\(\text{Ta có: }\)

\(x=4y=\frac{z-9}{125}=\frac{x+y+z-9}{1+0,25+125}=\frac{2029-9}{126,25}=\frac{2020}{126,25}=16\)

=>x=16

=>4y=16 Vậy y=4

=>z-9/125=16 Vậy z=2009

Giải:
Đặt \(x=4y=\dfrac{z-9}{125}=k\Rightarrow\left\{{}\begin{matrix}x=k\\y=\dfrac{1}{4}k\\z=125k+9\end{matrix}\right.\)

Mà \(x+y+z=2029\)

\(\Rightarrow k+\dfrac{1}{4}k+125k+9=2029\)

\(\Rightarrow\dfrac{505}{4}k=2020\)

\(\Rightarrow k=16\)

\(\Rightarrow\left\{{}\begin{matrix}x=16\\y=4\\z=2009\end{matrix}\right.\)

Vậy \(x=16;y=4;z=2009\)

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

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x+y-z}{2\cdot4+3-2}=\dfrac{9}{9}=1\)

Do đó: x=4; y=3; z=2

a.

$7x-2y=5x-3y$

$\Leftrightarrow 2x=-y$. Thay vào điều kiện số 2 ta có:

$-y+3y=20$

$2y=20$

$\Rightarrow y=10$. 

$x=\frac{-y}{2}=\frac{-10}{2}=-5$

b.

$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}$

$3y=4z-2y\Rightarrow 5y=4z\Rightarrow \frac{y}{4}=\frac{z}{5}$

$\Rightarrow \frac{x}{6}=\frac{y}{4}=\frac{z}{5}$

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{6}=\frac{y}{4}=\frac{z}{5}=\frac{x+y+z}{6+4+5}=\frac{45}{15}=3$

$\Rightarrow x=6.3=18; y=4.3=12; z=5.3=15$ 

c.

$3x=4y-2x$

$\Rightarrow 5x=4y\Rightarrow x=\frac{4}{5}y$

$3x=7z-4y$

$\Leftrightarrow \frac{12}{5}y=7z-4y$

$\Leftrightarrow \frac{32}{5}y=7z\Rightarrow z=\frac{32}{35}y$

Khi đó:

$x+y-2z=10$

$\frac{4}{5}y+y-2.\frac{32}{35}y=10$

$y.\frac{-1}{35}=10$

$y=-350$

$x=\frac{4}{5}y=\frac{4}{5}.(-350)=-280$

$z=\frac{32}{35}y=\frac{32}{35}.(-350)=-320$

1) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{7}=\frac{2x-4y+3z}{2.2-4.3+3.7}=\frac{-39}{13}=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=-3.2=-6\\y=-3.3=-9\\z=-3.7=-21\end{cases}}\)

2) \(9x=10y\Leftrightarrow\frac{x}{10}=\frac{y}{9},4y=3z\Leftrightarrow\frac{y}{9}=\frac{z}{12}\)

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

\(\Leftrightarrow\hept{\begin{cases}x=6.10=60\\y=6.9=54\\z=6.12=72\end{cases}}\)

3) \(3x=4y=6z\Leftrightarrow\frac{x}{4}=\frac{y}{3}=\frac{z}{2}=\frac{x-y+z}{4-3+2}=\frac{-9}{3}=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=-3.4=-12\\y=-3.3=-9\\z=-3.2=-6\end{cases}}\)

cố làm giùm mik dk

a)6x=4y=>x/4=y/6

4y=3z=>y/3=z/4=>y/6=z/8

x+y+z/4+6+8=18/18=1

x=4;y=6;z=8

b)3x=2y=>x/2=y/3

2y=z=>y/1=z/2=>y/3=z/6

x+y+z/2+3+6=99/11=9

x=18;y=27;z=54

tick dung nha

cách 1:

1. T/có: \(\frac{x}{4}\)=\(\frac{y}{6}\) ; \(\frac{y}{3}\)=\(\frac{z}{4}\) => \(\frac{x}{4}\)=\(\frac{y}{6}\) ; \(\frac{y}{6}\)=\(\frac{z}{8}\)=> Đặt \(\frac{x}{4}\)=\(\frac{y}{6}\)=\(\frac{z}{8}\)=k

=>x=4k, y=6k, z=8k

=> x+y+z=4k+6k+8k=18k=18

=>k=1

=>x=4k=4.1=4

Tương tự vs y và z nhak 

Câu 2 cũng dạng tuoeng tự

\(\Rightarrow\dfrac{3x}{60}=\dfrac{4y}{60}=\dfrac{5z}{60}\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x+y+z}{20+15+12}=\dfrac{141}{47}=3\\ \Rightarrow\left\{{}\begin{matrix}x=60\\y=45\\z=36\end{matrix}\right.\)

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

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x+y+z}{20+15+12}=\dfrac{141}{47}=3\)

Do đó: x=60; y=45; z=36

\(3x=4y=5z\text{ và }x+y+z=141\)

\(\Rightarrow\dfrac{3x}{60}=\dfrac{4y}{60}=\dfrac{5z}{60}\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}\text{ và }x+y+z=141\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau:}\)

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x+y+z}{20+15+12}=\dfrac{141}{47}=3\)

\(\Rightarrow x=20.3=60\)

\(y=15.3=45\)

\(z=12.3=36\)