Question

Sa se scrie nr 100 ca o suma de trei termeni : a) invers proportionali cu 2/5 ; 4/13 ; 6/25

Answer 1

insemna ca vrem sa gasim a, b, c invers prop cu 2/5, 4/13 si 6/25
a+b+c=100

a*2/5=b*4/13=c*6/25=k
⇒ a=5k/2, b=13k/4 si c=25k/6
5k/2+13k/4+25k/6=100
aduc la acelasi numitor (care e 12)
amplific cu 6 , 3 , 2 respectiv 12
30k+39k+50k=1200
119k=1200
k=1200/119
a=5*1200/2*119=3000/119
b=13*1200/4*119=3900/119
c=25*1200/6*119=5000/119

Answer 2

a·2/5=b·4/13=c·6/25=k
⇒a=k/2/5⇒a=k·5/2⇒a=5k/2;
⇒b=k/4/13⇒b=k·13/4⇒b=13k/4;
⇒c=k/6/25⇒c=k·25/6⇒c=25k/6;
⇒a+b+c=100
⇒5k/2+13/4+25k/6=100
⇒30k/12+39k/12+50k/12=100
⇒119k/12=100⇒119k=12·100⇒119k=1200⇒k=1200/119;
⇒a=5·1200/119/2⇒a=6000/119/2⇒a=6000/119·1/2⇒a=6000/238;
⇒b=13·1200/119/2⇒b=15600/119/2⇒b=15600/119·1/2⇒b=7800/238;
⇒c=25·1200/119/2⇒c=30000/119/2⇒c=30000/119·1/2⇒c=10000/238;