Lời giải:
Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-4}{4}=t\Rightarrow \left\{\begin{matrix} x=2t+1\\ y=3t+2\\ z=4t+4\end{matrix}\right.\)
Khi đó:
\(2x+3y-z=49\)
\(\Leftrightarrow 2(2t+1)+3(3t+2)-(4t+4)=49\)
\(\Leftrightarrow 9t=45\Rightarrow t=5\)
\(\Rightarrow \left\{\begin{matrix} x=2t+1=2.5+1=11\\ y=3t+2=3.5+2=17\\ z=4t+4=4.5+4=24\end{matrix}\right.\)
Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-4}{4}.\)
=> \(\frac{2.\left(x-1\right)}{4}=\frac{3.\left(y-2\right)}{9}=\frac{z-4}{4}\)
=> \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-4}{4}\) và \(2x+3y-z=49.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-4}{4}=\frac{2x-2+3y-6-z+4}{4+9-4}=\frac{\left(2x+3y-z\right)-\left(2+6-4\right)}{9}=\frac{49-4}{9}=5.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x-1}{2}=5\Rightarrow x-1=10\Rightarrow x=11\\\frac{y-2}{3}=5\Rightarrow y-2=15\Rightarrow y=17\\\frac{z-4}{4}=5\Rightarrow z-4=20\Rightarrow z=24\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(11;17;24\right).\)
Chúc bạn học tốt!
