\(\frac{x}{y}=\frac{2}{3}\Rightarrow\frac{x}{2}=\frac{y}{3};\frac{y}{z}=\frac{1}{4}\Rightarrow\frac{y}{1}=\frac{z}{4}\Rightarrow\frac{y}{3}=\frac{z}{12}\)
=>x=2k;y=3k;z=12k
thay vào ta có:
\(\frac{1}{2k}+\frac{1}{3k}+\frac{1}{12k}=1\)
\(\Rightarrow\frac{1}{2}.\frac{1}{k}+\frac{1}{3}.\frac{1}{k}+\frac{1}{12}.\frac{1}{k}=1\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{12}\right)\frac{1}{k}=1\)
\(\Rightarrow\frac{11}{12}.\frac{1}{k}=1\Rightarrow\frac{1}{k}=\frac{1}{\frac{11}{12}}\)
\(\Rightarrow x=\frac{11}{6};y=\frac{11}{4};z=11\)
\(\frac{x}{y}=\frac{2}{3}\Rightarrow\frac{x}{2}=\frac{y}{3};\frac{y}{z}=\frac{1}{4}\Rightarrow\frac{y}{1}=\frac{z}{4}\Rightarrow\frac{y}{3}=\frac{z}{12}\)
\(\Rightarrow x=2k;y=3k;z=12k\)
Thay vào ta có:
\(\frac{1}{2k}+\frac{1}{3k}+\frac{1}{12k}=1\)
\(\Rightarrow\frac{1}{2}.\frac{1}{k}+\frac{1}{3}.\frac{1}{k}+\frac{1}{12}.\frac{1}{k}=1\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{12}\right)\frac{1}{k}=1\)
\(\Rightarrow\frac{11}{12}.\frac{1}{k}=1\Rightarrow\frac{1}{k}=\frac{1}{\frac{11}{12}}\)
\(\Rightarrow x=\frac{11}{6};y=\frac{11}{4};z=11\)
