\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
\u00c0 direita, est\u00e1 completada cada uma das igualdades.<\/p>\n\n\n\n\n\n\n\n\n
a)<\/td>\n \\(5a + 5b = 5\\left( { \\ldots + \\ldots } \\right)\\)<\/td>\n \\(5a + 5b = 5\\left( {a + b} \\right)\\)<\/td>\n<\/tr>\n
b)<\/td>\n \\(3y – 2xy = \\ldots \\left( {3 – 2x} \\right)\\)<\/td>\n \\(3y – 2xy = y\\left( {3 – 2x} \\right)\\)<\/td>\n<\/tr>\n
c)<\/td>\n \\(x + {x^2} – 6{x^3} = x\\left( { \\ldots + \\ldots – \\ldots } \\right)\\)<\/td>\n \\(x + {x^2} – 6{x^3} = x\\left( {1 + x – 6{x^2}} \\right)\\)<\/td>\n<\/tr>\n
d)<\/td>\n \\(8x – 6y = 2\\left( { \\ldots – \\ldots } \\right)\\)<\/td>\n \\(8x – 6y = 2\\left( {4x – 3y} \\right)\\)<\/td>\n<\/tr>\n
e)<\/td>\n \\(6ab + 2b = \\ldots \\left( { \\ldots + \\ldots } \\right)\\)<\/td>\n \\(6ab + 2b = 2b\\left( {3a + 1} \\right)\\)<\/td>\n<\/tr>\n
f)<\/td>\n \\(2ab + 6{a^2}b – 4a{b^2} = \\ldots \\left( { \\ldots + \\ldots – \\ldots } \\right)\\)<\/td>\n \\(2ab + 6{a^2}b – 4a{b^2} = 2ab\\left( {1 + 3a – 2b} \\right)\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\u00a0<\/p>\n
<< Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"
Enunciado Resolu\u00e7\u00e3o Enunciado Copia e completa cada uma das seguintes igualdades. a) \\(5a + 5b = 5\\left( { \\ldots + \\ldots } \\right)\\) b) \\(3y – 2xy = \\ldots \\left( {3 – 2x} \\right)\\)...<\/p>\n","protected":false},"author":1,"featured_media":19173,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,700],"tags":[424,197,705],"series":[],"class_list":["post-24570","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-monomios-e-polinomios","tag-8-o-ano","tag-decomposicao-em-factores","tag-polinomios"],"views":150,"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\/24570","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=24570"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/24570\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19173"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=24570"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=24570"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=24570"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=24570"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}