{"id":11722,"date":"2014-01-30T15:34:17","date_gmt":"2014-01-30T15:34:17","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=11722"},"modified":"2021-12-26T02:59:34","modified_gmt":"2021-12-26T02:59:34","slug":"resolve-em-mathbbr-as-equacoes","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=11722","title":{"rendered":"Resolva, em $\\mathbb{R}$, as equa\u00e7\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_11722' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_11722' class='GTTabs_curr'><a  id=\"11722_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_11722' ><a  id=\"11722_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_11722'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolva, em $\\mathbb{R}$, as equa\u00e7\u00f5es:<\/p>\n<ol>\n<li>$a &#8211; \\frac{5}{a} = 4$<\/li>\n<li>$\\frac{9}{{x + 5}} = \\frac{3}{{x &#8211; 3}}$<\/li>\n<li>$\\frac{{x + 4}}{x} + \\frac{3}{{x + 3}} =\u00a0 &#8211; \\frac{{16}}{{{x^2} &#8211; 4x}}$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_11722' onClick='GTTabs_show(1,11722)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_11722'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Tem-se sucessivamente:<br \/>\n\\[\\begin{array}{*{20}{l}}<br \/>\n{\\mathop a\\limits_{\\left( a \\right)}\u00a0 &#8211; \\frac{5}{a} = \\mathop 4\\limits_{\\left( a \\right)} }&amp; \\Leftrightarrow &amp;{\\frac{{{a^2} &#8211; 5 &#8211; 4a}}{a} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{{a^2} &#8211; 4a &#8211; 5 = 0}&amp; \\wedge &amp;{a \\ne 0}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{a = \\frac{{4 \\pm \\sqrt {16 + 20} }}{2}}&amp; \\wedge &amp;{a \\ne 0}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{\\left( {\\begin{array}{*{20}{c}}<br \/>\n{a =\u00a0 &#8211; 1}&amp; \\vee &amp;{a = 5}<br \/>\n\\end{array}} \\right)}&amp; \\wedge &amp;{a \\ne 0}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{a =\u00a0 &#8211; 1}&amp; \\vee &amp;{a = 5}<br \/>\n\\end{array}}<br \/>\n\\end{array}\\]<\/li>\n<li>Tem-se sucessivamente:<br \/>\n\\[\\begin{array}{*{20}{l}}<br \/>\n{\\frac{9}{{\\mathop {x + 5}\\limits_{\\left( {x &#8211; 3} \\right)} }} = \\frac{3}{{\\mathop {x &#8211; 3}\\limits_{\\left( {x + 5} \\right)} }}}&amp; \\Leftrightarrow &amp;{\\frac{{9x &#8211; 27 &#8211; 3x &#8211; 15}}{{\\left( {x + 5} \\right)\\left( {x &#8211; 3} \\right)}} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{6x &#8211; 42 = 0}&amp; \\wedge &amp;{\\left( {x + 5} \\right)\\left( {x &#8211; 3} \\right) \\ne 0}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{x = 7}&amp; \\wedge &amp;{\\left( {\\begin{array}{*{20}{c}}<br \/>\n{x \\ne\u00a0 &#8211; 5}&amp; \\wedge &amp;{x \\ne 3}<br \/>\n\\end{array}} \\right)}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = 7}<br \/>\n\\end{array}\\]<\/li>\n<li>Tem-se sucessivamente:<br \/>\n\\[\\begin{array}{*{20}{l}}<br \/>\n{\\frac{{x + 4}}{{\\mathop x\\limits_{\\left( {\\left( {x + 3} \\right)\\left( {x &#8211; 4} \\right)} \\right)} }} + \\frac{3}{{\\mathop {x + 3}\\limits_{\\left( {x\\left( {x &#8211; 4} \\right)} \\right)} }} =\u00a0 &#8211; \\frac{{16}}{{\\mathop {{x^2} &#8211; 4x}\\limits_{\\left( {x + 3} \\right)} }}}&amp; \\Leftrightarrow &amp;{\\frac{{\\left( {x + 3} \\right)\\left( {x &#8211; 4} \\right)\\left( {x + 4} \\right) + 3x\\left( {x &#8211; 4} \\right) + 16\\left( {x + 3} \\right)}}{{x\\left( {x + 3} \\right)\\left( {x &#8211; 4} \\right)}} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\frac{{\\left( {x + 3} \\right)\\left( {{x^2} &#8211; 16} \\right) + 3x\\left( {x &#8211; 4} \\right) + 16\\left( {x + 3} \\right)}}{{x\\left( {x + 3} \\right)\\left( {x &#8211; 4} \\right)}} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\frac{{\\left( {x + 3} \\right){x^2} &#8211; 16\\left( {x + 3} \\right) + 3x\\left( {x &#8211; 4} \\right) + 16\\left( {x + 3} \\right)}}{{x\\left( {x + 3} \\right)\\left( {x &#8211; 4} \\right)}} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\frac{{\\left( {x + 3} \\right){x^2} + 3x\\left( {x &#8211; 4} \\right)}}{{x\\left( {x + 3} \\right)\\left( {x &#8211; 4} \\right)}} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\frac{{x\\left[ {x\\left( {x + 3} \\right) + 3\\left( {x &#8211; 4} \\right)} \\right]}}{{x\\left( {x + 3} \\right)\\left( {x &#8211; 4} \\right)}} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\frac{{x\\left( {{x^2} + 6x &#8211; 12} \\right)}}{{x\\left( {x + 3} \\right)\\left( {x &#8211; 4} \\right)}} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{x\\left( {{x^2} + 6x &#8211; 12} \\right) = 0}&amp; \\wedge &amp;{x\\left( {x + 3} \\right)\\left( {x &#8211; 4} \\right) \\ne 0}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{\\left( {\\begin{array}{*{20}{c}}<br \/>\n{x = 0}&amp; \\vee &amp;{{x^2} + 6x &#8211; 12 = 0}<br \/>\n\\end{array}} \\right)}&amp; \\wedge &amp;{\\left( {\\begin{array}{*{20}{l}}<br \/>\n{x \\ne 0}&amp; \\wedge &amp;{x \\ne\u00a0 &#8211; 3}&amp; \\wedge &amp;{x \\ne 4}<br \/>\n\\end{array}} \\right)}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{\\left( {\\begin{array}{*{20}{c}}<br \/>\n{x = 0}&amp; \\vee &amp;{x = \\frac{{ &#8211; 6 \\pm \\sqrt {36 + 48} }}{2}}<br \/>\n\\end{array}} \\right)}&amp; \\wedge &amp;{\\left( {\\begin{array}{*{20}{l}}<br \/>\n{x \\ne 0}&amp; \\wedge &amp;{x \\ne\u00a0 &#8211; 3}&amp; \\wedge &amp;{x \\ne 4}<br \/>\n\\end{array}} \\right)}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{\\left( {\\begin{array}{*{20}{l}}<br \/>\n{x = 0}&amp; \\vee &amp;{x =\u00a0 &#8211; 3 &#8211; \\sqrt {21} }&amp; \\vee &amp;{x = }<br \/>\n\\end{array} &#8211; 3 + \\sqrt {21} } \\right)}&amp; \\wedge &amp;{\\left( {\\begin{array}{*{20}{l}}<br \/>\n{x \\ne 0}&amp; \\wedge &amp;{x \\ne\u00a0 &#8211; 3}&amp; \\wedge &amp;{x \\ne 4}<br \/>\n\\end{array}} \\right)}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x =\u00a0 &#8211; 3 &#8211; \\sqrt {21} }&amp; \\vee &amp;{x =\u00a0 &#8211; 3 + \\sqrt {21} }<br \/>\n\\end{array}}<br \/>\n\\end{array}\\]<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_11722' onClick='GTTabs_show(0,11722)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Resolva, em $\\mathbb{R}$, as equa\u00e7\u00f5es: $a &#8211; \\frac{5}{a} = 4$ $\\frac{9}{{x + 5}} = \\frac{3}{{x &#8211; 3}}$ $\\frac{{x + 4}}{x} + \\frac{3}{{x + 3}} =\u00a0 &#8211; \\frac{{16}}{{{x^2} &#8211; 4x}}$ Resolu\u00e7\u00e3o &gt;&gt;&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14113,"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-11722","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":2199,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat55.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11722","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=11722"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11722\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14113"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=11722"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=11722"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=11722"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=11722"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}