\nEnunciado<\/b><\/span><\/p>\n
Que n\u00fameros devemos colocar no lugar de $\\square $ ?<\/p>\n
\n
$\\left( \\square\u00a0 \\right) \\div \\left( { – 3} \\right) = 8$<\/li>\n
$\\left( \\square\u00a0 \\right) \\div \\left( { – 12} \\right) =\u00a0 – 10$<\/li>\n
$\\left( \\square\u00a0 \\right) \\div \\left( { – 25} \\right) =\u00a0 – 1$<\/li>\n
$\\left( { – 352} \\right) \\div \\left( \\square\u00a0 \\right) =\u00a0 – 8$<\/li>\n
$\\left( { – 38} \\right) \\div \\left( \\square\u00a0 \\right) = 19$<\/li>\n
$\\left( \\square\u00a0 \\right) \\div \\left( { – 7} \\right) = 0$<\/li>\n<\/ol>\n
Resolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n
\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
\n
Na divis\u00e3o
\n$$\\begin{array}{*{20}{c}} \u00a0 {\\begin{array}{*{20}{c}} \u00a0 1&0&0&0 \\\\ \u00a0 {}&4&0&{} \\\\ \u00a0 {}&{}&4&0 \\\\ \u00a0 {}&{}&{}&4 \\end{array}}&{\\begin{array}{*{20}{c}} \u00a0 6&{}&{}&{} \\\\ \\hline \u00a0 1&6&6&{} \\\\ \u00a0 {}&{}&{}&. \\\\ \u00a0 {}&{}&{}&. \\end{array}} \\end{array}$$<\/p>\n
\n
$D$ – dividendo: $1000$<\/li>\n
$d$ – divisor: $6$<\/li>\n
$q$ – quociente: $166$<\/li>\n
$r$ – resto: $4$<\/li>\n<\/ul>\n
Nesta divis\u00e3o verifica-se:
\n$$1000 = 6 \\times 166 + 4$$<\/p>\n
Em qualquer divis\u00e3o, tem-se:
\n$$\\begin{array}{*{20}{c}} \u00a0 {D = d \\times q + r}&{{\\text{com}}}&{0 \\leqslant r < d} \\end{array}$$<\/p>\n
Se a divis\u00e3o \u00e9 exata, ent\u00e3o $r = 0$ e, por isso, ${D = d \\times q}$, ou seja, $\\frac{D}{d} = q$, com $d \\ne 0$.<\/p>\n<\/blockquote>\n
\u00ad<\/p>\n
Recordado o que est\u00e1 acima, vem:<\/p>\n
\n
Ora,
\n$$\\begin{array}{*{20}{c}} \u00a0 {\\left( \\square\u00a0 \\right) \\div \\left( { – 3} \\right) = 8}& \\Rightarrow &{\\begin{array}{*{20}{l}} \u00a0 \\square & = &{\\left( { – 3} \\right) \\times 8} \\\\ \u00a0 {}& = &{ – 24} \\end{array}} \\end{array}$$<\/li>\n
Ora,
\n$$\\begin{array}{*{20}{c}} \u00a0 {\\left( \\square\u00a0 \\right) \\div \\left( { – 12} \\right) =\u00a0 – 10}& \\Rightarrow &{\\begin{array}{*{20}{l}} \u00a0 \\square & = &{\\left( { – 12} \\right) \\times \\left( { – 10} \\right)} \\\\ \u00a0 {}& = &{120} \\end{array}} \\end{array}$$<\/li>\n
Ora,
\n$$\\begin{array}{*{20}{c}} \u00a0 {\\left( \\square\u00a0 \\right) \\div \\left( { – 25} \\right) =\u00a0 – 1}& \\Rightarrow &{\\begin{array}{*{20}{l}} \u00a0 \\square & = &{\\left( { – 25} \\right) \\times \\left( { – 1} \\right)} \\\\ \u00a0 {}& = &{25} \\end{array}} \\end{array}$$<\/li>\n
Ora,
\n$$\\begin{array}{*{20}{c}} \u00a0 {\\left( { – 352} \\right) \\div \\left( \\square\u00a0 \\right) =\u00a0 – 8}& \\Rightarrow &{\\begin{array}{*{20}{l}} \u00a0 \\square & = &{\\left( { – 352} \\right) \\div \\left( { – 8} \\right)} \\\\ \u00a0 {}& = &{44} \\end{array}} \\end{array}$$<\/li>\n
Ora,
\n$$\\begin{array}{*{20}{c}} \u00a0 {\\left( { – 38} \\right) \\div \\left( \\square\u00a0 \\right) = 19}& \\Rightarrow &{\\begin{array}{*{20}{l}} \u00a0 \\square & = &{\\left( { – 38} \\right) \\div 19} \\\\ \u00a0 {}& = &{ – 2} \\end{array}} \\end{array}$$<\/li>\n
Ora,
\n$$\\begin{array}{*{20}{c}} \u00a0 {\\left( \\square\u00a0 \\right) \\div \\left( { – 7} \\right) = 0}& \\Rightarrow &{\\begin{array}{*{20}{l}} \u00a0 \\square & = &{\\left( { – 7} \\right) \\times 0} \\\\ \u00a0 {}& = &0 \\end{array}} \\end{array}$$<\/li>\n<\/ol>\n
<< Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"
Enunciado Resolu\u00e7\u00e3o Enunciado Que n\u00fameros devemos colocar no lugar de $\\square $ ? $\\left( \\square\u00a0 \\right) \\div \\left( { – 3} \\right) = 8$ $\\left( \\square\u00a0 \\right) \\div \\left( { – 12} \\right) =\u00a0...<\/p>\n","protected":false},"author":1,"featured_media":19271,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[317,97,318],"tags":[428,330,319],"series":[],"class_list":["post-10491","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-divisao","tag-numeros-inteiros-2"],"views":2178,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat92.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/10491","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=10491"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/10491\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19271"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=10491"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=10491"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=10491"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=10491"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}