Đáp án:
$\\$
Đặt : `(x-16)/9 = (y-25)/16 = (z+9)/25 = k`
`↔` \(\left\{ \begin{array}{l}\dfrac{x-16}{9}=k\\ \dfrac{y-25}{16}=k\\ \dfrac{z+9}{25}=k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x+16=9k\\y-25=16k\\z+9=25k\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=9k -16\\y=16k + 25\\z=25k - 9\end{array} \right.\) `(1)`
Có : `2x^3 - 1 = 15`
Thay `x=9k-16` vào ta được :
`↔ 2 . (9k + 16)^3 - 1 = 15`
`↔ 2 (9k - 16)^3 = 15 + 6`
`↔ 2 . (9k - 16)^3 = 16`
`↔ (9k - 16)^3 = 16 : 2`
`↔ (9k - 16)^3 = 8`
`↔ (9k - 16)^3 = 2^3`
`↔ 9k - 16 = 2`
`↔ 9k = 18`
`↔ k = 2`
Với `k = 2` thay vào `(1)` ta được :
`↔` \(\left\{ \begin{array}{l}x=9 . 2 - 16\\y = 16 . 2 + 15\\z= 25 . 2 - 9\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=18-16\\y=32 + 25\\z=50-9\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=2\\y=57\\z=41\end{array} \right.\)
Có : `x + y + z`
`= 2 + 57 + 41`
`= 100`
Vậy `x + y + z = 100`
