{"id":11782,"date":"2014-02-08T23:15:33","date_gmt":"2014-02-08T23:15:33","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=11782"},"modified":"2021-12-26T01:06:44","modified_gmt":"2021-12-26T01:06:44","slug":"resolva-em-mathbbr-a-equacao","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=11782","title":{"rendered":"Resolva, em $\\mathbb{R}$, a equa\u00e7\u00e3o"},"content":{"rendered":"<p><ul id='GTTabs_ul_11782' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_11782' class='GTTabs_curr'><a  id=\"11782_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_11782' ><a  id=\"11782_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_11782'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolva, em $\\mathbb{R}$, a equa\u00e7\u00e3o seguinte: \\[{\\frac{{2x + 4}}{{x &#8211; 3}} = \\frac{{x &#8211; 2}}{{x + 5}}}\\]<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_11782' onClick='GTTabs_show(1,11782)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_11782'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p>\\[\\begin{array}{*{20}{l}}{\\frac{{2x + 4}}{{\\mathop {x{\\rm{ }} &#8211; {\\rm{ }}3}\\limits_{\\left( {x + 5} \\right)} }} = \\frac{{x &#8211; 2}}{{\\mathop {x{\\rm{ }} + {\\rm{ }}5}\\limits_{\\left( {x &#8211; 3} \\right)} }}}&amp; \\Leftrightarrow &amp;{\\frac{{2{x^2} + 10x + 4x + 20 &#8211; {x^2} + 3x + 2x &#8211; 6}}{{\\left( {x &#8211; 3} \\right)\\left( {5 + 5} \\right)}} = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\frac{{{x^2} + 19x + 14}}{{\\left( {x &#8211; 3} \\right)\\left( {5 + 5} \\right)}} = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}{{x^2} + 19x + 14 = 0}&amp; \\wedge &amp;{\\left( {x &#8211; 3} \\right)\\left( {5 + 5} \\right) \\ne 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}{x = \\frac{{ &#8211; 19 \\pm \\sqrt {361 &#8211; 56} }}{2}}&amp; \\wedge &amp;{\\begin{array}{*{20}{c}}{x \\ne 3}&amp; \\wedge &amp;{x \\ne \u00a0&#8211; 5}\\end{array}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}{x = \\frac{{ &#8211; 19 &#8211; \\sqrt {305} }}{2}}&amp; \\vee &amp;{x = \\frac{{ &#8211; 19 + \\sqrt {305} }}{2}}\\end{array}}\\end{array}\\]<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_11782' onClick='GTTabs_show(0,11782)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Resolva, em $\\mathbb{R}$, a equa\u00e7\u00e3o seguinte: \\[{\\frac{{2x + 4}}{{x &#8211; 3}} = \\frac{{x &#8211; 2}}{{x + 5}}}\\] Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":19177,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,125],"tags":[422,160,126],"series":[],"class_list":["post-11782","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-funcoes-racionais","tag-11-o-ano","tag-equacao","tag-funcao-racional"],"views":3468,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat68.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11782","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=11782"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11782\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19177"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=11782"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=11782"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=11782"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=11782"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}