Đáp án:
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Đặt `(x+99)/(-1) = (y-98)/2 = (z+97)/(-3)=k`
`↔` \(\left\{ \begin{array}{l}\dfrac{x+99}{-1}=k\\ \dfrac{y-98}{2}=k\\ \dfrac{z+97}{-3}=k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x+99=-k\\y-98=2k\\z+97=-3k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=-k-99\\y=2k+98\\z=-3k-97\end{array} \right.\) `(1)`
Có : `x-y+3=99`
Thay `(1)` vào ta được :
`↔ (-k-99) - (2k+98)+3=99`
`↔ -k-99-2k-98+3=99`
`↔ (-k-2k) + (-99-98+3)=99`
`↔ -3k -194=99`
`↔ -3k=99+194`
`↔ -3k=293`
`↔k=293÷(-3)`
`↔k=(-293)/3`
Với `k=(-293)/3` thay vào `(1)` ta được :
`↔` \(\left\{ \begin{array}{l}x=- (\dfrac{-293}{3}) - 99\\y=2 × \dfrac{-293}{3}+98\\z=-3 × (\dfrac{-293}{3})-97\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=\dfrac{293}{3}-99\\y=\dfrac{-586}{3}+98\\z=293 - 97\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=\dfrac{-4}{3}\\y=\dfrac{-292}{3}\\z=196\end{array} \right.\)
Vậy `(x;y;z) = ( (-4)/3; (-292)/3; 196)`
