Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\) = \(\frac{2x-2+3y-6-z+3}{4+9-4}\) = \(\frac{\left(2x+3y-z\right)-\left(2+6-3\right)}{9}=\frac{50-5}{9}=5\)
Do \(\left[\begin{matrix}\frac{2x-2}{4}=5\\\frac{3y-6}{9}=5\\\frac{z-3}{4}=5\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}2x-2=20\\3y-6=45\\z-3=20\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}2x=22\\3y=51\\z=20+3\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)
Khi đó \(x+y+z=11+17+23=51\)
Vậy \(x+y+z=51.\)
