{"id":99062,"date":"2018-03-02T21:20:28","date_gmt":"2018-03-03T00:20:28","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=99062"},"modified":"2023-04-15T20:28:45","modified_gmt":"2023-04-15T23:28:45","slug":"problema-verdadeiro-ou-falso-3","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-verdadeiro-ou-falso-3\/","title":{"rendered":".Problema: Verdadeiro ou Falso (3)"},"content":{"rendered":"

\n

\n

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

\nDetermine, em cada item, se as afirmativas s\u00e3o verdadeiras ou falsas. Justifique sua resposta.<\/p>\n

[tex]\\textcolor{#800000}{(a)} \\, \\boxed{2^{2017}-2^{2016}=2^{2016}}[\/tex]
\n\t [tex]\\textcolor{#800000}{(b)} \\, \\boxed{ \\dfrac{3}{7}+\\dfrac{7}{3}=1}[\/tex]
\n\t [tex]\\textcolor{#800000}{(c)} \\, \\boxed{ \\sqrt{16\\%}=4\\%}[\/tex]
\n\t [tex]\\textcolor{#800000}{(d)} \\, \\boxed{ \\sqrt{81}=\\pm 9}[\/tex]
\n\t [tex]\\textcolor{#800000}{(e)} \\, \\boxed{ \\dfrac{2}{\\sqrt[4]{3}}=\\dfrac{2\\sqrt[4]{3}}{3}}[\/tex]<\/p>\n

\n<\/div>\n

\n

\n

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

\n [tex]\\textcolor{#800000}{(a)}[\/tex] Verdadeiro<\/strong>.
\nColocando [tex]2^{2016}[\/tex] em evid\u00eancia, obtemos:
\n [tex]\\qquad 2^{2017}-2^{2016}=2^{2016}(2^1-1)=2^{2016}.[\/tex]<\/p>\n

\n

[tex]\\textcolor{#800000}{(b)}[\/tex] Falso<\/strong>.
\nPara somar fra\u00e7\u00f5es, devemos reduzi-las a um denominador comum. Geralmente, escolhemos o [tex]mmc[\/tex] dos denominadores iniciais, no caso, [tex]mmc(3,7)=21[\/tex].
\nAssim:
\n [tex]\\qquad \\boxed{\\dfrac{3}{7}+\\dfrac{7}{3}}=\\dfrac{9}{21}+\\dfrac{49}{21}=\\dfrac{9+49}{21}=\\boxed{\\dfrac{58}{21}}.[\/tex]<\/p>\n

\n

[tex]\\textcolor{#800000}{(c)}[\/tex] Falso<\/strong>.
\nObserve que:
\n [tex]\\qquad \\boxed{\\sqrt{16\\%}}=\\sqrt{\\dfrac{16}{100}}=\\dfrac{4}{10}=\\boxed{40\\%}[\/tex].<\/p>\n

\n

[tex]\\textcolor{#800000}{(d)}[\/tex] Falso<\/strong>.
\nNo conjunto dos n\u00fameros reais, se [tex]a[\/tex] \u00e9 um real n\u00e3o negativo, o s\u00edmbolo [tex]\\sqrt{a}[\/tex] indica o n\u00famero n\u00e3o negativo cujo quadrado \u00e9 [tex]a[\/tex].
\nDessa forma,
\n[tex]\\qquad \\boxed{\\sqrt{81}=9} \\, [\/tex],
\npois [tex]9[\/tex] \u00e9 n\u00e3o negativo (na verdade ele \u00e9 positivo)<\/span> e [tex]9^2=81.[\/tex]<\/p>\n

\n

[tex]\\textcolor{#800000}{(e)}[\/tex] Falso<\/strong>.
\nObserve que:
\n[tex]\\qquad \\begin{align*}
\n\\boxed{\\dfrac{2}{\\sqrt[4]{3}}}&=\\dfrac{2}{\\sqrt[4]{3}}\\times 1\\\\
\n&=\\dfrac{2}{\\sqrt[4]{3}}\\times\\dfrac{\\sqrt[4]{3^3}}{\\sqrt[4]{3^3}}\\\\
\n&=\\dfrac{2}{3^{\\frac{1}{4}}}\\times\\dfrac{\\sqrt[4]{3^3}}{3^{\\frac{3}{4}}}\\\\
\n&=\\dfrac{2 \\times \\sqrt[4]{3^3}}{3^{\\frac{1}{4}} \\times 3^{\\frac{3}{4}}}\\\\
\n&=\\dfrac{2\\sqrt[4]{3}}{3^{\\left(\\frac{1}{4}+\\frac{3}{4}\\right)}}\\\\
\n&=\\dfrac{2\\sqrt[4]{3^3}}{3^1}\\\\
\n&=\\boxed{\\dfrac{2\\sqrt[4]{3^3}}{3}}
\n\\end{align*}[\/tex] <\/p>\n

\n

e, assim, [tex]\\boxed{\\dfrac{2}{\\sqrt[4]{3}}=\\dfrac{2\\sqrt[4]{3^3}}{3}}.[\/tex]
\nComo [tex]\\sqrt[4]{3^3}\\neq \\sqrt[4]{3} \\, [\/tex], ent\u00e3o [tex]\\dfrac{2\\sqrt[4]{3^3}}{3}\\neq \\dfrac{2\\sqrt[4]{3}}{3} \\, [\/tex] e, portanto, [tex]\\dfrac{2}{\\sqrt[4]{3}}\\neq\\dfrac{2\\sqrt[4]{3}}{3}.[\/tex]<\/p>\n

\n

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

\n\n\n
\"\"<\/td>\n\n

Para aprender mais, clique AQUI<\/strong><\/a>.<\/span><\/p>\n

<\/td>\n<\/tr>\n<\/table>\n

\n

\nParticipou da discuss\u00e3o o Clube: CEMIR<\/strong>.<\/div>\n","protected":false},"excerpt":{"rendered":"

Problema (Indicado a partir do 9\u00ba ano do E. F.) Determine, em cada item, se as afirmativas s\u00e3o verdadeiras ou falsas. Justifique sua resposta. [tex]\\textcolor{#800000}{(a)} \\, \\boxed{2^{2017}-2^{2016}=2^{2016}}[\/tex] [tex]\\textcolor{#800000}{(b)} \\, \\boxed{ \\dfrac{3}{7}+\\dfrac{7}{3}=1}[\/tex] [tex]\\textcolor{#800000}{(c)} \\, \\boxed{ \\sqrt{16\\%}=4\\%}[\/tex] [tex]\\textcolor{#800000}{(d)} \\, \\boxed{ \\sqrt{81}=\\pm 9}[\/tex] [tex]\\textcolor{#800000}{(e)} \\, \\boxed{ \\dfrac{2}{\\sqrt[4]{3}}=\\dfrac{2\\sqrt[4]{3}}{3}}[\/tex] Solu\u00e7\u00e3o [tex]\\textcolor{#800000}{(a)}[\/tex] Verdadeiro. Colocando [tex]2^{2016}[\/tex] em evid\u00eancia, obtemos: [tex]\\qquad 2^{2017}-2^{2016}=2^{2016}(2^1-1)=2^{2016}.[\/tex] [tex]\\textcolor{#800000}{(b)}[\/tex] … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":620,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-99062","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/99062","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\/620"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=99062"}],"version-history":[{"count":10,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/99062\/revisions"}],"predecessor-version":[{"id":187238,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/99062\/revisions\/187238"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=99062"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}