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

=> x=8,y=6,z=18

b, \(\hept{\begin{cases}\frac{x}{y}=\frac{9}{7}\Rightarrow\frac{x}{9}=\frac{y}{7}\\\frac{y}{z}=\frac{7}{3}\Rightarrow\frac{y}{7}=\frac{z}{3}\end{cases}\Rightarrow\frac{x}{9}=\frac{y}{7}=\frac{z}{3}=\frac{x-y+z}{9-7+3}=\frac{-15}{5}=-3}\)

=> x=-27,y=-21,z=-9

c, \(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\Rightarrow\frac{x}{33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

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

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

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

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

a, Ta có: \(\frac{x}{4}=\frac{y}{3}=\frac{z}{9}\Rightarrow\frac{x}{4}=\frac{3y}{9}=\frac{4z}{36}\)

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

\(\Rightarrow x=2.4=8\)

\(\Rightarrow y=2.3=6\)

\(\Rightarrow z=2.9=18\)

b, Ta có: \(\frac{x}{y}=\frac{10}{9}\Rightarrow\frac{x}{10}=\frac{y}{9}\)

\(\frac{y}{z}=\frac{3}{4}\Rightarrow\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{y}{9}=\frac{z}{12}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)

\(\Rightarrow x=6.10=60\)

\(\Rightarrow y=6.9=54\)

\(\Rightarrow z=6.12=72\)

c, Ta có: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\)

\(\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\)

\(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\Rightarrow\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}\)

\(\Rightarrow\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

\(\Rightarrow x=3.9=27\)

\(\Rightarrow y=3.12=36\)

\(\Rightarrow z=3.20=60\)

các bạn giúp mình vs

Ta có :

\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)

\(\Leftrightarrow\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}\)

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

\(\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}=\frac{12\left(x+y+z\right)}{18+16+15}=\frac{12\cdot49}{49}=12\) ( do \(x+y+z=49\) )

\(\Rightarrow\hept{\begin{cases}\frac{12x}{18}=12\\\frac{12y}{16}=12\\\frac{12z}{15}=12\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=18\\y=16\\z=15\end{cases}}\) ( thỏa mãn )

Vậy : \(\left(x,y,z\right)=\left(18,16,15\right)\)

\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)

\(\Rightarrow\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}\)

\(\Rightarrow\frac{12x+12y+12z}{18+16+15}=\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)     

\(\Rightarrow\frac{12\left(x+y+z\right)}{49}=\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\) có  x + y + z = 49

\(\Rightarrow\frac{12\cdot49}{49}=12=\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)

\(\Rightarrow\hept{\begin{cases}2x=36\\3y=48\\4z=60\end{cases}\Rightarrow\hept{\begin{cases}x=18\\y=16\\z=15\end{cases}}}\)

a) Ta có: x - 3y + 4z = 62

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

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

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

\(\Rightarrow\begin{cases}\frac{x}{4}=2\Rightarrow x=2\cdot4=8\\\frac{y}{3}=2\Rightarrow y=2\cdot3=6\\\frac{z}{9}=2\Rightarrow z=2\cdot9=18\end{cases}\)

Vậy x = 8; y = 6 và z = 18

a) \(\left(\frac{1}{3}.x\right):\frac{2}{3}=4\frac{3}{8}\)

⇒ \(\left(\frac{1}{3}.x\right)=\frac{35}{8}.\frac{2}{3}\)

⇒ \(\frac{1}{3}.x=\frac{35}{12}\)

⇒ \(x=\frac{35}{12}:\frac{1}{3}\)

⇒ \(x=\frac{35}{4}\)

Vậy \(x=\frac{35}{4}.\)

c) \(\left|x-\frac{2}{5}\right|+\frac{3}{4}=\frac{11}{4}\)

⇒ \(\left|x-\frac{2}{5}\right|=\frac{11}{4}-\frac{3}{4}\)

⇒ \(\left|x-\frac{2}{5}\right|=2\)

⇒ \(\left[{}\begin{matrix}x-\frac{2}{5}=2\\x-\frac{2}{5}=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=2+\frac{2}{5}\\x=\left(-2\right)+\frac{2}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{12}{5}\\x=-\frac{8}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{12}{5};-\frac{8}{5}\right\}.\)

d) Ta có \(x:y:z=2:3:4\)

=> \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}.\)

=> \(\frac{x}{2}=\frac{y}{3}=\frac{2z}{8}\) và \(x+y-2z=3.\)

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

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

\(\left\{{}\begin{matrix}\frac{x}{2}=-1=>x=\left(-1\right).2=-2\\\frac{y}{3}=-1=>y=\left(-1\right).3=-3\\\frac{z}{4}=-1=>z=\left(-1\right).4=-4\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(-2;-3;-4\right).\)

Còn câu e) thì bạn làm tương tự như câu trên nhé.

Chúc bạn học tốt!

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

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

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

Theo đề ta có:

\(\frac{x}{y}=\frac{9}{7};\frac{y}{z}=\frac{7}{3}\)

=>  \(\frac{x}{9}=\frac{y}{7};\frac{y}{7}=\frac{z}{3}\)

Hay: \(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}\)

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

\(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}=\frac{x-y+z}{9-7+3}=\frac{-15}{5}=-3\)

=> \(\frac{x}{9}=-3\)

  \(\frac{y}{7}=-3\)

\(\frac{z}{3}=-3\)

=> x = -27

  y = -21

x= -9

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