Estas pilhas de cubos com n\u00fameros foram montadas seguindo um padr\u00e3o.<\/p>\n
Descobre-o. Depois completa cada uma das pilhas de cubos.<\/p>\n
\u00ad<\/p>\n Conclui-se, ent\u00e3o, que essa opera\u00e7\u00e3o \u00e9 a subtra\u00e7\u00e3o<\/strong>: $\\left( { – 3} \\right) – \\left( { + 2} \\right) = \\left( { – 3} \\right) + \\left( { – 2} \\right) = \\left( { – 5} \\right)$.![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/09\/pilhascubos.jpg\" "\\"Pilhas")
<\/a><\/p>\n<\/a><\/p>\n\n
\nConclui-se, ent\u00e3o, que essa opera\u00e7\u00e3o \u00e9 a adi\u00e7\u00e3o<\/strong>: $\\left( { + 6} \\right) + \\left( { – 2} \\right) = \\left( { + 4} \\right)$.
\nAssim, temos:
\n$$\\begin{array}{*{20}{l}} {\\left( { – 2} \\right) + \\left( { – 6} \\right) = \\left( { – 8} \\right)}&{}&{\\left( { + 4} \\right) + \\left( { – 8} \\right) = \\left( { – 4} \\right)}&{}&{\\left( { + 5} \\right) + \\left( { – 4} \\right) = \\left( { + 1} \\right)} \\end{array}$$
\nPor \u00faltimo, vem:
\n$$\\begin{array}{*{20}{l}} {\\left( ? \\right) + \\left( { + 4} \\right) = \\left( { + 5} \\right)}& \\Rightarrow &{\\begin{array}{*{20}{l}} {\\left( ? \\right)}& = &{\\left( { + 5} \\right) – \\left( { + 4} \\right)}\\\\ {}& = &{\\left( { + 1} \\right)} \\end{array}}\\\\ {}&{}&{}\\\\ {\\left( ? \\right) + \\left( { + 6} \\right) = \\left( { + 1} \\right)}& \\Rightarrow &{\\begin{array}{*{20}{l}} {\\left( ? \\right)}& = &{\\left( { + 1} \\right) – \\left( { + 6} \\right)}\\\\ {}& = &{\\left( { – 5} \\right)} \\end{array}} \\end{array}$$
\n\u00ad<\/a><\/li>\n
\nConclui-se, ent\u00e3o, que essa opera\u00e7\u00e3o \u00e9 a multiplica\u00e7\u00e3o<\/strong>: $\\left( { – 1} \\right) \\times \\left( { + 2} \\right) = \\left( { – 2} \\right)$.
\nAssim, temos:
\n$$\\begin{array}{*{20}{l}} {\\left( { – 3} \\right) \\times \\left( { – 1} \\right) = \\left( { + 3} \\right)}&{}&{\\left( { + 2} \\right) \\times \\left( { – 2} \\right) = \\left( { – 4} \\right)}&{}&{\\left( { + 3} \\right) \\times \\left( { – 2} \\right) = \\left( { – 6} \\right)}&{}&{\\left( { – 2} \\right) \\times \\left( { – 4} \\right) = \\left( { + 8} \\right)} \\end{array}$$
\nVerificando-se ainda: $\\left( { – 6} \\right) \\times \\left( { + 8} \\right) = \\left( { – 48} \\right)$.
\n\u00ad<\/a><\/li>\n
\nAssim, temos:
\n$$\\begin{array}{*{20}{l}} {\\left( { + 2} \\right) – \\left( { + 4} \\right) = \\left( { – 2} \\right)}&{}&{\\left( { + 4} \\right) – \\left( { – 1} \\right) = \\left( { + 5} \\right)}&{}&{\\left( { – 5} \\right) – \\left( { – 2} \\right) = \\left( { – 3} \\right)}&{}&{\\left( { – 2} \\right) – \\left( { + 5} \\right) = \\left( { – 7} \\right)} \\end{array}$$
\nVerificando-se ainda: $\\left( { – 3} \\right) – \\left( { – 7} \\right) = \\left( { – 3} \\right) + \\left( { + 7} \\right) = \\left( { + 4} \\right)$.<\/p>\n<\/li>\n<\/ol>\n