\(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{6}\)và\(y+z+x=11\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)

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

\(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=\frac{x+y+z}{15+20+24}=\frac{11}{59}\)

\(suy\)\(ra\)\(\frac{x}{15}=\frac{11}{59}\Rightarrow x=\frac{11.15}{59}=\frac{165}{59}\)

\(\frac{y}{20}=\frac{11}{59}\Rightarrow y=\frac{11.20}{59}=\frac{220}{59}\)

\(\frac{z}{24}=\frac{11}{59}\Rightarrow z=\frac{24.11}{59}=\frac{264}{59}\)