Răspunsuri

2014-10-03T09:32:13+03:00
a+2b+c=51 
2a+b+2c=42
adunam cele 2 relatii  si avem  a+2a +2b+b +c+2c = 51+42
3a +3b + 3c =93    3 x (a+b+c)=93 /:3        a+b+c=31
poti ajuta?
Daca abc sunt nr. naturale si a+b+c=18 iar 2a+b+2c=25, atunci calculati (a+c)*(3a+2b+3c)
(a+c)*(3a+2b+3c)=(a+c)*[(2a+b+2c)+(a+b+c)]= (a+c) * (25 + 18)=43 * (a+c)
a+b+c=18 -1) si 2a+b+2c=25 - 2) se scade relatia 1 din 2 si avem: 2a+b+2c-a-b-c=25-18 a+c=7 deci, a+c)*(3a+2b+3c)=
(a+c)*(3a+2b+3c)= 43 * (a+c)=43 * 7=301
2014-10-03T16:36:03+03:00
A+2b+c+2a+b+2c=51+42=93
3a+3b+3c=93|impartim pe 93 la 3 si ne va iesii a+b+c
a+b+c=93:3=31