\(\frac{x-1}{2}\)=\(\frac{y-2}{3}\)=\(\frac{z-3}{4}\)=\(\frac{2x-2}{4}\)=\(\frac{3y-6}{9}\)=\(\frac{2x-2+3y-6-z+3}{4+9-4}\)=\(\frac{50-2-6+3}{9}\)
=\(\frac{45}{9}\)=5
=>x=5.2+1=11 ; y = 11.3+2=35 ; z=11.4+3=47
Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Leftrightarrow\frac{2\left(x-1\right)}{4}=\frac{3\left(y-2\right)}{9}=\frac{z-3}{4}\)
\(\Leftrightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
Ta có: \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\) và 2x+3y-z=50
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{\left(2x+3y-z\right)+\left(-2-6+3\right)}{4+9-4}=\frac{50-5}{9}=5\)
Do đó:
\(\left\{{}\begin{matrix}\frac{2x-2}{4}=5\\\frac{3y-6}{9}=5\\\frac{z-3}{4}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-2=20\\3y-6=45\\z-3=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=22\\3y=51\\z=23\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)
Vậy: x=11; y=17 và z=23
