`@` `\text {Ans}`

`\downarrow`

Ta có:

`x/2 = y/3 = z/4`

`=>`\(\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{z}{4}\)

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

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

`=>`\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=3\)

`=>`\(x=2\cdot3=6,\) `y = 3*3 = 9, z = 4*3=12`

mình cần gấp ạ :)))

\(\Rightarrow\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}=\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

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

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{2x}{10.2}=\frac{3y}{15.3}=\frac{z}{21}=\frac{2x}{20}=\frac{3y}{45}=\frac{z}{21}=\frac{2x+3y+z}{20+45+21}=\frac{172}{86}=2\)

\(\frac{x}{10}=2\Rightarrow x=2.10=20\)

\(\frac{y}{15}=2\Rightarrow y=2.15=30\)

\(\frac{z}{21}=2\Rightarrow z=2.21=42\)

Vậy x=20 ; y=30 và z=42

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405

Áp dụng t/c dãy tỉ số bằng nhau:

a.

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\dfrac{14}{13}=\dfrac{52}{13}\\y=5.\dfrac{14}{13}=\dfrac{70}{13}\end{matrix}\right.\)

(Em có nhầm đề 26 thành 28 ko nhỉ, số xấu quá)

b.

\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{3x}{15}=\dfrac{-2y}{-8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=4.2=20\end{matrix}\right.\)

c.

\(\dfrac{x}{-3}=\dfrac{y}{-7}=\dfrac{2x}{-6}=\dfrac{4y}{-28}=\dfrac{2x+4y}{-6-28}=\dfrac{68}{-34}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.\left(-2\right)=6\\y=-7.\left(-2\right)=14\end{matrix}\right.\)

d.

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{-3y}{9}=\dfrac{-2z}{-8}=\dfrac{4x-3y-2z}{8+9-8}=\dfrac{16}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{16}{9}=\dfrac{32}{9}\\y=-3.\dfrac{16}{9}=-\dfrac{48}{9}\\z=4.\dfrac{16}{9}=\dfrac{64}{9}\end{matrix}\right.\)

\(\dfrac{x}{2}=\dfrac{y}{5};\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{20}\)

Áp dụng tc dstbn:

\(\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{20}=\dfrac{2x+3y-2z}{6\cdot2+3\cdot15-2\cdot20}=\dfrac{34}{17}=2\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=30\\z=40\end{matrix}\right.\)

Lời giải:

$\frac{x}{2}=\frac{y}{5}; \frac{y}{3}=\frac{z}{4}$

$\Rightarrow \frac{x}{6}=\frac{y}{15}=\frac{z}{20}$

Áp dụng TCDTSBN:

$\frac{x}{6}=\frac{y}{15}=\frac{z}{20}$

$=\frac{2x}{12}=\frac{3y}{45}=\frac{2z}{40}=\frac{2x+3y-2z}{12+45-40}=\frac{34}{17}=2$

$\Rightarrow x=2.6=12; y=2.15=30; z=2.20=40$

ta có \(\frac{x}{y}=\frac{7}{20}\Rightarrow\frac{x}{7}=\frac{y}{20}\Rightarrow\frac{x}{14}=\frac{y}{40}\Rightarrow\frac{2x}{28}=\frac{5y}{200}\left(1\right)\)

       \(\frac{y}{z}=\frac{5}{8}\Rightarrow\frac{y}{5}=\frac{z}{8}\Rightarrow\frac{y}{40}=\frac{z}{64}\Rightarrow\frac{5y}{200}=\frac{2z}{128}\left(2\right)\)

          \(\left(1\right)\&\left(2\right)\Rightarrow\frac{2x+5y-2z}{28+200-128}=\frac{100}{100}=1\)

                       \(\frac{2x}{28}=1\Rightarrow x=\frac{28.1}{2}=14\)

                          \(\frac{5y}{200}=1\Rightarrow y=\frac{200.1}{5}=40\)

                          \(\frac{2z}{128}=1\Rightarrow z=\frac{128.1}{2}=64\)

\(\frac{x}{y}=\frac{7}{20};\frac{y}{z}=\frac{5}{8}\Rightarrow\frac{x}{7}=\frac{y}{20};\frac{y}{5}=\frac{z}{8}\Rightarrow\frac{x}{35}=\frac{y}{100};\frac{y}{100}=\frac{z}{160}\Rightarrow\frac{x}{35}=\frac{y}{100}=\frac{z}{160}\)

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

\(\frac{x}{35}=\frac{y}{100}=\frac{z}{160}=\frac{2x+5y-2z}{2.35+5.100-2.160}=\frac{100}{250}\)= số lẽ sai đề     

Bài 1:

a) \(\frac{x-1}{0-2}=\frac{1,2}{1,5}\)

\(\Leftrightarrow\frac{1-x}{2}=\frac{4}{5}\)

\(\Leftrightarrow5-5x=8\)

\(\Leftrightarrow x=-\frac{3}{5}\)

b) Ta có: \(x=\frac{y}{2}=\frac{z}{3}=\frac{4x-3y+2z}{4-6+6}=\frac{16}{4}=4\)

\(\Rightarrow\hept{\begin{cases}x=4\\y=8\\z=12\end{cases}}\)

Bài 1:

c) \(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{21}=\frac{y}{14}\)

\(5y=7z\Leftrightarrow\frac{y}{7}=\frac{z}{5}\Leftrightarrow\frac{y}{14}=\frac{z}{10}\)

\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)

\(\Rightarrow\hept{\begin{cases}x=42\\y=28\\z=20\end{cases}}\)

d) \(x:y:z=3:5:2\Leftrightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{2}=\frac{5x-7y+5z}{15-35+10}=\frac{124}{-10}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{186}{5}\\y=-62\\z=-\frac{124}{5}\end{cases}}\)

Bài 1:

e) \(\frac{x}{3}=\frac{y}{5}=k\left(k\inℝ\right)\)

\(\Rightarrow\hept{\begin{cases}x=3k\\y=5k\end{cases}}\)

Ta có: \(x\cdot y=240\Leftrightarrow15k^2=240\)

\(\Leftrightarrow k^2=16\Rightarrow k=\pm4\)

=> \(\hept{\begin{cases}x=\pm12\\y=\pm20\end{cases}}\)