{"id":6903,"date":"2011-06-08T22:21:17","date_gmt":"2011-06-08T21:21:17","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6903"},"modified":"2022-01-05T15:53:27","modified_gmt":"2022-01-05T15:53:27","slug":"determina-o-conjunto-solucao-de-cada-uma-das-equacoes","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6903","title":{"rendered":"Determina o conjunto-solu\u00e7\u00e3o de cada uma das equa\u00e7\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_6903' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6903' class='GTTabs_curr'><a  id=\"6903_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6903' ><a  id=\"6903_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_6903'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Determina o conjunto-solu\u00e7\u00e3o de cada uma das equa\u00e7\u00f5es:<\/p>\n<ol>\n<li>${{x}^{2}}-6x+9=0$<\/li>\n<li>${{x}^{3}}-2{{x}^{2}}+x=0$<\/li>\n<li>${{x}^{2}}-16=0$<\/li>\n<li>$x({{x}^{2}}-25)=0$<\/li>\n<li>$8{{x}^{3}}-2x=0$<\/li>\n<li>$4{{x}^{2}}+4x+1=0$<\/li>\n<li>${{x}^{2}}-36=0$<\/li>\n<li>${{x}^{2}}-{{(3x+1)}^{2}}=0$<\/li>\n<li>${{(x+1)}^{2}}-(x+1)=0$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6903' onClick='GTTabs_show(1,6903)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6903'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{x}^{2}}-6x+9=0 &amp; \\Leftrightarrow\u00a0 &amp; {{(x-3)}^{2}}=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; (x-3)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; x=3\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Portanto, o conjunto-solu\u00e7\u00e3o da equa\u00e7\u00e3o \u00e9 $S=\\left\\{ 3 \\right\\}$.<\/p>\n<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{x}^{3}}-2{{x}^{2}}+x=0 &amp; \\Leftrightarrow\u00a0 &amp; x({{x}^{2}}-2x+1)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; x{{(x-1)}^{2}}=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x-1=0\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x=1\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Portanto, o conjunto-solu\u00e7\u00e3o da equa\u00e7\u00e3o \u00e9 $S=\\left\\{ 0,1 \\right\\}$.<\/p>\n<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{x}^{2}}-16=0 &amp; \\Leftrightarrow\u00a0 &amp; (x+4)(x-4)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nx+4=0 &amp; \\vee\u00a0 &amp; x-4=0\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nx=-4 &amp; \\vee\u00a0 &amp; x=4\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Portanto, o conjunto-solu\u00e7\u00e3o da equa\u00e7\u00e3o \u00e9 $S=\\left\\{ -4,4 \\right\\}$.<\/p>\n<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\nx({{x}^{2}}-25)=0 &amp; \\Leftrightarrow\u00a0 &amp; x(x+5)(x-5)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x+5=0 &amp; \\vee\u00a0 &amp; x-5=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x=-5 &amp; \\vee\u00a0 &amp; x=5\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Portanto, o conjunto-solu\u00e7\u00e3o da equa\u00e7\u00e3o \u00e9 $S=\\left\\{ -5,0,5 \\right\\}$.<\/p>\n<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n8{{x}^{3}}-2x=0 &amp; \\Leftrightarrow\u00a0 &amp; 2x(4{{x}^{2}}-1)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 2x(2x+1)(2x-1)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\n2x=0 &amp; \\vee\u00a0 &amp; 2x+1=0 &amp; \\vee\u00a0 &amp; 2x-1=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x=-\\frac{1}{2} &amp; \\vee\u00a0 &amp; x=\\frac{1}{2}\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Portanto, o conjunto-solu\u00e7\u00e3o da equa\u00e7\u00e3o \u00e9 $S=\\left\\{ -\\frac{1}{2},0,\\frac{1}{2} \\right\\}$.<\/p>\n<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n4{{x}^{2}}+4x+1=0 &amp; \\Leftrightarrow\u00a0 &amp; {{(2x+1)}^{2}}=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 2x+1=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; x=-\\frac{1}{2}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Portanto, o conjunto-solu\u00e7\u00e3o da equa\u00e7\u00e3o \u00e9 $S=\\left\\{ -\\frac{1}{2} \\right\\}$.<\/p>\n<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{x}^{2}}-36=0 &amp; \\Leftrightarrow\u00a0 &amp; (x+6)(x-6)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nx+6=0 &amp; \\vee\u00a0 &amp; x-6=0\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nx=-6 &amp; \\vee\u00a0 &amp; x=6\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Portanto, o conjunto-solu\u00e7\u00e3o da equa\u00e7\u00e3o \u00e9 $S=\\left\\{ -6,6 \\right\\}$.<\/p>\n<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{x}^{2}}-{{(3x+1)}^{2}}=0 &amp; \\Leftrightarrow\u00a0 &amp; \\left[ x+(3x+1) \\right]\\left[ x-(3x+1) \\right]=0\\,\\,\\,(Porqu\\hat{e}?)\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; (4x+1)(-2x-1)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\n4x+1=0 &amp; \\vee\u00a0 &amp; -2x-1=0\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nx=-\\frac{1}{4} &amp; \\vee\u00a0 &amp; x=-\\frac{1}{2}\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Portanto, o conjunto-solu\u00e7\u00e3o da equa\u00e7\u00e3o \u00e9 $S=\\left\\{ -\\frac{1}{2},-\\frac{1}{4} \\right\\}$.<\/p>\n<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{(x+1)}^{2}}-(x+1)=0 &amp; \\Leftrightarrow\u00a0 &amp; (x+1)\\left[ (x+1)-1 \\right]=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; (x+1)x=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nx+1=0 &amp; \\vee\u00a0 &amp; x=0\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nx=-1 &amp; \\vee\u00a0 &amp; x=0\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Portanto, o conjunto-solu\u00e7\u00e3o da equa\u00e7\u00e3o \u00e9 $S=\\left\\{ -1,0 \\right\\}$.<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6903' onClick='GTTabs_show(0,6903)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Determina o conjunto-solu\u00e7\u00e3o de cada uma das equa\u00e7\u00f5es: ${{x}^{2}}-6x+9=0$ ${{x}^{3}}-2{{x}^{2}}+x=0$ ${{x}^{2}}-16=0$ $x({{x}^{2}}-25)=0$ $8{{x}^{3}}-2x=0$ $4{{x}^{2}}+4x+1=0$ ${{x}^{2}}-36=0$ ${{x}^{2}}-{{(3x+1)}^{2}}=0$ ${{(x+1)}^{2}}-(x+1)=0$ Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/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":[100,97,193],"tags":[198],"series":[],"class_list":["post-6903","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-equacoes-de-grau-superior-ao-1-","tag-lei-do-anulamento-do-produto"],"views":3246,"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\/6903","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=6903"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6903\/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=6903"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6903"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6903"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6903"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}