Đáp án:
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Đặt `3/4x = 4/5y = 6/7z=k`
`↔` \(\left\{ \begin{array}{l}\dfrac{3}{4}x=k\\ \dfrac{4}{5}y=k\\ \dfrac{6}{7}z=k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=k÷\dfrac{3}{4}\\y=k÷\dfrac{4}{5}\\z=k÷\dfrac{6}{7}\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=\dfrac{4}{3}k\\ y=\dfrac{5}{4}k\\z=\dfrac{7}{6}k\end{array} \right.\) `(1)`
Có : `x + y + z=-45`
Thay `(1)` vào ta được :
`↔ 4/3k + 5/4k + 7/6k = -45`
`↔ (4/3 + 5/4 + 7/6) k =-45`
`↔ 15/4k =-45`
`↔k=-45÷15/4`
`↔k=-45 × 4/15`
`↔k=-12`
Với `k=-12` thay vào `(1)` ta được :
`↔` \(\left\{ \begin{array}{l}x=\dfrac{4}{3}×(-12)\\y=\dfrac{5}{4}×(-12)\\z=\dfrac{7}{6}×(-12)\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=-16\\y=-15\\z=-14\end{array} \right.\)
Vậy `(x;y;z) = (-16;-15;-14)`
