Đặt \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=m\)
\(\Rightarrow x=2m+1;y=3m+2;z=4m+3\)
mà \(2x+3y-z=50\)
\(\Rightarrow2.\left(2m+1\right)+3.\left(3m+2\right)-\left(4m+3\right)=50\)
\(\Rightarrow4m+2+9m+6-4m-3=50\)
\(\Rightarrow4m+9m-4m+2+6-3=50\)
\(\Rightarrow9m+5=50\)
\(\Rightarrow9m=45\Rightarrow m=5\)
Với \(m=5\)
\(\Rightarrow\) \(x=2.5+1=11\)
\(y=3.5+2=17\)
\(z=4.5+3=23\)
Vậy ....
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
\(\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\)
=\(\dfrac{\left(2x+3y-z\right)-2-6+3}{9}=\dfrac{45}{9}=5\)
\(\Rightarrow\dfrac{x-1}{2}=5\Rightarrow x=11\)
\(\Rightarrow\dfrac{y-2}{3}=5\Rightarrow y=17\)
\(\Rightarrow\dfrac{z-3}{4}=5\Rightarrow z=23\)
Thanks Đẹp Trai Không Bao Giờ Sai and Lưu Hạ Vy