{"id":10135,"date":"2012-10-05T23:40:36","date_gmt":"2012-10-05T22:40:36","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=10135"},"modified":"2021-12-26T19:33:14","modified_gmt":"2021-12-26T19:33:14","slug":"resolva-as-equacoes","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=10135","title":{"rendered":"Resolva as equa\u00e7\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_10135' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_10135' class='GTTabs_curr'><a  id=\"10135_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_10135' ><a  id=\"10135_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_10135'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolva, em $\\mathbb{R}$, as seguintes equa\u00e7\u00f5es:<\/p>\n<ol>\n<li>$$\\frac{{2\\left( {x + 1} \\right)}}{3} + 5\\left( {x + 2} \\right) = 8 &#8211; 3x$$<\/li>\n<li>$$3\\left( {\\frac{{x + 1}}{2} + \\frac{{x &#8211; 1}}{3}} \\right) = 5x &#8211; 2$$<\/li>\n<li>$$5 &#8211; \\frac{{2\\left( {x + 1} \\right)}}{4} = \\frac{{3x &#8211; 1}}{7}$$<\/li>\n<li>$$\\frac{{x + 4}}{6} &#8211; \\frac{{2\\left( {x + 1} \\right)}}{9} = \\frac{{x &#8211; 2}}{6} + \\frac{{11 &#8211; 2x}}{{18}}$$<\/li>\n<li>$$\\left( {3x &#8211; \\frac{2}{3}} \\right)\\left( {3x + \\frac{2}{3}} \\right) &#8211; 4 = {\\left( {3x &#8211; 5} \\right)^2} + \\frac{5}{9}$$<\/li>\n<li>$$5{\\left( {x &#8211; 2} \\right)^2} &#8211; 500 = 0$$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_10135' onClick='GTTabs_show(1,10135)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_10135'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\frac{{2\\left( {x + 1} \\right)}}{3}}\\limits_{(1)}\u00a0 + \\mathop {5\\left( {x + 2} \\right)}\\limits_{(3)}\u00a0 = \\mathop 8\\limits_{(3)}\u00a0 &#8211; \\mathop {3x}\\limits_{(3)} }&amp; \\Leftrightarrow &amp;{2\\left( {x + 1} \\right) + 15\\left( {x + 2} \\right) = 24 &#8211; 9x} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{2x + 2 + 15x + 30 = 24 &#8211; 9x} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{26x =\u00a0 &#8211; 8} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{x =\u00a0 &#8211; \\frac{4}{{13}}} \\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}} \u00a0 {3\\left( {\\frac{{x + 1}}{2} + \\frac{{x &#8211; 1}}{3}} \\right) = 5x &#8211; 2}&amp; \\Leftrightarrow &amp;{\\frac{{3x + 3}}{2} + x &#8211; 1 = 5x &#8211; 2} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{3x + 3 + 2x &#8211; 2 = 10x &#8211; 4} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{ &#8211; 5x =\u00a0 &#8211; 5} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{x = 1} \\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop 5\\limits_{(28)}\u00a0 &#8211; \\mathop {\\frac{{2\\left( {x + 1} \\right)}}{4}}\\limits_{(7)}\u00a0 = \\mathop {\\frac{{3x &#8211; 1}}{7}}\\limits_{(4)} }&amp; \\Leftrightarrow &amp;{140 &#8211; 14\\left( {x + 1} \\right) = 12x &#8211; 4} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{140 &#8211; 14x &#8211; 14 = 12x &#8211; 4} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{ &#8211; 26x =\u00a0 &#8211; 130} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{x = 5} \\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\frac{{x + 4}}{6}}\\limits_{(3)}\u00a0 &#8211; \\mathop {\\frac{{2\\left( {x + 1} \\right)}}{9}}\\limits_{(2)}\u00a0 = \\mathop {\\frac{{x &#8211; 2}}{6}}\\limits_{(3)}\u00a0 + \\mathop {\\frac{{11 &#8211; 2x}}{{18}}}\\limits_{(1)} }&amp; \\Leftrightarrow &amp;{3\\left( {x + 4} \\right) &#8211; 4\\left( {x + 1} \\right) = 3\\left( {x &#8211; 2} \\right) + 11 &#8211; 2x} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{3x + 12 &#8211; 4x &#8211; 4 = 3x &#8211; 6 + 11 &#8211; 2x} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{ &#8211; 2x =\u00a0 &#8211; 3} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{x = \\frac{3}{2}} \\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}} \u00a0 {\\left( {3x &#8211; \\frac{2}{3}} \\right)\\left( {3x + \\frac{2}{3}} \\right) &#8211; 4 = {{\\left( {3x &#8211; 5} \\right)}^2} + \\frac{5}{9}}&amp; \\Leftrightarrow &amp;{{{\\left( {3x} \\right)}^2} &#8211; {{\\left( {\\frac{2}{3}} \\right)}^2} &#8211; 4 = 9{x^2} &#8211; 30x + 25 + \\frac{5}{9}} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{9{x^2} &#8211; \\frac{4}{9} &#8211; 4 = 9{x^2} &#8211; 30x + 25 + \\frac{5}{9}} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{30x = 29 + 1} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{x = 1} \\end{array}$$<\/p>\n<p><strong>RECORDE<\/strong>:<br \/>\n<strong>$$\\left( {A + B} \\right)\\left( {A &#8211; B} \\right) = {A^2} &#8211; B{}^2$$<\/strong><br \/>\n<strong>$${\\left( {A + B} \\right)^2} = {A^2} + 2AB + {B^2}$$<\/strong><\/p>\n<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}} \u00a0 {5{{\\left( {x &#8211; 2} \\right)}^2} &#8211; 500 = 0}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; 2} \\right)}^2} &#8211; 100 = 0} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 4x + 4 &#8211; 100 = 0} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\pm \\sqrt {16 &#8211; 4 \\times 1 \\times \\left( { &#8211; 96} \\right)} }}{2}} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\pm \\sqrt {400} }}{2}} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\pm 20}}{2}} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}} \u00a0 {x =\u00a0 &#8211; 8}&amp; \\vee &amp;{x = 12} \\end{array}} \\end{array}$$<\/p>\n<p><strong>ALTERNATIVA<\/strong>:<br \/>\n$$\\begin{array}{*{20}{l}} \u00a0 {5{{\\left( {x &#8211; 2} \\right)}^2} &#8211; 500 = 0}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; 2} \\right)}^2} = \\frac{{500}}{5}} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; 2} \\right)}^2} = 100} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}} \u00a0 {x &#8211; 2 =\u00a0 &#8211; 10}&amp; \\vee &amp;{x &#8211; 2 =\u00a0 + 10} \\end{array}} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}} \u00a0 {x =\u00a0 &#8211; 8}&amp; \\vee &amp;{x = 12} \\end{array}} \\end{array}$$<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_10135' onClick='GTTabs_show(0,10135)'>&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 equa\u00e7\u00f5es: $$\\frac{{2\\left( {x + 1} \\right)}}{3} + 5\\left( {x + 2} \\right) = 8 &#8211; 3x$$ $$3\\left( {\\frac{{x + 1}}{2} + \\frac{{x &#8211; 1}}{3}} \\right) =&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14061,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[321,97,322],"tags":[429,306,425,430],"series":[],"class_list":["post-10135","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-10-o-ano","category-aplicando","category-modulo-inicial","tag-10-o-ano","tag-equacao-do-2-o-grau","tag-equacoes","tag-modulo-inicial"],"views":2386,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat06.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/10135","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=10135"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/10135\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14061"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=10135"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=10135"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=10135"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=10135"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}