`**`
Theo bài ra, `2a = 3b = 4c`
`⇒ (2a)/12 = (3b)/12 = (4c)/12`
`⇒ a/6 = b/4 = c/3`
Đặt `a/6 = b/4 = c/3 = k`
`⇒` $\left\{ \begin{array}{l}a = 6k\\b = 4k\\c = 3k\end{array} \right.$
Khi đó, `(a - b + c)/(a + 2b - c) = (6k - 4k + 3k)/(6k + 8k - 3k) = (5k)/(11k) = 5/11`
Vậy `A = 5/11`
`***`
Đặt `a/2 = b/5 = c/7 = k`
`⇒` $\left\{ \begin{array}{l}a = 2k\\b = 5k\\c = 7k\end{array} \right.$
Khi đó, `(a - b + c)/(a + 2b - c) = (2k - 5k + 7k)/(2k + 10k - 7k) = (4k)/(5k) = 4/5`
Vậy `A = 4/5`