{"id":11755,"date":"2014-02-05T01:17:16","date_gmt":"2014-02-05T01:17:16","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=11755"},"modified":"2021-12-26T01:35:16","modified_gmt":"2021-12-26T01:35:16","slug":"resolva-em-mathbbr-as-seguintes-inequacoes-2","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=11755","title":{"rendered":"Resolva, em $\\mathbb{R}$, as seguintes inequa\u00e7\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_11755' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_11755' class='GTTabs_curr'><a  id=\"11755_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_11755' ><a  id=\"11755_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_11755'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolva, em $\\mathbb{R}$, as seguintes inequa\u00e7\u00f5es:<\/p>\n<ol>\n<li>$\\frac{{x + 1}}{{x &#8211; 2}} &gt; 0$<\/li>\n<li>$\\frac{{ &#8211; 5}}{{1 &#8211; 2x}} &lt; 0$<\/li>\n<li>$\\frac{2}{{{x^2} + 2x}} &#8211; \\frac{{x + 1}}{{x + 2}} &lt; 0$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_11755' onClick='GTTabs_show(1,11755)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_11755'>\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{\\frac{{x + 1}}{{x &#8211; 2}} &gt; 0}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{x + 1 &lt; 0} \\\\<br \/>\n{x &#8211; 2 &lt; 0}<br \/>\n\\end{array}} \\right.}&amp; \\vee &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{x + 1 &gt; 0} \\\\<br \/>\n{x &#8211; 2 &gt; 0}<br \/>\n\\end{array}} \\right.}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{x &lt;\u00a0 &#8211; 1} \\\\<br \/>\n{x &lt; 2}<br \/>\n\\end{array}} \\right.}&amp; \\vee &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{x &gt;\u00a0 &#8211; 1} \\\\<br \/>\n{x &gt; 2}<br \/>\n\\end{array}} \\right.}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x &lt;\u00a0 &#8211; 1}&amp; \\vee &amp;{x &gt; 2}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x \\in \\left] { &#8211; \\infty , &#8211; 1} \\right[ \\cup \\left] {2, + \\infty } \\right[}<br \/>\n\\end{array}\\]<\/li>\n<li>Tem-se sucessivamente:<br \/>\n\\[\\begin{array}{*{20}{l}}<br \/>\n{\\frac{{ &#8211; 5}}{{1 &#8211; 2x}} &lt; 0}&amp; \\Leftrightarrow &amp;{1 &#8211; 2x &gt; 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x &lt; \\frac{1}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x \\in \\left] { &#8211; \\infty ,\\frac{1}{2}} \\right[}<br \/>\n\\end{array}\\]<\/li>\n<li>Tem-se sucessivamente:<br \/>\n\\[\\begin{array}{*{20}{l}}<br \/>\n{\\frac{2}{{{x^2} + 2x}} &#8211; \\frac{{x + 1}}{{\\mathop {x + 2}\\limits_{\\left( x \\right)} }} &lt; 0}&amp; \\Leftrightarrow &amp;{\\frac{{2 &#8211; {x^2} &#8211; x}}{{x\\left( {x + 2} \\right)}} &lt; 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\frac{{{x^2} + x &#8211; 2}}{{x\\left( {x + 2} \\right)}} &gt; 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\frac{{\\left( {x &#8211; 1} \\right)\\left( {x + 2} \\right)}}{{x\\left( {x + 2} \\right)}} &gt; 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{\\frac{{\\left( {x &#8211; 1} \\right)}}{x} &gt; 0}&amp; \\wedge &amp;{x \\ne\u00a0 &#8211; 2}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{\\left( {\\begin{array}{*{20}{c}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{x &#8211; 1 &lt; 0} \\\\<br \/>\n{x &lt; 0}<br \/>\n\\end{array}} \\right.}&amp; \\vee &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{x &#8211; 1 &gt; 0} \\\\<br \/>\n{x &gt; 0}<br \/>\n\\end{array}} \\right.}<br \/>\n\\end{array}} \\right)}&amp; \\wedge &amp;{x \\ne\u00a0 &#8211; 2}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{\\left( {\\begin{array}{*{20}{c}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{x &lt; 1} \\\\<br \/>\n{x &lt; 0}<br \/>\n\\end{array}} \\right.}&amp; \\vee &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{x &gt; 1} \\\\<br \/>\n{x &gt; 0}<br \/>\n\\end{array}} \\right.}<br \/>\n\\end{array}} \\right)}&amp; \\wedge &amp;{x \\ne\u00a0 &#8211; 2}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{\\left( {\\begin{array}{*{20}{c}}<br \/>\n{x &lt; 0}&amp; \\vee &amp;{x &gt; 1}<br \/>\n\\end{array}} \\right)}&amp; \\wedge &amp;{x \\ne\u00a0 &#8211; 2}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x \\in \\left] { &#8211; \\infty , &#8211; 2} \\right[ \\cup \\left] { &#8211; 2,0} \\right[ \\cup \\left] {1, + \\infty } \\right[}<br \/>\n\\end{array}\\]<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_11755' onClick='GTTabs_show(0,11755)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Resolva, em $\\mathbb{R}$, as seguintes inequa\u00e7\u00f5es: $\\frac{{x + 1}}{{x &#8211; 2}} &gt; 0$ $\\frac{{ &#8211; 5}}{{1 &#8211; 2x}} &lt; 0$ $\\frac{2}{{{x^2} + 2x}} &#8211; \\frac{{x + 1}}{{x + 2}} &lt; 0$&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19178,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,125],"tags":[422,126,270],"series":[],"class_list":["post-11755","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-funcoes-racionais","tag-11-o-ano","tag-funcao-racional","tag-inequacao"],"views":2681,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat69.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11755","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=11755"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11755\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=11755"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=11755"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=11755"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=11755"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}