Question

determinați numerele x,y și z invers proporționale cu 3, 6 și 9 , știind că: ...​

Answer 1

Răspuns:

Explicație pas cu pas:

{x ; y ; z} i.p {3 ; 6 ; 9}

3x = 6y = 9z = k, unde k = coeficient de propoționalitate

$x = \frac{k}{3}$

$y = \frac{k}{6}$

$z = \frac{k}{9}$

a)

x + y + z = 28

$\frac{k}{3} + \frac{k}{6} + \frac{k}{9} = 28 \: \: \Big| \times 18$

6k + 3k + 2k = 504

11k = 504

$k = \frac{504}{11}$

$x = \frac{ \frac{504}{11} }{3} = \frac{504}{11} \times \frac{1}{3} = \frac{504 {}^{(3} }{33} = \boxed{\frac{168}{11} }$

$y = \frac{ \frac{504}{11} }{6} = \frac{504}{11} \times \frac{1}{6} = \frac{504 {}^{(6} }{66} = \boxed{\frac{84}{11}}$

$z = \frac{ \frac{504}{11} }{9} = \frac{504}{11} \times \frac{1}{9} = \frac{504 {}^{(9} }{99} = \boxed{\frac{56}{11}}$

b)

x + y + z = 42

$\frac{k}{3} + \frac{k}{6} + \frac{k}{9} = 42 \: \: \Big| \times 18$

6k + 3k + 2k = 756

11k = 756

$k = \frac{756}{11}$

$x = \frac{ \frac{756}{11} }{3} = \frac{756}{11} \times \frac{1}{3} = \frac{756 {}^{(3} }{33} = \boxed{\frac{252}{11} }$

$y = \frac{ \frac{756}{11} }{6} = \frac{756}{11} \times \frac{1}{6} = \frac{756 {}^{(6} }{66} = \boxed{ \frac{126}{11}}$

$z = \frac{ \frac{756}{11} }{9} = \frac{756}{11} \times \frac{1}{9} = \frac{756 {}^{(9} }{99} = \boxed{\frac{84}{11} }$

c)

x + y + z = 84

$\frac{k}{3} + \frac{ k}{6} + \frac{k}{9} = 84 \: \: \Big| \times 18$

6k + 3k + 2k = 1512

11k = 1512

$k = \frac{1512}{11}$

$x = \frac{ \frac{1512}{11} }{3} = \frac{1512}{11} \times \frac{1}{3} = \frac{1512 {}^{(3} }{33} = \boxed{\frac{504}{11}}$

$y= \frac{ \frac{1512}{11} }{6} = \frac{1512}{11} \times \frac{1}{6} = \frac{1512 {}^{(6} }{66} = \boxed{ \frac{252}{11} }$

$z = \frac{ \frac{1512}{11} }{9} = \frac{1512}{11} \times \frac{1}{9} = \frac{1512 {}^{(9} }{99} = \boxed{ \frac{168}{11} }$

Answer 2

Răspuns:

Explicație pas cu pas:

Determinați numerele x,y și z invers proporționale cu 3, 6 și 9 , știind că: ...​

x+y+z=28

x·3=y·6=z·9=k  ⇒x·3=k  ⇒x=k/3

                           y·6=k  ⇒y=k/6

                           z·9=k   ⇒z=k/9   Inlocuim valorile in suma data.  ⇒

a) x+y+z=28  ⇒k/3+k/6+k/9=28  Aducem la acelasi numitor.  ⇒(6k+3k+2k)/18=28  ⇒11k=18·28  ⇒11k=504  ⇒k=504/11  ⇒

x=k/3=504/(11·3)=168/11

y=k/6=504/(11·6)=84/11

z=k/9=504/(11·9)=56/11

Verificare: 168/11+84/11+56/11=308/11=28

b) x+y+z=42

k/3+k/6+k/9=42  Aducem la acelasi numitor.  ⇒(6k+3k+2k)/18=42  ⇒11k=18·42  ⇒11k= 756  ⇒k=756/11

x=k/3=756/(11·3)=252/11

y=k/6=756/(11·6)=126/11

z=k/9=756/(11·9)=84/11

Verificare: 252/11+126/11+84/11=462/11=42

c) x+y+z=84

k/3+k/6+k/9=84  Aducem la acelasi numitor.  ⇒(6k+3k+2k)/18=84  ⇒11k=18·84  ⇒11k= 1512  ⇒k=1512/11

x=k/3=1512/(11·3)=504/11

y=k/6=1512/(11·6)=252/11

z=k/9=1512/(11·9)=168/11

Verificare: 504/11+252/11+168/11=924/11=84