{"id":7269,"date":"2011-12-10T17:15:46","date_gmt":"2011-12-10T17:15:46","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7269"},"modified":"2022-01-08T16:41:47","modified_gmt":"2022-01-08T16:41:47","slug":"calcula-o-valor-das-expressoes","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7269","title":{"rendered":"Calcula o valor das express\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_7269' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7269' class='GTTabs_curr'><a  id=\"7269_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7269' ><a  id=\"7269_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_7269'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Calcula o valor das express\u00f5es:<\/p>\n<ol>\n<li>$2\\sqrt{3}+4\\sqrt{3}-5\\sqrt{3}$<\/li>\n<li>${{\\left( \\sqrt{2}+2 \\right)}^{2}}$<\/li>\n<li>$\\frac{1}{3}\\pi -\\pi +3\\pi $<\/li>\n<li>$(5-\\sqrt{5})(5+\\sqrt{5})$<\/li>\n<li>${{\\left( \\sqrt{7}-1 \\right)}^{2}}$<\/li>\n<li>$(2+\\sqrt{3})(7-\\sqrt{3})$<\/li>\n<li>$\\sqrt{5}-\\sqrt{6}+2\\sqrt{5}-2\\sqrt{6}$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7269' onClick='GTTabs_show(1,7269)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7269'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p><strong>Quadrado de uma soma:<\/strong> $${{(A+B)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$$<\/p>\n<p><strong>Diferen\u00e7a de dois quadrados:<\/strong> $${{A}^{2}}-{{B}^{2}}=(A+B)(A-B)$$<\/p>\n<\/blockquote>\n<p>\u00ad<\/p>\n<ol>\n<li>Ora,<br \/>\n$$\\begin{array}{*{35}{l}}<br \/>\n2\\sqrt{3}+4\\sqrt{3}-5\\sqrt{3} &amp; = &amp; (2+4-5)\\sqrt{3}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\sqrt{3}\u00a0 \\\\<br \/>\n\\end{array}$$<br \/>\n\u00ad<\/li>\n<li>Aplicando o caso not\u00e1vel:\u00a0$$\\begin{array}{*{35}{l}}<br \/>\n{{\\left( \\sqrt{2}+2 \\right)}^{2}} &amp; = &amp; {{\\left( \\sqrt{2} \\right)}^{2}}+2\\times \\sqrt{2}\\times 2+{{2}^{2}}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 2+4\\sqrt{2}+4\u00a0 \\\\<br \/>\n{} &amp; = &amp; 6+4\\sqrt{2}\u00a0 \\\\<br \/>\n\\end{array}$$<br \/>\nN\u00e3o aplicando o caso not\u00e1vel: $$\\begin{array}{*{35}{l}}<br \/>\n{{\\left( \\sqrt{2}+2 \\right)}^{2}} &amp; = &amp; (\\sqrt{2}+2)(\\sqrt{2}+2)\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\sqrt{2}\\times \\sqrt{2}+\\sqrt{2}\\times 2+2\\times \\sqrt{2}+2\\times 2\u00a0 \\\\<br \/>\n{} &amp; = &amp; 2+2\\sqrt{2}+2\\sqrt{2}+4\u00a0 \\\\<br \/>\n{} &amp; = &amp; 6+4\\sqrt{2}\u00a0 \\\\<br \/>\n\\end{array}$$<br \/>\nComo se pode verificar, n\u00e3o h\u00e1 vantagem em n\u00e3o aplicar o caso not\u00e1vel.<br \/>\n\u00ad<\/li>\n<li>Ora,<br \/>\n$$\\begin{array}{*{35}{l}}<br \/>\n\\frac{1}{3}\\pi -\\pi +3\\pi\u00a0 &amp; = &amp; \\frac{\\pi }{3}-\\frac{3\\pi }{3}+\\frac{9\\pi }{3}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{7\\pi }{3}\u00a0 \\\\<br \/>\n\\end{array}$$<br \/>\n\u00ad<\/li>\n<li>Aplicando o caso not\u00e1vel: $$\\begin{array}{*{35}{l}}<br \/>\n(5-\\sqrt{5})(5+\\sqrt{5}) &amp; = &amp; {{5}^{2}}-{{\\left( \\sqrt{5} \\right)}^{2}}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 25-5\u00a0 \\\\<br \/>\n{} &amp; = &amp; 20\u00a0 \\\\<br \/>\n\\end{array}$$<br \/>\nN\u00e3o aplicando o caso not\u00e1vel: $$\\begin{array}{*{35}{l}}<br \/>\n(5-\\sqrt{5})(5+\\sqrt{5}) &amp; = &amp; 5\\times 5+5\\times \\sqrt{5}-\\sqrt{5}\\times 5-\\sqrt{5}\\times \\sqrt{5}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 25+5\\sqrt{5}-5\\sqrt{5}-5\u00a0 \\\\<br \/>\n{} &amp; = &amp; 20\u00a0 \\\\<br \/>\n\\end{array}$$<br \/>\nComo se pode verificar, n\u00e3o h\u00e1 vantagem em n\u00e3o aplicar o caso not\u00e1vel.<br \/>\n\u00ad<\/li>\n<li>Aplicando o caso not\u00e1vel: $$\\begin{array}{*{35}{l}}<br \/>\n{{\\left( \\sqrt{7}-1 \\right)}^{2}} &amp; = &amp; {{\\left( \\sqrt{7} \\right)}^{2}}+2\\times \\sqrt{7}\\times (-1)+{{(-1)}^{2}}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 7-2\\sqrt{7}+1\u00a0 \\\\<br \/>\n{} &amp; = &amp; 8-2\\sqrt{7}\u00a0 \\\\<br \/>\n\\end{array}$$<br \/>\nN\u00e3o aplicando o caso not\u00e1vel: $$\\begin{array}{*{35}{l}}<br \/>\n{{\\left( \\sqrt{7}-1 \\right)}^{2}} &amp; = &amp; \\left( \\sqrt{7}-1 \\right)\\left( \\sqrt{7}-1 \\right)\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\sqrt{7}\\times \\sqrt{7}+\\sqrt{7}\\times (-1)-1\\times \\sqrt{7}-1\\times (-1)\u00a0 \\\\<br \/>\n{} &amp; = &amp; 7-\\sqrt{7}-\\sqrt{7}+1\u00a0 \\\\<br \/>\n{} &amp; = &amp; 8-2\\sqrt{7}\u00a0 \\\\<br \/>\n\\end{array}$$<br \/>\nComo se pode verificar, n\u00e3o h\u00e1 vantagem em n\u00e3o aplicar o caso not\u00e1vel.<br \/>\n\u00ad<\/li>\n<li>Nota que a situa\u00e7\u00e3o n\u00e3o configura um caso not\u00e1vel.<br \/>\n$$\\begin{array}{*{35}{l}}<br \/>\n(2+\\sqrt{3})(7-\\sqrt{3}) &amp; = &amp; 2\\times 7+2\\times (-\\sqrt{3})+\\sqrt{3}\\times 7+\\sqrt{3}\\times (-\\sqrt{3})\u00a0 \\\\<br \/>\n{} &amp; = &amp; 14-2\\sqrt{3}+7\\sqrt{3}-3\u00a0 \\\\<br \/>\n{} &amp; = &amp; 11+5\\sqrt{3}\u00a0 \\\\<br \/>\n\\end{array}$$<br \/>\n\u00ad<\/li>\n<li>Ora,<br \/>\n$$\\begin{array}{*{35}{l}}<br \/>\n\\sqrt{5}-\\sqrt{6}+2\\sqrt{5}-2\\sqrt{6} &amp; = &amp; (1+2)\\sqrt{5}+(-1-2)\\sqrt{6}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 3\\sqrt{5}-3\\sqrt{6}\u00a0 \\\\<br \/>\n\\end{array}$$<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_7269' onClick='GTTabs_show(0,7269)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Calcula o valor das express\u00f5es: $2\\sqrt{3}+4\\sqrt{3}-5\\sqrt{3}$ ${{\\left( \\sqrt{2}+2 \\right)}^{2}}$ $\\frac{1}{3}\\pi -\\pi +3\\pi $ $(5-\\sqrt{5})(5+\\sqrt{5})$ ${{\\left( \\sqrt{7}-1 \\right)}^{2}}$ $(2+\\sqrt{3})(7-\\sqrt{3})$ $\\sqrt{5}-\\sqrt{6}+2\\sqrt{5}-2\\sqrt{6}$ Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":14095,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,258],"tags":[426,259],"series":[],"class_list":["post-7269","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9--ano","category-aplicando","category-os-numeros-reais","tag-9-o-ano","tag-numeros-irracionais"],"views":2050,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat40.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7269","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=7269"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7269\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14095"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7269"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7269"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7269"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7269"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}