Ta có : \(\frac{x+2}{2}=\frac{y+3}{3}=\frac{z+4}{4}=>\frac{2x+4}{4}=\frac{y+3}{3}=\frac{z+4}{4}\)\(=\frac{2x+4+y+3+z+4}{4+3+4}=\frac{2x+y+z+11}{11}=\frac{11+11}{11}=2\)

+)\(\frac{x+2}{2}=2=>x+2=4=>x=2\)

+)\(\frac{y+3}{3}=2=>y+3=6=>y=3\)

+)\(\frac{z+4}{4}=2=>z+4=8=>z=4\)

Vậy x = 2 ; y = 3 ; z = 4