Question

impartind pe 55 ip
3,5,7afla partile

Answer 1

x·3=y·5=z·7=k
⇒x·3=k⇒x=k/3;
⇒y·5=k⇒y=k/5;
⇒z·7=k⇒z=k/7;
⇒x+y+z=55
⇒k/3+k/5+k/7=55
⇒35k/105+21k/105+15k/105=55
⇒71k/105=55
⇒k=105·55/71
⇒k=81,33;
⇒x=27,11;
⇒y=16,26;
⇒z=11,61;