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sa se afle x,y,z daca sunt invers proportionale cu 2,3,4 iar suma lor este 13

Răspuns :

Baiatul122001

2x=3y=4z=k

x+y+z=13

2x=k=>x=k/2

3y=k=>y=k/3

4z=k=>z=k/4

k/2+k/3+k/4=(6k+4k+3k)/12=13k/12

13k/12=13=>k=12

x=12/2=6

x=6

y=12/4=3

y=3

z=12/3=4

z=4

Silionioana
x,y,z i.p 2,3,4
x+y+z= 13
2x=3y=4z not k -> x=k/2
y=k/3
z=k/4
k/2+k/3+k/4=13 (aducem la accelasi numitor)
6k+4k+3k=156
13k=156
k=12
x=6
y=4
z=3
Proba: 6+4+3=13

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