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

\(\frac{x-1}{2}=\frac{2\left(x-1\right)}{2}=\frac{2x-2}{2}\)

\(\frac{y-2}{3}=\frac{3\left(y-2\right)}{3}=\frac{3y-6}{3}\)

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

\(=\frac{2x+3y-z+3-2-6}{9}=\frac{50-5}{9}=\frac{45}{9}=5\)

=>x-1=5.2=10

=>x=11

y-2=5.3=15

=>y=17

z-3=5.4=20

=>z=23

vậy (x;y;z)=(11;17;23)

đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)

=>\(x=2x+1;y=3k+2;z=4k+3\)

thay \(x=2k+1;y=3k+2;z=4k+3\)vào 2x+3y-z=50 ta được:

2(2k+1)+3(3k+2)-(4k+3)=50

<=>4k+2+9k+6-4k-3=50

<=>9k+5=50

<=>9k=45

<=>k=5

=>x=2.5+1=11

y=3.5+2=17

z=4.5+3=23

x=3;y=5;z=7

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\Rightarrow x=2k+1;y=3k+2;z=4k+3\)

ta có: 2(2k+1)+3(3k+2)-(4k+3)=50

4k+2+9k+6-4k-3=50

9k+5=50

9k=50-5=45

k=45:9=5

x=5x2+1=11

y=5x3+2=17

z=5x4+3=23

\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{5}\end{cases}}\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{x+y+z}{2+3+5}=\frac{50}{10}=5\)

\(\Rightarrow\hept{\begin{cases}x=10\\y=15\\z=25\end{cases}}\)

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

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

\(\frac{x}{4}=\frac{y}{6}=\frac{z}{15}=\frac{x+y+z}{4+6+15}=\frac{50}{25}=2\)

Suy ra: \(\frac{x}{4}=2\Rightarrow x=8\)

\(\frac{y}{6}=3\Rightarrow y=12\)

\(\frac{z}{15}=2\Rightarrow z=30\)

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

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

\(\frac{x}{4}=\frac{y}{6}=\frac{z}{15}=\frac{x+y+z}{4+6+15}=\frac{50}{25}=2\)

suy ra:

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

\(\frac{x}{6}=2\Rightarrow x=12\)

\(\frac{z}{15}=2\Rightarrow z=30\)

x/4=y/6=z/15

áp dụng tính chất dãy tỉ số bằng nhau ta có 

x/4=y/6=z/15=x+y+z/4+6+15=50/25=2

=>x/4=2=>x=8

=>y/6=2=>y=12

=>z/15=2=>z=30

\(\frac{x}{2}=\frac{y}{3};\frac{y}{2}=\frac{z}{5}\) va xy+z=50

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

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

\(\frac{x}{4}=\frac{y}{6}=\frac{z}{15}=\frac{x+y+z}{4+6+15}=\frac{50}{25}=2\)

Suy ra : \(\frac{x}{4}=2\Rightarrow x=2.4=8\)

\(\frac{y}{6}=2\Rightarrow y=2.6=12\)

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

Vay : x=8 ; y=12 va z=30