{"id":6904,"date":"2011-06-08T23:01:04","date_gmt":"2011-06-08T22:01:04","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6904"},"modified":"2022-01-05T15:55:29","modified_gmt":"2022-01-05T15:55:29","slug":"resolve-as-seguintes-equacoes","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6904","title":{"rendered":"Resolve as seguintes equa\u00e7\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_6904' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6904' class='GTTabs_curr'><a  id=\"6904_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6904' ><a  id=\"6904_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_6904'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolve as seguintes equa\u00e7\u00f5es:<\/p>\n<ol>\n<li>$x(x-1)=0$<\/li>\n<li>$(a-1)(a+1)=0$<\/li>\n<li>${{x}^{2}}-2x=0$<\/li>\n<li>${{a}^{2}}-6a+9=0$<\/li>\n<li>$4{{y}^{2}}+25=20y$<\/li>\n<li>${{c}^{2}}-0,25=0$<\/li>\n<li>$0,04{{x}^{2}}-0,4x+1=0$<\/li>\n<li>${{x}^{2}}=0,01$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6904' onClick='GTTabs_show(1,6904)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6904'>\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 \/>\nx(x-1)=0 &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x-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=1\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n(a-1)(a+1)=0 &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\na-1=0 &amp; \\vee\u00a0 &amp; a+1=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\na=1 &amp; \\vee\u00a0 &amp; a=-1\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{x}^{2}}-2x=0 &amp; \\Leftrightarrow\u00a0 &amp; x(x-2)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x-2=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=2\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{a}^{2}}-6a+9=0 &amp; \\Leftrightarrow\u00a0 &amp; {{(a-3)}^{2}}=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; a-3=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; a=3\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n4{{y}^{2}}+25=20y &amp; \\Leftrightarrow\u00a0 &amp; 4{{y}^{2}}-20y+25=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; {{(2y-5)}^{2}}=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 2y-5=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; y=\\frac{5}{2}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{c}^{2}}-0,25=0 &amp; \\Leftrightarrow\u00a0 &amp; (c+0,5)(c-0,5)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nc+0,5=0 &amp; \\vee\u00a0 &amp; c-0,5=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nc=-0,5 &amp; \\vee\u00a0 &amp; c=0,5\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n0,04{{x}^{2}}-0,4x+1=0 &amp; \\Leftrightarrow\u00a0 &amp; {{(0,2x-1)}^{2}}=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 0,2x-1=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; x=5\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{x}^{2}}=0,01 &amp; \\Leftrightarrow\u00a0 &amp; {{x}^{2}}-0,01=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; (x+0,1)(x-0,1)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx+0,1=0 &amp; \\vee\u00a0 &amp; x-0,1=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=-0,1 &amp; \\vee\u00a0 &amp; x=0,1\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{y}^{3}}-4y=0 &amp; \\Leftrightarrow\u00a0 &amp; y({{y}^{2}}-4)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; y(y+2)(y-2)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\ny=0 &amp; \\vee\u00a0 &amp; y+2=0 &amp; \\vee\u00a0 &amp; y-2=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\ny=0 &amp; \\vee\u00a0 &amp; y=-2 &amp; \\vee\u00a0 &amp; y=2\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{x}^{3}}+4{{x}^{2}}+4x=0 &amp; \\Leftrightarrow\u00a0 &amp; x({{x}^{2}}+4x+4)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; x{{(x+2)}^{2}}=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x+2=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=-2\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6904' onClick='GTTabs_show(0,6904)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Resolve as seguintes equa\u00e7\u00f5es: $x(x-1)=0$ $(a-1)(a+1)=0$ ${{x}^{2}}-2x=0$ ${{a}^{2}}-6a+9=0$ $4{{y}^{2}}+25=20y$ ${{c}^{2}}-0,25=0$ $0,04{{x}^{2}}-0,4x+1=0$ ${{x}^{2}}=0,01$ Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":19173,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_post_was_ever_published":false},"categories":[100,97,193],"tags":[198],"series":[],"class_list":["post-6904","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":2971,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat64.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6904","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=6904"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6904\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6904"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6904"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6904"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6904"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}