Ta có:\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{3}=\dfrac{z}{4}\)và x+y+z=46

Suy ra:\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{46}{9}\)

Nên \(x=\dfrac{46}{9}.2=\dfrac{92}{9}\)

\(y=\dfrac{46}{9}.3=\dfrac{46}{3}\)

\(z=\dfrac{46}{9}.4=\dfrac{184}{9}\)

Vậy \(x=\dfrac{92}{9};y=\dfrac{46}{3};z=\dfrac{184}{9}\)