Răspuns:
Explicație pas cu pas:
{x ; y ; z} i.p {3 ; 6 ; 9}
3x = 6y = 9z = k, unde k = coeficient de propoționalitate
[tex]x = \frac{k}{3} \\ [/tex]
[tex]y = \frac{k}{6} \\ [/tex]
[tex]z = \frac{k}{9} \\ [/tex]
a)
x + y + z = 28
[tex] \frac{k}{3} + \frac{k}{6} + \frac{k}{9} = 28 \: \: \Big| \times 18 \\ [/tex]
6k + 3k + 2k = 504
11k = 504
[tex]k = \frac{504}{11} \\ [/tex]
[tex]x = \frac{ \frac{504}{11} }{3} = \frac{504}{11} \times \frac{1}{3} = \frac{504 {}^{(3} }{33} = \boxed{\frac{168}{11} }\\ [/tex]
[tex]y = \frac{ \frac{504}{11} }{6} = \frac{504}{11} \times \frac{1}{6} = \frac{504 {}^{(6} }{66} = \boxed{\frac{84}{11}} \\ [/tex]
[tex]z = \frac{ \frac{504}{11} }{9} = \frac{504}{11} \times \frac{1}{9} = \frac{504 {}^{(9} }{99} = \boxed{\frac{56}{11}} \\ [/tex]
b)
x + y + z = 42
[tex] \frac{k}{3} + \frac{k}{6} + \frac{k}{9} = 42 \: \: \Big| \times 18 \\ [/tex]
6k + 3k + 2k = 756
11k = 756
[tex]k = \frac{756}{11} \\ [/tex]
[tex]x = \frac{ \frac{756}{11} }{3} = \frac{756}{11} \times \frac{1}{3} = \frac{756 {}^{(3} }{33} = \boxed{\frac{252}{11} } \\ [/tex]
[tex]y = \frac{ \frac{756}{11} }{6} = \frac{756}{11} \times \frac{1}{6} = \frac{756 {}^{(6} }{66} = \boxed{ \frac{126}{11}} \\ [/tex]
[tex]z = \frac{ \frac{756}{11} }{9} = \frac{756}{11} \times \frac{1}{9} = \frac{756 {}^{(9} }{99} = \boxed{\frac{84}{11} } \\ [/tex]
c)
x + y + z = 84
[tex] \frac{k}{3} + \frac{ k}{6} + \frac{k}{9} = 84 \: \: \Big| \times 18 \\ [/tex]
6k + 3k + 2k = 1512
11k = 1512
[tex]k = \frac{1512}{11} \\ [/tex]
[tex] x = \frac{ \frac{1512}{11} }{3} = \frac{1512}{11} \times \frac{1}{3} = \frac{1512 {}^{(3} }{33} = \boxed{\frac{504}{11}} \\ [/tex]
[tex]y= \frac{ \frac{1512}{11} }{6} = \frac{1512}{11} \times \frac{1}{6} = \frac{1512 {}^{(6} }{66} = \boxed{ \frac{252}{11} } \\ [/tex]
[tex]z = \frac{ \frac{1512}{11} }{9} = \frac{1512}{11} \times \frac{1}{9} = \frac{1512 {}^{(9} }{99} = \boxed{ \frac{168}{11} } \\ [/tex]
