Question

\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}biet\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\)

Answer 1

\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=>4x=6y=>x=\dfrac{3y}{2}\)\(=>4z=3y=>z=\dfrac{3y}{4}\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3< =>\dfrac{1}{\dfrac{3y}{2}}+\dfrac{1}{y}+\dfrac{1}{\dfrac{3y}{4}}=3\)

\(< =>\dfrac{2}{3y}+\dfrac{1}{y}+\dfrac{4}{3y}=3< =>\dfrac{2}{3y}+\dfrac{3}{3y}+\dfrac{4}{3y}=3\)

\(< =>\dfrac{9}{3y}=3< =>\dfrac{3}{y}=3< =>3=3y=>y=1\)

\(=>x=\dfrac{3y}{2}=\dfrac{3}{2};z=\dfrac{3y}{4}=\dfrac{3}{4}\)