Purceamario Purceamario 27-05-2020 Matematică a fost răspuns Calculati: [2002-(1/2+2/3+3/4+...+2002/2003)]:(1/2+1/3+1/4+...1/2003) n=(3/2+4/3+5/4+...+1008/1007):[2012-(1/2+2/3+3/4+...+1006/1007)]Dau coroana , mersi