problemao-valor-de-texab-tex","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-valor-de-texab-tex\/","title":{"rendered":".Problem\u00e3o: Valor de [tex]a+b[\/tex]"},"content":{"rendered":"

\n

\n

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

\n

Dados os n\u00fameros reais [tex]a[\/tex] e [tex]b[\/tex], onde <\/p>\n

[tex]\\qquad a^2+b^2=\\left(\\dfrac{56}{13}\\right)^2[\/tex] e [tex]a+\\dfrac{5b}{12}=\\dfrac{56}{12},[\/tex]<\/p>\n

determine o valor de [tex]a+b[\/tex].<\/p>\n

\n<\/div>\n

\n

\n

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

\n

De [tex]a+\\dfrac{5b}{12}=\\dfrac{56}{12}[\/tex], vem [tex]a=\\dfrac{56-5b}{12}.[\/tex]<\/p>\n

Da\u00ed, <\/p>\n

[tex]\\qquad a^2+b^2=\\left(\\dfrac{56}{13}\\right)^2[\/tex]<\/p>\n

[tex]\\qquad \\left(\\dfrac{56-5b}{12}\\right)^2+b^2=\\dfrac{56^2}{13^2}[\/tex]<\/p>\n

[tex]\\qquad \\left(\\dfrac{56}{12}\\right)^2-\\dfrac{560b}{144}+\\left(\\dfrac{5b}{12}\\right)^2+b^2=\\dfrac{56^2}{13^2}[\/tex]<\/p>\n

[tex]\\qquad \\left(\\dfrac{14}{3}\\right)^2-\\dfrac{35b}{9}+\\left(\\dfrac{5b}{12}\\right)^2+b^2=\\dfrac{56^2}{13^2}[\/tex]<\/p>\n

[tex]\\qquad \\dfrac{196}{9}-\\dfrac{35b}{9}+\\dfrac{25b^2}{144}+b^2=\\dfrac{3136}{169}[\/tex]<\/p>\n

[tex]\\qquad \\dfrac{196}{9}-\\dfrac{35b}{9}+\\dfrac{169b^2}{144}=\\dfrac{3136}{169}[\/tex]<\/p>\n

[tex]\\qquad \\dfrac{169b^2}{144}-\\dfrac{35b}{9}+\\dfrac{4900}{1521}=0[\/tex]<\/p>\n

[tex]\\qquad 169b^2-560b+\\dfrac{78400}{169}=0[\/tex]<\/p>\n

[tex]\\qquad b = \\dfrac{560\\pm\\sqrt{560^2-4\\cdot \\cancel{169}\\cdot \\frac{78400}{\\cancel{169}}}}{2\\cdot 169}[\/tex]<\/p>\n

[tex]\\qquad b = \\dfrac{560\\pm\\sqrt{560^2-4\\cdot 78400}}{2\\cdot 169}[\/tex]<\/p>\n

[tex]\\qquad b = \\dfrac{560\\pm\\sqrt{0}}{2\\cdot 169}[\/tex]<\/p>\n

[tex]\\qquad b = \\dfrac{560}{2\\cdot 169}[\/tex]<\/p>\n

[tex]\\qquad b = \\dfrac{280}{169}.[\/tex]<\/p>\n

Assim, <\/p>\n

[tex]\\qquad a=\\dfrac{56-5b}{12}=\\dfrac{56-5\\cdot \\frac{280}{169}}{12}=\\dfrac{\\frac{8064}{169}}{12}=\\dfrac{672}{169}.[\/tex]<\/p>\n

Finalmente, <\/p>\n

[tex]\\qquad a+b=\\dfrac{672}{169}+\\dfrac{280}{169}=\\dfrac{952}{169}[\/tex].<\/p>\n

\n

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

Problema (Indicado a partir do 9\u00ba ano do E. F.) Dados os n\u00fameros reais [tex]a[\/tex] e [tex]b[\/tex], onde [tex]\\qquad a^2+b^2=\\left(\\dfrac{56}{13}\\right)^2[\/tex] e [tex]a+\\dfrac{5b}{12}=\\dfrac{56}{12},[\/tex] determine o valor de [tex]a+b[\/tex]. Solu\u00e7\u00e3o De [tex]a+\\dfrac{5b}{12}=\\dfrac{56}{12}[\/tex], vem [tex]a=\\dfrac{56-5b}{12}.[\/tex] Da\u00ed, [tex]\\qquad a^2+b^2=\\left(\\dfrac{56}{13}\\right)^2[\/tex] [tex]\\qquad \\left(\\dfrac{56-5b}{12}\\right)^2+b^2=\\dfrac{56^2}{13^2}[\/tex] [tex]\\qquad \\left(\\dfrac{56}{12}\\right)^2-\\dfrac{560b}{144}+\\left(\\dfrac{5b}{12}\\right)^2+b^2=\\dfrac{56^2}{13^2}[\/tex] [tex]\\qquad \\left(\\dfrac{14}{3}\\right)^2-\\dfrac{35b}{9}+\\left(\\dfrac{5b}{12}\\right)^2+b^2=\\dfrac{56^2}{13^2}[\/tex] [tex]\\qquad \\dfrac{196}{9}-\\dfrac{35b}{9}+\\dfrac{25b^2}{144}+b^2=\\dfrac{3136}{169}[\/tex] [tex]\\qquad \\dfrac{196}{9}-\\dfrac{35b}{9}+\\dfrac{169b^2}{144}=\\dfrac{3136}{169}[\/tex] [tex]\\qquad \\dfrac{169b^2}{144}-\\dfrac{35b}{9}+\\dfrac{4900}{1521}=0[\/tex] [tex]\\qquad 169b^2-560b+\\dfrac{78400}{169}=0[\/tex] [tex]\\qquad b = \\dfrac{560\\pm\\sqrt{560^2-4\\cdot \\cancel{169}\\cdot \\frac{78400}{\\cancel{169}}}}{2\\cdot … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":164,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-226164","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/226164","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=226164"}],"version-history":[{"count":7,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/226164\/revisions"}],"predecessor-version":[{"id":226393,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/226164\/revisions\/226393"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=226164"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}