Question

aflati nr naturale a.b.c.stiindca ma a nr a si b este 43 ma a nr b si c este 68 si ma a nr a si c este 97.

Answer 1

(a+b)/2=43=> a+b=86 |
(b+c)/2=68=> b+c=136|(+)
(a+c)/2=97=> a+c=194|
 
2a+2b+2c= 416| :2
a+b+c=208
86+c=208=> c=122
a+136=208=> a=72
194+b=208=> b=14
 

Answer 2

[tex] \frac{a+b}{2} =43=\ \textgreater \ a+b=86\\ \frac{b+c}{2} =(b+c)/2=68=\ \textgreater \ b+c=136\\ \frac{a+c}{2}(a+c)/2=97=\ \textgreater \ a+c=194\\ a+b+c=208\\ a+136=208=\ \textgreater \ a=208-103=\ \textgreater \ a=72\\ b+194=208=\ \textgreater \ b=208-194=\ \textgreater \ b=14\\ c+86=208=\ \textgreater \ c=208-86=\ \textgreater \ c=122[/tex]