Đáp án:
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Đặt `(x-1)/2 = (y-2)/3 = (z-3)/4 = k`
`↔` \(\left\{ \begin{array}{l}\dfrac{x-1}{2}=k\\ \dfrac{y-2}{3}=k\\ \dfrac{z-3}{4}=k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x-1=2k\\y-2=3k\\z-3=4k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=2k+1\\ y=3k+2\\z=4k+3\end{array} \right.\) `(1)`
Có : `2x + 3y - z=50`
Thay `(1)` vào ta được :
`↔2 (2k + 1) + 3 (3k + 2) - (4k+3)=50`
`↔4k + 2 + 9k + 6 - 4k - 3 = 50`
`↔ (4k + 9k - 4k) + (2 + 6 - 3) = 50`
`↔ 9k+5=50`
`↔9k=50-5`
`↔9k=45`
`↔k=45÷9`
`↔k=5`
Với `k=5` thay vào `(1)` ta được :
`↔` \(\left\{ \begin{array}{l}x=2×5+1\\y=3×5+2\\z=4×5+3\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=10+1\\y=15+2\\z=20+3\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=11\\y=17\\z=23\end{array} \right.\)
Vậy `(x;y;z) = (11;17;23)`
