\nEnunciado<\/b><\/span><\/p>\n
Calcula, por dois processos diferentes:<\/p>\n
\n
$2 \\times \\left( { – 5 – 1} \\right)$<\/li>\n
$ – 3 \\times \\left( { – 2 + 7} \\right)$<\/li>\n
$4 \\times \\left( {10 – 2} \\right)$<\/li>\n
$ – 5 \\times \\left( { – 3 + 6} \\right)$<\/li>\n<\/ol>\n
Resolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n
\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
\n
Calculando o par\u00eantesis, vem:
\n$$\\begin{array}{*{20}{l}} \u00a0 {2 \\times \\left( { – 5 – 1} \\right)}& = &{2 \\times \\left( { – 6} \\right)} \\\\ \u00a0 {}& = &{ – 12} \\end{array}$$
\nAplicando a propriedade distributiva da multiplica\u00e7\u00e3o em rela\u00e7\u00e3o \u00e0 adi\u00e7\u00e3o, vem:
\n$$\\begin{array}{*{20}{l}} \u00a0 {2 \\times \\left( { – 5 – 1} \\right)}& = &{2 \\times \\left( { – 5} \\right) + 2 \\times \\left( { – 1} \\right)} \\\\ \u00a0 {}& = &{ – 10 + \\left( { – 2} \\right)} \\\\ \u00a0 {}& = &{ – 10 – 2} \\\\ \u00a0 {}& = &{ – 12} \\end{array}$$
\nou, simplificando a escrita:
\n$$\\begin{array}{*{20}{l}}\u00a0\u00a0 {2 \\times \\left( { – 5 – 1} \\right)}& = &{2 \\times \\left( { – 5} \\right) + 2 \\times \\left( { – 1} \\right)} \\\\\u00a0\u00a0 {}& = &{ – 10 – 2} \\\\\u00a0\u00a0 {}& = &{ – 12} \\end{array}$$
\n\u00ad<\/li>\n
Calculando o par\u00eantesis, vem:
\n$$\\begin{array}{*{20}{l}} \u00a0 { – 3 \\times \\left( { – 2 + 7} \\right)}& = &{ – 3 \\times 5} \\\\ \u00a0 {}& = &{ – 15} \\end{array}$$
\nAplicando a propriedade distributiva da multiplica\u00e7\u00e3o em rela\u00e7\u00e3o \u00e0 adi\u00e7\u00e3o, vem:
\n$$\\begin{array}{*{20}{l}} \u00a0 { – 3 \\times \\left( { – 2 + 7} \\right)}& = &{ – 3 \\times \\left( { – 2} \\right) + \\left( { – 3} \\right) \\times 7} \\\\ \u00a0 {}& = &{6 + \\left( { – 21} \\right)} \\\\ \u00a0 {}& = &{6 – 21} \\\\ \u00a0 {}& = &{ – 15} \\end{array}$$
\nou, simplificando a escrita:
\n$$\\begin{array}{*{20}{l}} \u00a0 { – 3 \\times \\left( { – 2 + 7} \\right)}& = &{ – 3 \\times \\left( { – 2} \\right) – 3 \\times 7} \\\\ \u00a0 {}& = &{6 – 21} \\\\ \u00a0 {}& = &{ – 15} \\end{array}$$
\n\u00ad<\/li>\n
Calculando o par\u00eantesis, vem:
\n$$\\begin{array}{*{20}{l}} \u00a0 {4 \\times \\left( {10 – 2} \\right)}& = &{4 \\times 8} \\\\ \u00a0 {}& = &{32} \\end{array}$$
\nAplicando a propriedade distributiva da multiplica\u00e7\u00e3o em rela\u00e7\u00e3o \u00e0 adi\u00e7\u00e3o (e simplificando a escrita em simult\u00e2neo), vem:
\n$$\\begin{array}{*{20}{l}} \u00a0 {4 \\times \\left( {10 – 2} \\right)}& = &{4 \\times 10 + 4 \\times \\left( { – 2} \\right)} \\\\ \u00a0 {}& = &{40 – 8} \\\\ \u00a0 {}& = &{32} \\end{array}$$
\n\u00ad<\/li>\n
Calculando o par\u00eantesis, vem:
\n$$\\begin{array}{*{20}{l}} \u00a0 { – 5 \\times \\left( { – 3 + 6} \\right)}& = &{ – 5 \\times 3} \\\\ \u00a0 {}& = &{ – 15} \\end{array}$$
\nAplicando a propriedade distributiva da multiplica\u00e7\u00e3o em rela\u00e7\u00e3o \u00e0 adi\u00e7\u00e3o (e simplificando a escrita em simult\u00e2neo), vem:
\n$$\\begin{array}{*{20}{l}} \u00a0 { – 5 \\times \\left( { – 3 + 6} \\right)}& = &{ – 5 \\times \\left( { – 3} \\right) – 5 \\times 6} \\\\ \u00a0 {}& = &{15 – 30} \\\\ \u00a0 {}& = &{ – 15} \\end{array}$$<\/li>\n<\/ol>\n
<< Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"
Enunciado Resolu\u00e7\u00e3o Enunciado Calcula, por dois processos diferentes: $2 \\times \\left( { – 5 – 1} \\right)$ $ – 3 \\times \\left( { – 2 + 7} \\right)$ $4 \\times \\left( {10 – 2}...<\/p>\n","protected":false},"author":1,"featured_media":19301,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[317,97,318],"tags":[428,320,327],"series":[],"class_list":["post-10268","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-7-o-ano","category-aplicando","category-numeros-inteiros","tag-7-o-ano","tag-multiplicacao","tag-propriedades-da-multiplicacao"],"views":2084,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat111.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/10268","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=10268"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/10268\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19301"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=10268"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=10268"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=10268"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=10268"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}