Răspunsuri

2014-11-03T15:09:45+02:00

(a+b+c)/3=79 de unde a+b+c=79*3, a+b+c=237

(a+b)/2=105

a:b=1 rest 66, de unde a=b*1+66, a=b+66

din (a+b)/2=105 avem (b+66+b)/2=105, 2*b+66=105*2, 2*b=210-66=144, b=144/2=72

a=b+66=72+66=138

a+b+c=237, 72+138+c=237, de unde c=237-72-138=27

Cel mai inteligent răspuns!
2014-11-03T15:20:00+02:00
(a+b+c):3=79; (a+b):2=105; a:b=1rest 66=> a=b+66; Inlocuim (b+66+b):2=105; 2b+66=105x2; 2b=2-10-66; 2b=144=> b=72 ; a=72+66=138 a+b+c=79x3=237
138+72+c=237; c= 237-138-72 ; c=27