Question

Sa se afle nr x,y,z apartine Q ,stiind ca sunt invers proportionale cu numerele 0,8;0,6;0,4 si
x la puterea 2+y la puterea 2+z la puterea 2=1525 .
dau coroana+15 puncte
va rogggg

Answer 1

x*0,8=y*0,6=z*0,4=k
x*8/10=y*6/10=z*4/10=k/10 pt ca scap de numitor
x=k/8,y=k/6, z=k/4
(k/8)^2+(k/6)^2+(k/4)^2=1525
k^2/64+k^2/36+k^2/16=1525
numitorul comun e 576
amplific cu 9, 16 respectiv 36
9k^2/576+16k^2/576+36k^2/576=1525*576/576
renunt la numitor si adun termenii cu k^2
61k^2=1525*576
k^2=1525*576/61
k^2=14400
rezulta k=120
atunci x=120/8=15, y=120/6=20, z=120/4=30