{"id":6770,"date":"2011-04-13T22:11:38","date_gmt":"2011-04-13T21:11:38","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6770"},"modified":"2022-01-05T16:22:22","modified_gmt":"2022-01-05T16:22:22","slug":"equacoes-com-denominadores","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6770","title":{"rendered":"Equa\u00e7\u00f5es com denominadores"},"content":{"rendered":"<p><ul id='GTTabs_ul_6770' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6770' class='GTTabs_curr'><a  id=\"6770_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6770' ><a  id=\"6770_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_6770'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolve as equa\u00e7\u00f5es:<\/p>\n<ol>\n<li>$1+\\frac{x-3}{2}=1$<\/li>\n<li>$\\frac{x-2}{4}+\\frac{2x}{3}=1$<\/li>\n<li>$\\frac{y+1}{4}-\\frac{5+y}{2}=\\frac{3}{2}$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6770' onClick='GTTabs_show(1,6770)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6770'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>$1+\\frac{x-3}{2}=1$\n<p>Escrevendo uma equa\u00e7\u00e3o equivalente onde todas as fra\u00e7\u00f5es figurem com igual denominador (igual ao m.m.c. dos denominadores), temos:<br \/>\n\\[\\begin{matrix}<br \/>\n\\underset{(2)}{\\mathop{1}}\\,+\\frac{x-3}{\\underset{(1)}{\\mathop{2}}\\,}=\\underset{(2)}{\\mathop{1}}\\,\u00a0 \\\\<br \/>\n\\frac{2}{2}+\\frac{x-3}{2}=\\frac{2}{2}\u00a0 \\\\<br \/>\n\\end{matrix}\\]<br \/>\nSe multiplicarmos por 2 ambos os membros da equa\u00e7\u00e3o\u00a0obtemos uma equa\u00e7\u00e3o equivalente. Seguidamente, aplicando a propriedade distributiva da multiplica\u00e7\u00e3o em ambos os membros, obt\u00e9m-se uma equa\u00e7\u00e3o equivalente sem denominadores:<br \/>\n\\[\\begin{matrix}<br \/>\n2\\times \\left( \\frac{2}{2}+\\frac{x-3}{2} \\right)=\\left( \\frac{2}{2} \\right)\\times 2\u00a0 \\\\<br \/>\n2+x-3=2\u00a0 \\\\<br \/>\n\\end{matrix}\\]<br \/>\nEsta sequ\u00eancia de procedimentos pode ser abreviada: suprimindo a 2.\u00aa e 3.\u00aa equa\u00e7\u00f5es.<br \/>\nPara isso, bastar\u00e1 multiplicar apenas os numeradores das fra\u00e7\u00f5es da 1.\u00aa equa\u00e7\u00e3o pelos valores indicados dentro dos par\u00eantesis curvos:<br \/>\n\\[\\begin{matrix}<br \/>\n\\underset{(2)}{\\mathop{1}}\\,+\\frac{x-3}{\\underset{(1)}{\\mathop{2}}\\,}=\\underset{(2)}{\\mathop{1}}\\,\u00a0 \\\\<br \/>\n2+x-3=2\u00a0 \\\\<br \/>\nx=2-2+3\u00a0 \\\\<br \/>\nx=3\u00a0 \\\\<br \/>\n\\end{matrix}\\]<\/p>\n<\/li>\n<li>$\\frac{x-2}{4}+\\frac{2x}{3}=1$\n<p>Escrevendo uma equa\u00e7\u00e3o equivalente onde todas as fra\u00e7\u00f5es figurem com igual denominador (igual ao m.m.c. dos denominadores), temos:<br \/>\n\\[\\begin{matrix}<br \/>\n\\frac{x-2}{\\underset{(3)}{\\mathop{4}}\\,}+\\frac{2x}{\\underset{(4)}{\\mathop{3}}\\,}=\\underset{(12)}{\\mathop{1}}\\,\u00a0 \\\\<br \/>\n\\frac{3x-6}{12}+\\frac{8x}{12}=\\frac{12}{12}\u00a0 \\\\<br \/>\n\\end{matrix}\\]<br \/>\nSe multiplicarmos por 12 ambos os membros da equa\u00e7\u00e3o\u00a0obtemos uma equa\u00e7\u00e3o equivalente. Seguidamente, aplicando a propriedade distributiva da multiplica\u00e7\u00e3o em ambos os membros, obt\u00e9m-se uma equa\u00e7\u00e3o equivalente sem denominadores:<br \/>\n\\[\\begin{matrix}<br \/>\n12\\times \\left( \\frac{3x-6}{12}+\\frac{8x}{12} \\right)=\\left( \\frac{12}{12} \\right)\\times 12\u00a0 \\\\<br \/>\n3x-6+8x=12\u00a0 \\\\<br \/>\n\\end{matrix}\\]<br \/>\nEsta sequ\u00eancia de procedimentos pode ser abreviada: suprimindo a 2.\u00aa e 3.\u00aa equa\u00e7\u00f5es.<br \/>\nPara isso, bastar\u00e1 multiplicar apenas os numeradores das fra\u00e7\u00f5es da 1.\u00aa equa\u00e7\u00e3o pelos valores indicados dentro dos par\u00eantesis curvos:<br \/>\n\\[\\begin{matrix}<br \/>\n\\frac{x-2}{\\underset{(3)}{\\mathop{4}}\\,}+\\frac{2x}{\\underset{(4)}{\\mathop{3}}\\,}=\\underset{(12)}{\\mathop{1}}\\,\u00a0 \\\\<br \/>\n3x-6+8x=12\u00a0 \\\\<br \/>\n11x=18\u00a0 \\\\<br \/>\nx=\\frac{18}{11}\u00a0 \\\\<br \/>\n\\end{matrix}\\]<\/p>\n<\/li>\n<li>$\\frac{y+1}{4}-\\frac{5+y}{2}=\\frac{3}{2}$\n<p>Escrevendo uma equa\u00e7\u00e3o equivalente onde todas as fra\u00e7\u00f5es figurem com igual denominador (igual ao m.m.c. dos denominadores), temos:<br \/>\n\\[\\begin{matrix}<br \/>\n\\frac{y+1}{\\underset{(1)}{\\mathop{4}}\\,}-\\frac{5+y}{\\underset{(2)}{\\mathop{2}}\\,}=\\frac{3}{\\underset{(2)}{\\mathop{2}}\\,}\u00a0 \\\\<br \/>\n\\frac{y+1}{4}-\\frac{10+2y}{4}=\\frac{6}{4}\u00a0 \\\\<br \/>\n\\end{matrix}\\]<br \/>\nSe multiplicarmos por\u00a04 ambos os membros da equa\u00e7\u00e3o\u00a0obtemos uma equa\u00e7\u00e3o equivalente. Seguidamente, aplicando a propriedade distributiva da multiplica\u00e7\u00e3o em ambos os membros, obt\u00e9m-se uma equa\u00e7\u00e3o equivalente sem denominadores:<br \/>\n\\[\\begin{matrix}<br \/>\n4\\times \\left( \\frac{y+1}{4}-\\frac{10+2y}{4} \\right)=\\left( \\frac{6}{4} \\right)\\times 4\u00a0 \\\\<br \/>\ny+1-(10+2y)=6\u00a0 \\\\<br \/>\n\\end{matrix}\\]<br \/>\nEsta sequ\u00eancia de procedimentos pode ser abreviada: suprimindo a 2.\u00aa e 3.\u00aa equa\u00e7\u00f5es.<br \/>\nPara isso, bastar\u00e1 multiplicar apenas os numeradores das fra\u00e7\u00f5es da 1.\u00aa equa\u00e7\u00e3o pelos valores indicados dentro dos par\u00eantesis curvos:<br \/>\n\\[\\begin{matrix}<br \/>\n\\frac{y+1}{\\underset{(1)}{\\mathop{4}}\\,}-\\frac{5+y}{\\underset{(2)}{\\mathop{2}}\\,}=\\frac{3}{\\underset{(2)}{\\mathop{2}}\\,}\u00a0 \\\\<br \/>\ny+1-(10+2y)=6\u00a0 \\\\<br \/>\ny+1-10-2y=6\u00a0 \\\\<br \/>\n-y=15\u00a0 \\\\<br \/>\ny=-15\u00a0 \\\\<br \/>\n\\end{matrix}\\]<\/p>\n<\/li>\n<\/ol>\n<p>Esta forma abreviada de procedimentos \u00e9 conhecida por <strong>DESEMBARA\u00c7AR DE PAR\u00caNTESIS<\/strong> e , frequentemente, \u00e9 apresentada como segue:<\/p>\n<p>\\[\\begin{array}{*{35}{l}}<br \/>\n\\frac{y+1}{\\underset{(1)}{\\mathop{4}}\\,}-\\frac{5+y}{\\underset{(2)}{\\mathop{2}}\\,}=\\frac{3}{\\underset{(2)}{\\mathop{2}}\\,} &amp; \\Leftrightarrow\u00a0 &amp; y+1-(10+2y)=6\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; y+1-10-2y=6\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; -y=15\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; y=-15\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6770' onClick='GTTabs_show(0,6770)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Resolve as equa\u00e7\u00f5es: $1+\\frac{x-3}{2}=1$ $\\frac{x-2}{4}+\\frac{2x}{3}=1$ $\\frac{y+1}{4}-\\frac{5+y}{2}=\\frac{3}{2}$ Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":14114,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,159],"tags":[425],"series":[],"class_list":["post-6770","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-equacoes-do-1--grau","tag-equacoes"],"views":18007,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat56.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6770","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=6770"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6770\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14114"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6770"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6770"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6770"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6770"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}