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

\(\frac{z}{x}=\frac{1}{6}\Rightarrow\frac{x}{z}=6\)

\(\Rightarrow\frac{t}{x}.\frac{x}{z}=\frac{t}{z}=8\)

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

\(\Rightarrow\frac{t}{z}.\frac{z}{y}=\frac{t}{y}=\frac{16}{3}\)

thank you nha !!

Bài 1: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12};\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\)

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

Áp dụng tính chất của dãy tỉ số bằng nhau: \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2z}{18}=\frac{3y}{36}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

=>x=27;z=36;z=60

Bài 2: \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=5k\end{cases}}\Rightarrow xy=2k.5k=10k^2=40\Rightarrow k^2=4\Rightarrow\hept{\begin{cases}k=-2\\k=2\end{cases}}\)

+)k=-2 => x=-4;y=-5

+)k=2 => x=4;y=5

Vậy x=-4;y=-5 hoặc x=4;y=5

Ta có: \(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\)

\(=\frac{3\left(x-1\right)}{6}=\frac{4\left(y+3\right)}{16}=\frac{5\left(z-5\right)}{30}\)

\(=\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}\)\(=\frac{5z-25-3x+3-4y-12}{6-16-30}\)\(=\frac{\left(5z-3x-4y\right)-\left(25-3+12\right)}{-40}\)\(=\frac{50-34}{-40}=\frac{16}{-40}=\frac{2}{-5}\)

+) \(\frac{x-1}{2}=\frac{-2}{5}\Rightarrow5\left(x-1\right)=-4\Rightarrow x-1=\frac{-4}{5}\)\(\Rightarrow x=\frac{-4}{5}+1=\frac{1}{5}\)

+)\(\frac{y+3}{4}=\frac{-2}{5}\Rightarrow5\left(y+3\right)=-8\Rightarrow y+3=\frac{-8}{5}\)\(\Rightarrow y=\frac{-8}{5}-3=\frac{-23}{5}\)

+)\(\frac{z-5}{6}=\frac{-2}{5}\Rightarrow5\left(z-5\right)=-12\Rightarrow z-5=\frac{-12}{5}\)\(\Rightarrow z=\frac{-12}{5}+5=\frac{13}{5}\)

Vậy...

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