Question

1) Sa se scrie numarul 100 ca o suma de trei termeni invers proportionali cu
$\frac{2}{5} , \frac{4}{13} si \frac{6}{25}$ .
2)Sa se afle numerele x si y stiind ca sunt invers proportionale cu 0,(3) si 0.25 si ca x*y=300

Answer 1

prob.1
a+b+c=100
{a,b,c}ip{2/5,4/13,6/25}
2a/5=4b/13=6c/25=k
a=5k/2
b=13k/4
c=25k/6
inlocuim in a+b+c=100
5k/2+13k/4+25k/6=100
(30k+39k+50k)/12=100
119k=100·12
k=1200/119
a=5k/2=(5·1200/119)/2=6000/238=3000/119
b=13k/4=(13·1200/119)/4=3900/119
c=25k/6=(25·1200/119)/6=5000/119
verificare
3000/119+3900/119+5000+/19=100
11900/119=100
100=100

prob.2
0,(3)=3/9=1/3
0,25=25/100=1/4
xy=300
{x,y}ip{1/3,1/4}
x/3=y/4=k
x=3k
y=4k
xy=300
3k·4k=300
12k=300
k=300/12
k=25
x=3k=3·25=75
y=4k=4·25=100