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Ta có:\(\frac{z}{x}=\frac{1}{6}\Rightarrow\frac{z}{1}=\frac{x}{6}\Rightarrow\frac{z}{2}=\frac{x}{12}\left(1\right)\)
\(\frac{y}{z}=\frac{3}{2}\Rightarrow\frac{y}{3}=\frac{z}{2}\left(2\right)\)
\(\frac{t}{x}=\frac{4}{3}\Rightarrow\frac{t}{4}=\frac{x}{3}\Rightarrow\frac{t}{16}=\frac{x}{12}\left(3\right)\)
Từ (1),(2) và (3)\(\Rightarrow\frac{z}{2}=\frac{x}{12}=\frac{y}{3}=\frac{t}{16}\)
\(\Rightarrow\frac{t}{y}=\frac{16}{3}\)
Vậy \(\frac{t}{y}=\frac{16}{3}\)
y/z .z/x=3/2.1/6=1/4 nên y/x bằng 1/4 hay x/y bằng 4
t/x .x/y=4/3.4 nên t/y =16/3
\(\frac{x-1}{3}=\frac{2y-1}{4}=\frac{z+2}{5}=\frac{y+t+3}{6}\)\(=\frac{x-1+2y-1+z+2-y-t-3}{3+4+5-6}\)
\(=\frac{x+y+z-t-3}{6}=\frac{1-3}{6}=-\frac{1}{3}\)
=> \(x-1=-1;2y-1=-\frac{4}{3};z+2=-\frac{5}{3};y+t+3=-2\)
=> \(x=0;y=-\frac{1}{6};z=-\frac{11}{3};t=-\frac{29}{6}\)
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 \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)
=> \(x=1\cdot15=15\)
\(y=1\cdot7=7\)
\(z=1\cdot3=3\)
\(t=1\cdot1=1\)
Ta có:
\(\frac{t}{x}=\frac{4}{3}\Rightarrow\frac{x}{t}=\frac{3}{4}\)
Vậy ta có:
\(\frac{x}{t}.\frac{y}{z}.\frac{z}{x}=\frac{3.3.1}{4.2.6}\)\(\Rightarrow\frac{y}{t}=\frac{3}{16}\Rightarrow\frac{t}{y}=\frac{16}{3}\)
Vậy \(\frac{t}{y}=\frac{16}{3}\)
Ta có: \(\frac{y}{z}=\frac{3}{2}\) => \(\frac{z}{y}=\frac{2}{3}\)
=> \(\frac{t}{x}.\frac{z}{y}.\frac{z}{x}=\frac{4}{3}.\frac{2}{3}.\frac{1}{6}\)
=> \(\frac{t}{y}=\frac{4}{27}\)