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Daryus9
a fost răspuns

det nr x y z stiind ca x/3=y/5=z/9 si xy+yz+zx=1392 plss nu stiu cum se fac astea

Răspuns :

Albatran
x/3=y/4=z/9=k
x=3k
y=5k
z=9k
 xy+yz+zx=1392

 3k*5k+5k*9k+9k*3k=1392

15k²+45k²+27k²=1392

87k²=1392
k²=1292:87=16
k²-16=0
(k-4)(k+4)=0
k1=4
k2=-4
pt k1=4
 x=3*4=12
y1=5*4=20
z1=9*4=36

ptk2=-4

x1=-12
y1=-20
z2=-36


(x;y;z) ∈{ (-12;-20;-36);(12;20;36)}

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