標題:
maths......急急急急急急急
發問:
bc + cd + db - dbc How to simplify it?? thanks~~/\\/\\**
最佳解答:
其他解答:
bc + cd + db - dbc = c (b + d) + db (1 - c)|||||bc + cd + db - dbc =bc+cd+(db-dbc) =c(b+d)+ db(1-c)|||||bc + cd + db - dbc=c(b+d)+db(1+c)
maths......急急急急急急急
發問:
bc + cd + db - dbc How to simplify it?? thanks~~/\\/\\**
最佳解答:
此文章來自奇摩知識+如有不便請留言告知
Simply: bc + cd + db - dbc =bc+cd+db-dbc Answer: The no. is bc+cd+db-dbc. P.S.: It can't simply because there are not the equal unknown.其他解答:
bc + cd + db - dbc = c (b + d) + db (1 - c)|||||bc + cd + db - dbc =bc+cd+(db-dbc) =c(b+d)+ db(1-c)|||||bc + cd + db - dbc=c(b+d)+db(1+c)
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