Giải:
Ta có: \(3\left(x-1\right)=2\left(y-2\right)\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}\)
\(4\left(y-2\right)=3\left(z-3\right)\Rightarrow\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{2x-2+3y-6}{4+9}=\frac{\left(2x+3y\right)-\left(2+6\right)}{13}=\frac{50-8}{13}=\frac{42}{13}\)
+) \(\frac{x-1}{2}=\frac{42}{13}\Rightarrow x-1=\frac{84}{13}\Rightarrow x=\frac{97}{13}\)
+) \(\frac{y-2}{3}=\frac{42}{13}\Rightarrow y-2=\frac{126}{13}\Rightarrow y=\frac{152}{13}\)
+) \(\frac{z-3}{4}=\frac{42}{13}\Rightarrow z-3=\frac{168}{13}\Rightarrow z=\frac{207}{13}\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{97}{13};\frac{152}{13};\frac{207}{13}\right)\)