1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)

\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)

=>\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)

=>\(x=3\cdot20=60\)

    \(y=3\cdot24=72\)

    \(z=3\cdot21=63\)

3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)

=> \(x=1\cdot15=15\)

     \(y=1\cdot7=7\)

     \(z=1\cdot3=3\)

     \(t=1\cdot1=1\)

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

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)(đk x+y+z\(\ne0\)

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

\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=0,5\)

\(\Rightarrow y+z=0,5-x,x+z=0,5-y,x+y=0,5-z\)

\(\Rightarrow\frac{0,5-x+1}{x}=2\Rightarrow\frac{1,5-x}{x}=2\Rightarrow1,5-x=2x\Rightarrow3x=1,5\Rightarrow x=\frac{1}{2}\)

\(\Rightarrow\frac{0,5-y+2}{y}=2\Rightarrow\frac{2,5-y}{y}=2\Rightarrow2,5-y=2y\Rightarrow3y=2,5\Rightarrow y=\frac{5}{6}\)

\(\Rightarrow z=0,5-\frac{1}{2}-\frac{5}{6}=-\frac{5}{6}\)

Vậy \(x=\frac{1}{2},y=\frac{5}{6},z=-\frac{5}{6}\)

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\)\(\frac{x+y-3}{z}=\frac{2\left(x+y+z\right)}{x+y+z}=\frac{1}{x+y+z}\)

\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)

\(\frac{y+z+1}{x}+1=\frac{\frac{3}{2}}{x}=3\Rightarrow x=\frac{1}{2}\)

Tương tự suy ra \(y=\frac{5}{6},z=-\frac{5}{6}\)

k cho mình nha!

Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+y-3}{z}=\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{1}{x+y+z}=\frac{x+y-3+y+z+1+x+z+2}{x+y+z}=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\frac{1}{x+y+z}=2\)

\(\Rightarrow x+y+z=\frac{1}{2}\)

Xét \(\frac{x+y-3}{z}=2\)

\(\Rightarrow x+y-3=2z\)

\(\Rightarrow x+y+z-3=3z\)

\(\Rightarrow\frac{1}{2}-3=3z\)

\(\Rightarrow\frac{-5}{2}=3z\)

\(\Rightarrow z=\frac{-5}{6}\)

Xét \(\frac{y+z+1}{x}=2\)

\(\Rightarrow y+z+1=2x\)

\(\Rightarrow x+y+z+1=3x\)

\(\Rightarrow\frac{1}{2}+1=3x\)

\(\Rightarrow\frac{3}{2}=3x\)

\(\Rightarrow x=\frac{1}{2}\)

Xét \(\frac{x+z+2}{y}=2\)

\(\Rightarrow x+z+2=2y\)

\(\Rightarrow x+y+z+2=3y\)

\(\Rightarrow\frac{1}{2}+2=3y\)

\(\Rightarrow\frac{5}{2}=3y\)

\(\Rightarrow y=\frac{5}{6}\)

Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{1}{2};\frac{5}{6};\frac{-5}{6}\right)\)

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

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

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

= \(\frac{2\left(x+y+z\right)}{x+y+z}=2\)(Do x + y + z \(\ne\)0)

=> \(\frac{1}{x+y+z}=2\) => x + y+ z = 1/2 

=> \(\frac{z+y+1}{x}=2\) => \(z+y+1=2x\)  => z + y + x = 3x - 1 => 3x - 1 = 1/2 => 3x = 3/2 => x = 1/2

=> \(\frac{x+z+2}{y}=2\) => x + z + 2 = 2y => x + y + z + 2 = 3y => 3y = 5/2 => y = 5/6

=> \(\frac{x+y-3}{z}=2\) => x +y - 3 = 2z => x + y + z - 3 = 3z => 3z = -5/2 => z = -5/6

Vậy ...

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

\(\frac{z+y+1+x+z+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\frac{1}{x+y+z}=\frac{2}{1}\)

hay \(\frac{x+y+z}{1}=\frac{1}{2}=0,5\)

\(\Rightarrow x+y+z=0,5\)

\(\Rightarrow y+z=0,5-x\)

        \(x+z=0,5-y\)

        \(x+y=0,5-z\)

+ Ta có : 

\(\frac{z+y+1}{x}=\frac{0,5-x+1}{x}=2\)

\(\Rightarrow1,5-x=2x\)

             \(3x=1,5\)

                \(x=0,5\)

+ Ta có : \(\frac{x+z+2}{y}=\frac{0,5-y+2}{y}=2\)

\(\Rightarrow2,5-y=2y\)

      \(3y=2,5\)

           \(y=\frac{5}{6}\)

+ Ta có : 

\(\frac{x+y-3}{z}=\frac{0,5-z-3}{z}=2\)

\(\Rightarrow-2,5-z=2z\)

\(3z=-2,5\)

     \(z=-\frac{5}{6}\)

Vậy \(x=0,5;y=\frac{5}{6};z=-\frac{5}{6}\)

Áp dụng tính chất dãy số bằng nhau ta có:

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

\(=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\frac{2\left(x+y+\right)}{x+y+z}=2\)

Mà: 

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

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

\(\Rightarrow\hept{\begin{cases}y+z+1=2x\left(1\right)\\x+z+2=2y\left(2\right)\\x+y-3=2z\left(3\right)\end{cases}}\)

\(x+y+z=\frac{1}{2}\left(4\right)\)

Ta có:

\(x+y+z=\frac{1}{2}\Rightarrow y+z=\frac{1}{2}-x\) (thay 1 vào được)

\(\frac{1}{2}-x+1=2x\Rightarrow\frac{3}{2}=3x\Rightarrow x=\frac{1}{2}\)

\(x+y+z=\frac{1}{2}\Rightarrow x+z=\frac{1}{2}-y\left(thay\left(2\right)\right)\)

\(\frac{1}{2}-y+2=\frac{2y\Rightarrow5}{2}=3y\Rightarrow y=\frac{5}{6}\)

\(x+y+z=\frac{1}{2}+\frac{5}{6}+z=\frac{1}{2}\)

\(\frac{\Rightarrow4}{3}+z=\frac{1}{2}\Leftrightarrow z=-\frac{5}{6}\)

Vậy.....

\(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)

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

\(=\frac{y+z+z+x+x+y+1+2-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)

\(\frac{y+z+1}{x}=2\)

\(\Rightarrow y+z+1=2x\)

\(x+y+z+1=3x\Rightarrow\frac{3}{2}=3x\)

Tương tự với mấy cái khác bạn tính được x,y,z

Áp dụng tính chất dãy tỉ số bằng nhau

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

\(\Rightarrow\frac{1}{x+y+z}=\frac{2x+2y+2z}{x+y+z}\)

\(\Rightarrow1=2\left(x+y+z\right)\)

\(\Rightarrow x+y+z=\frac{1}{2}\left(1\right)\)

Thay vào đề đc :

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

\(\Rightarrow\hept{\begin{cases}y+z+1=2x\left(2\right)\\z+x+2=2y\left(3\right)\\x+y-3=2z\left(4\right)\end{cases}}\)

Từ (2) => x + y + z + 1 = 3x

Thay (1) vào đc  \(\frac{1}{2}+1=3x\)

                   \(\Leftrightarrow3x=\frac{3}{2}\)

                  \(\Leftrightarrow x=\frac{1}{2}\)

Từ (3) => x + y + z + 2 =  3y

Thay (1) vào đc \(\frac{1}{2}+2=3y\)

                \(\Leftrightarrow y=\frac{5}{6}\)

Khi đó \(z=\frac{1}{2}-x-y=\frac{1}{2}-\frac{1}{2}-\frac{5}{6}=-\frac{5}{6}\)

Vậy \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)