{"id":197211,"date":"2023-08-11T19:16:48","date_gmt":"2023-08-11T22:16:48","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=197211"},"modified":"2023-09-09T19:50:02","modified_gmt":"2023-09-09T22:50:02","slug":"problema-que-radical-s","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-que-radical-s\/","title":{"rendered":".Desafio: Que radical!"},"content":{"rendered":"

✏ Link do problema para dispositivos da Apple<\/em>.<\/span><\/a><\/strong><\/p>\n

\n

Problem\u00e3o
\n (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n

\n

Calcule [tex]\\sqrt{(1000000)\\cdot(1000001)\\cdot (1000002)\\cdot (1000003) +1\\,}\\,.[\/tex]<\/p>\n

Extra\u00eddo de Olimp\u00edada Americana<\/strong>.<\/p>\n<\/div>\n

\n

\n

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

\nFa\u00e7amos [tex]x=10^6[\/tex]. Assim, segue que:<\/p>\n

[tex]\\quad \\sqrt{(1000000)\\cdot(1000001)\\cdot (1000002)\\cdot (1000003)+1}= \\\\
\n\\quad \\sqrt{x\\cdot(x+1)\\cdot (x+2)\\cdot (x+3)+1}=\\sqrt{(x^2+3x+2)\\cdot (x^2+3x)+1}=\\\\
\n\\quad \\sqrt{x^4+3x^3+2x^2+3x^3+9x^2+6x+1}=\\sqrt{x^4+6x^3+11x^2+6x+1}=\\\\
\n\\quad \\sqrt{x^2\\cdot\\left[\\left(x^2+\\dfrac{1}{x^2}\\right)+6\\cdot \\left(x+\\dfrac{1}{x}\\right)+11 \\right]\\,}. \\qquad \\quad \\textcolor{#800000}{(i)}[\/tex]
\nAgora, observe que se fizermos [tex]\\boxed{k=x+\\frac{1}{x}}[\/tex], teremos que:
\n[tex]\\quad k^2=\\left(x+\\dfrac{1}{x}\\right)^2\\\\
\n\\quad k^2=x^2+2+\\dfrac{1}{x^2}\\\\
\n\\quad x^2+\\dfrac{1}{x^2}=k^2-2.[\/tex]
\nDessa forma, a express\u00e3o [tex]\\textcolor{#800000}{(i)}[\/tex] pode ser reescrita como:
\n[tex]\\quad \\sqrt{x^2\\cdot\\left[k^2-2+6k+11\\right]}=\\\\
\n\\quad \\sqrt{x^2\\cdot \\left[k^2+6k+9\\right]}=\\\\
\n\\quad \\sqrt{x^2\\cdot \\left[k+3\\right]^2}= |x\\cdot(k+3)|.\\qquad \\quad \\textcolor{#800000}{(ii)}[\/tex]
\nMas, perceba que [tex]x\\cdot(k+3) \\gt 0[\/tex]; assim, [tex]|x\\cdot(k+3)|=x\\cdot(k+3)[\/tex]. Dessa forma, segue de [tex]\\textcolor{#800000}{(ii)}[\/tex] que
\n[tex]\\quad \\begin{align}\\sqrt{x^2\\cdot\\left[k^2-2+6k+11\\right]}&=|x\\cdot(k+3)|\\\\
\n&=x\\cdot(k+3);\\end{align}[\/tex]
\ne, como [tex]\\boxed{k=x+\\frac{1}{x}}[\/tex], temos ainda que:
\n[tex]\\quad \\begin{align}\\sqrt{x^2\\cdot[k^2-2+6k+11]}&= x\\cdot(k+3)\\\\
\n&=10^6\\cdot \\left(10^6+\\frac{1}{10^6}+3\\right)\\\\
\n&=10^{12}+3\\cdot 10^6+1.\\end{align}[\/tex]
\nPortanto, [tex]\\fcolorbox{black}{#eee0e5}{$\\sqrt{(1000000)\\cdot(1000001)\\cdot (1000002)\\cdot (1000003)+1\\,}=10^{12}+3\\cdot 10^6+1^{{~}^{~}}$}\\,.[\/tex]<\/p>\n

\n

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

✏ Link do problema para dispositivos da Apple. Problem\u00e3o (Indicado a partir do 9\u00ba ano do E. F.) Calcule [tex]\\sqrt{(1000000)\\cdot(1000001)\\cdot (1000002)\\cdot (1000003) +1\\,}\\,.[\/tex] Extra\u00eddo de Olimp\u00edada Americana. Solu\u00e7\u00e3o Fa\u00e7amos [tex]x=10^6[\/tex]. Assim, segue que: [tex]\\quad \\sqrt{(1000000)\\cdot(1000001)\\cdot (1000002)\\cdot (1000003)+1}= \\\\ \\quad \\sqrt{x\\cdot(x+1)\\cdot (x+2)\\cdot (x+3)+1}=\\sqrt{(x^2+3x+2)\\cdot (x^2+3x)+1}=\\\\ \\quad \\sqrt{x^4+3x^3+2x^2+3x^3+9x^2+6x+1}=\\sqrt{x^4+6x^3+11x^2+6x+1}=\\\\ \\quad \\sqrt{x^2\\cdot\\left[\\left(x^2+\\dfrac{1}{x^2}\\right)+6\\cdot \\left(x+\\dfrac{1}{x}\\right)+11 \\right]\\,}. \\qquad \\quad \\textcolor{#800000}{(i)}[\/tex] Agora, observe que … <\/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-197211","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/197211","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=197211"}],"version-history":[{"count":16,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/197211\/revisions"}],"predecessor-version":[{"id":199982,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/197211\/revisions\/199982"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=197211"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}