Matematică
Orhidee1
2015-11-10 04:39:42
8.Calculati: a) (2+4+6+...+2004)÷(1+2+3+...+1002) b) (1+2+3+...+2003)÷(3+6+9+...+6009) c) (7/5+7/10+7/15+...+7/200)÷(3/5+3/10+3/15+...+3/200) Va rog este urgent!!
Răspunsuri la întrebare
Gaby1247
2015-11-10 06:11:59

2(1+2+3+........1002):(1+2+3+.........1002)=2 (1+2+3+..............+2003):3x(1+2+3+........+2003)=1/3 7(1/5+1/10+1/15+........+1/200):3(1/5+1/10+1/15+...........+1/200)=7/3

smarta
2015-11-10 06:13:14

8.Calculati: a) (2+4+6+...+2004)÷(1+2+3+...+1002) =2(1+2+3+...+1002)(1+2+3+...+1002)=2 b) (1+2+3+...+2003)÷(3+6+9+...+6009) =(1+2+3+...+2003):3(1+2+3+...+2003)=1/3 c) (7/5+7/10+7/15+...+7/200)÷(3/5+3/10+3/15+...+3/200) =(1/5+1/10+...+1/200):3(1/5+1/10+...+1/200)=7/3

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