Question

Determinati numerele x,y ,z apartine lui Q+ daca ele sunt direct proporționale cu
a) 4,8,6 si diferenta dintre cel mai mare si cel mai mic este 12
b) 2,9,5 si produsul celor doua numere mai mici este 90
c) 1 supra 2,1 supra 4 si 1 supra 3 si suma dintre cele două nr mai mari este 25
d)0,6 ;0.(6);0.65 si media aritmetică a celor 3 nr este 5.75
e)1/2;2/3;0.25 si x*y*z=2 supra 81

Answer 1

a)
x=4k
y=8k
z=6k
y-x=12
8k-4k=12
4k=12
k=3
x=4k=4*3=12
y=8k=8*3=24
z=6k=6*3=18

b)
x=2k
y=9k
z=5k
x*z=90
2k*5k=90
10k²=90
k²=9
k=3
x=2k=2*3=6
y=9k=9*3=27
z=5k=5*3=15

c
x=k/2
y=k/4
z=k/3
x+z=25
k/2+k/3=25
3k+2k=150
5k=150
k=30
x=k/2=30/2=15
y=k/4=30/4=7,5
z=k/3=30/3=10

d
0,6=6/10=3/5
0,(6)=6/9=2/3
0,65=65/100=13/20
x=3k/5
y=2k/3
z=13k/20
(x+y+z)/3=5,75=575/100=23/4
x+y+z=69/4
3k/5+2k/3+13k/20=69/4
36k+40k+39k=1035
115k=1035
k=1035/115=9
x=3k/5=3*9/5=5,4
y=2k/3=2*9/3=6
z=13k/20=13*9/20=5,85

e
x=k/2
y=2k/3
z=k/4
x*y*z=2/81
k/2 *2k/3 * k/4=2/81
k³/12=2/81
k³=24/81
k³=24/81=8/27
k=2/3
x=k/2=2/3 *1/2=1/3
y=2k/3=2/3 *2/3=4/9
z=k/4=2/3* 1/4=2/12