{"id":220375,"date":"2024-07-30T17:55:54","date_gmt":"2024-07-30T20:55:54","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=220375"},"modified":"2024-07-30T17:55:54","modified_gmt":"2024-07-30T20:55:54","slug":"problemao-diferenca-de-cubos","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-diferenca-de-cubos\/","title":{"rendered":".Problem\u00e3o: Diferen\u00e7a de Cubos"},"content":{"rendered":"

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Problema
\n (Indicado a partir do 1\u00ba ano do E. M.)<\/span><\/p>\n

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Sabendo que [tex]x=2021^3-2019^3[\/tex], determine o valor de [tex]\\sqrt{\\dfrac{x-2}{6}}[\/tex].<\/p>\n

Adaptado de Instagram Mathceyhun.<\/strong><\/p>\n<\/div>\n

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 <\/p>\n

\n\"explicador_p\"<\/p>\n

Lembrete<\/p>\n

A fatora\u00e7\u00e3o not\u00e1vel “diferen\u00e7a de cubos” \u00e9 dada por [tex]a^3-b^3=(a-b)(a^2+ab+b^2).[\/tex]<\/p>\n<\/div>\n

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

\n

Vamos reescrever a express\u00e3o da seguinte forma:<\/p>\n

[tex]\\qquad x=2021^3-2019^3=(2021-2019) \\cdot (2021^2+2021 \\cdot 2019+2019^2)[\/tex]
\n[tex]\\qquad x=2 \\cdot ((2020+1)^2+(2020+1) \\cdot (2020-1)+(2020-1)^2)[\/tex].<\/p>\n

Em seguida, substitu\u00edmos [tex]2020[\/tex] por [tex]y[\/tex]. Logo, temos:
\n[tex]\\qquad x=2 \\cdot ((y+1)^2+(y+1) \\cdot (y-1)+(y-1)^2)[\/tex]
\n[tex]\\qquad x=2 \\cdot (y^2+2y+1+y^2-1+y^2-2y+1)[\/tex]
\n[tex]\\qquad x=2 \\cdot (3y^2+1)[\/tex].<\/p>\n

Agora, vamos substituir esta express\u00e3o e o valor de [tex]y[\/tex] na express\u00e3o buscada, obtendo:
\n[tex]\\qquad \\sqrt{\\dfrac{x-2}{6}}=\\sqrt{\\dfrac{2\\cdot(3y^2+1)-2}{6}}=\\sqrt{\\dfrac{6y^2+2-2}{6}}=\\sqrt{\\dfrac{6y^2}{6}}=\\sqrt{y^2}=\\sqrt{2020^2}=2020[\/tex].<\/p>\n

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

Problema (Indicado a partir do 1\u00ba ano do E. M.) Sabendo que [tex]x=2021^3-2019^3[\/tex], determine o valor de [tex]\\sqrt{\\dfrac{x-2}{6}}[\/tex]. Adaptado de Instagram Mathceyhun.   Lembrete A fatora\u00e7\u00e3o not\u00e1vel “diferen\u00e7a de cubos” \u00e9 dada por [tex]a^3-b^3=(a-b)(a^2+ab+b^2).[\/tex] Solu\u00e7\u00e3o Vamos reescrever a express\u00e3o da seguinte forma: [tex]\\qquad x=2021^3-2019^3=(2021-2019) \\cdot (2021^2+2021 \\cdot 2019+2019^2)[\/tex] [tex]\\qquad x=2 \\cdot ((2020+1)^2+(2020+1) \\cdot (2020-1)+(2020-1)^2)[\/tex]. Em … <\/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-220375","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/220375","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=220375"}],"version-history":[{"count":4,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/220375\/revisions"}],"predecessor-version":[{"id":221789,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/220375\/revisions\/221789"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=220375"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}