\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{x-1-2y+4+3z-9}{2-6+12}=\frac{x-2y+3z-6}{8}=\frac{14-6}{8}=1\)
=> x-1 = 2 ; y-2 = 3; z-3 = 4
=> x= 3 ; y= 5 ; z=7
Vậy x=3 ; y=5 ; z=7
đề phải là \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) chứ