\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

=>x=165,y=20,z=25

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

5x = 8y = 20z suy ra 5x/40 = 8y/40 = 20z/40 suy ra x/8 = y/5 = z/2 . ap dung tinh chat day ty so bang nhau ta co x/8 =y/5 =z/2 =          x-y-z/8-5-2 =3 /1 =3 . tu x/8 =3 suy ra x =24 . tu y /6=3 suy ra y=18 . tu z/2 =3 suy ra z =6 . vay x = 24 , y = 18 , z = 6

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

a, 5x = 8y => \(\frac{x}{8}=\frac{y}{5}\)

8y = 20z => 2y = 5z => \(\frac{y}{5}=\frac{z}{2}\)

=> \(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}=\frac{x-y-z}{8-5-2}=\frac{3}{1}=3\)

=> x = 24,y = 15,z = 6

b, \(\frac{6}{11}x=\frac{9}{2}y\)=> \(\frac{12x}{22}=\frac{99y}{22}\)=> 12x = 99y => 4x = 33y => \(\frac{x}{33}=\frac{y}{4}\)

\(\frac{9}{2}y=\frac{18}{5}z\)=> \(\frac{45y}{10}=\frac{36z}{10}\)=> 45y = 36z => 5y = 4z => \(\frac{y}{4}=\frac{z}{5}\)

=> \(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{120}{-24}=-5\)

=> x = -165 , y = -20 , z = -25

c, Đặt : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)=> x = 12k , y = 9k , z = 5k

=> xyz = 12k . 9k . 5k

=> xyz = 540k³

=> 540k³ =20

=> k³ = 20/540

=> k³ = 1/27

=> k = 1/3

Do đó : x= 4 , y = 3 , z = 5/3

mình mới học lớp 6

= 344 + 278 + 643 + 937 + 463 + 33

= 463 + 345 + 678 + 57 + 845

= 345 + 555 + 556

= 345

6x/11=9y/2=18z/5

=>-18x/-33=18y/4=18z/5

theo tính chất dãy các tỉ số = nhau, đẳng thức trên =

-18x+18y+18z/-33+4+5=18.(-x+y+z/-24)=18.(-5)=-90

=>x=-165;y=-20;z=-25

\(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\Leftrightarrow\frac{-6x}{-11}=\frac{9y}{2}=\frac{18z}{5}\Rightarrow\frac{-x}{\frac{-11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\)

\(\Rightarrow\frac{-x+y+z}{\frac{-11}{6}+\frac{2}{9}+\frac{5}{18}}=\frac{-120}{\frac{-4}{3}}=90\)

\(-x=90\times\frac{-11}{6}=-165\Rightarrow x=165\)

\(y=90\times\frac{2}{9}=20\)

\(z=90\times\frac{5}{18}=25\)

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{-x+y+z}{-\frac{11}{6}+\frac{2}{9}+\frac{5}{18}}=\frac{-120}{-\frac{4}{3}}=90\)

\(\frac{x}{\frac{11}{6}}=90\Rightarrow x=90\times\frac{11}{6}=165\)

\(\frac{y}{\frac{2}{9}}=90\Rightarrow y=90\times\frac{2}{9}=20\)

\(\frac{z}{\frac{5}{18}}=90\Rightarrow x=90\times\frac{5}{18}=25\)

Giải:

Ta có: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\Rightarrow\frac{-x}{\frac{-11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{-x}{\frac{-11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{-x+y+z}{\frac{-11}{6}+\frac{2}{9}+\frac{5}{18}}=\frac{-120}{\frac{-4}{3}}=90\)

+) \(\frac{x}{\frac{11}{6}}=90\Rightarrow x=165\)

+) \(\frac{y}{\frac{2}{9}}=90\Rightarrow y=20\)

+) \(\frac{z}{\frac{5}{18}}=20\Rightarrow z=25\)

Vậy bộ số \(\left(x,y,z\right)\) là: \(\left(165,20,25\right)\)

Tự nhiên máy mk bị restart nên mk gửi trả lời hơi chậm nhé!

\(\frac{x}{4}=\frac{y}{3}=\frac{z}{9}=\)\(\frac{x}{4}=\frac{3y}{9}=\frac{4z}{36}\)

Áp dụng tính chất của dãy tủ số bằng nhau ta có:

\(\frac{x}{4}=\frac{3y}{9}=\frac{4z}{36}=\frac{x-3y+4z}{4-9+36}=\frac{62}{31}=2\)

\(\frac{x}{4}=2=>x=8\)

\(\frac{3y}{9}=2=>y=6\)

\(\frac{4z}{36}=2=>z=18\)

Ta có: a) \(\hept{\begin{cases}\frac{x}{4}=\frac{y}{3}=\frac{z}{9}\\x-3y+4x=62\end{cases}\Rightarrow\frac{x-3y+4z}{4-9+36}=\frac{62}{31}=2}\)

\(\Rightarrow\hept{\begin{cases}x=2.4=8\\y=2.3=6\\z=2.9=18\end{cases}}\)

a) Ta có: x/4=y/3=z/9=x-3y+4z/4-9+36=62/31=2

 x/4=2 => x=2.4=8

 y/3=2 => y=2.3=6

 z/9=2 => z=2.9=18

Vậy x=8; y=6; z=18.

ta co : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\) va x.y.z=20

Dat : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)

x=12k³

y=9k³

z=5k³

x.y.z=540k³

20    = 540k³

k³     =27

k       = +-3

Voi : \(k=3\Rightarrow x=36;y=27;z=15\)

Voi :\(k=-3\Rightarrow x=-36;y=-27;z=-15\)

a) Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)

=>x=12k;y=9k;z=5k

Thay x=12k;y=9k;z=5k vào biểu thức x.y.z=20 ta được

(12k)(9k)(5k)=20

    12k.9k.5k=20

    540.\(k^3\)=20

           k\(^3\)=\(\frac{1}{27}\)

=>k=\(\frac{1}{3}\)

=>\(x=\frac{1}{3}.12=4\)

    \(y=\frac{1}{3}.9=3\)

   \(z=\frac{1}{3}.5=\frac{5}{3}\)

Vậy x=4;y=3;z=\(\frac{5}{3}\)

b)Ta có:

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{2}z\)=>\(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\)=>\(\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)=>\(\frac{6x}{198}=\frac{9y}{36}=\frac{18z}{90}\)

=>\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

=>\(\frac{x}{33}=5\)=>\(x=5.33=165\)

     \(\frac{y}{4}=5\)=>\(y=5.4=20\)

     \(\frac{z}{5}=5\)=>\(z=5.5=25\)

Vậy x=165;y=20;z=25

Ta có: \(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\Leftrightarrow\frac{-18x}{-33}=\frac{18y}{4}=\frac{18z}{5}\)

Áp dụng t/c của dãy tỉ số bằng nhau ta có:

\(\frac{-18x}{-33}=\frac{18y}{4}=\frac{18z}{5}=\frac{18\left(-x+y+z\right)}{-33+4+5}=\frac{18\cdot\left(-120\right)}{-24}=90\)

Do đó: 

 \(\frac{-18x}{-33}=90\Leftrightarrow x=165\)

\(\frac{18y}{4}=90\Leftrightarrow y=20\)

\(\frac{18z}{5}=90\Leftrightarrow z=25\)