Đáp án:
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Đặt `(x-3)/7 = (y+1)/2 = (z+3)/4=k`
`↔` \(\left\{ \begin{array}{l}\dfrac{x-3}{7}=k\\ \dfrac{y+1}{2}=k\\ \dfrac{z+3}{4}=k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x-3=7k\\y+1=2k\\z+3=4k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=7k+3\\y=2k-1\\z=4k-3\end{array} \right.\) `(1)`
Có : `x - y+3z =12`
Thay `(1)` vào ta được :
`↔ (7k + 3) - (2k - 1) + 3 (4k - 3) = 12`
`↔ 7k + 3 - 2k + 1 + 12k - 9 = 12`
`↔ (7k - 2k + 12k) + (3+1-9)=12`
`↔17k -5=12`
`↔17k=12+5`
`↔17k=17`
`↔k=17÷17`
`↔k=1`
Với `k=1` thay vào `(1)` ta được :
`↔` \(\left\{ \begin{array}{l}x=7 ×1+3\\y=2×1-1\\z=4×1-3\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=7+3\\y=2-1\\z=4-3\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=10\\y=1\\z=1\end{array} \right.\)
Vậy `(x;y;z) = (10;1;1)`
